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Ï"zS„Ż(#\(#\content.xmlIntermediate AlgebraIntermediate AlgebraTerrance BergKwantlen Polytechnic UniversitySurrey, B.C.Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.Â© 2020 Terrance BergThe CC licence permits you to retain, reuse, copy, redistribute, and revise this book for non-commercial purposesâin whole or in partâfor free providing the CC BY-NC-SA licence is retained and the author is attributed as follows:Intermediate Algebra by Terrance Berg is used under a CC BY-NC-SA 4.0 Licence.This text has been adapted from an original work by Tyler Wallace, Beginning and Intermediate Algebra [PDF], which is under a Creative Commons Attribution 3.0 Unported Licence.If you redistribute all or part of this book, it is recommended the following statement be added to the copyright page so readers can access the original book at no cost:Download for free from the B.C. Open Textbook Collection.Sample APA-style citation:This textbook can be referenced. In APA citation style, it would appear as follows:Berg, T. (2020). Intermediate Algebra. Victoria, B.C.: BCcampus. Retrieved from https://opentextbc.ca/intermediatealgebraberg/. CC BY-NC-SA.Cover image attribution: âMathematics Formula Physicsâ by geralt on Pixabay is used under the Pixabay Licence.Ebook ISBN: 978-1-77420-059-9Print ISBN: 978-1-77420-058-2Visit BCcampus Open Education to learn about open education in British Columbia.ContentsAccessibility StatementPreface Chapter 1: Algebra Review 1.1 Integers1.2 Fractions (Review)1.3 Order of Operations (Review)1.4 Properties of Algebra (Review)1.5 Terms and Definitions1.6 Unit Conversion Word Problems1.7 Puzzles for Homework Chapter 2: Linear Equations 2.1 Elementary Linear Equations2.2 Solving Linear Equations2.3 Intermediate Linear Equations2.4 Fractional Linear Equations2.5 Absolute Value Equations2.6 Working With Formulas2.7 Variation Word Problems2.8 The Mystery X Puzzle Chapter 3: Graphing 3.1 Points and Coordinates3.2 Midpoint and Distance Between Points3.3 Slopes and Their Graphs3.4 Graphing Linear Equations3.5 Constructing Linear Equations3.6 Perpendicular and Parallel Lines3.7 Numeric Word Problems3.8 The Newspaper Delivery Puzzle Chapter 4: Inequalities 4.1 Solve and Graph Linear Inequalities4.2 Compound Inequalities4.3 Linear Absolute Value Inequalities4.4 2D Inequality and Absolute Value Graphs4.5 Geometric Word Problems4.6 The Cook Oil Sharing Puzzle4.7Â Mathematics in Life: The Eiffel Tower Midterm 1 Preparation Midterm 1: Version AMidterm 1: Version BMidterm 1: Version CMidterm 1: Version DMidterm 1: Version E Chapter 5: Systems of Equations 5.1 Graphed Solutions5.2 Substitution Solutions5.3 Addition and Subtraction Solutions5.4 Solving for Three Variables5.5 Monetary Word Problems5.6 Solving for Four Variables5.7 Solving Internet Game Puzzles Chapter 6: Polynomials 6.1 Working with Exponents6.2 Negative Exponents6.3 Scientific Notation (Homework Assignment)6.4 Basic Operations Using Polynomials6.5 Multiplication of Polynomials6.6 Special Products6.7 Dividing Polynomials6.8 Mixture and Solution Word Problems6.9 Pascalâs Triangle and Binomial Expansion Chapter 7: Factoring 7.1 Greatest Common Factor7.2 Factoring by Grouping7.3 Factoring Trinomials where a = 17.4 Factoring Trinomials where a â 17.5 Factoring Special Products7.6 Factoring Quadratics of Increasing Difficulty7.7 Choosing the Correct Factoring Strategy7.8 Solving Quadriatic Equations by Factoring7.9 Age Word Problems7.10 The New Committee Memberâs Age Midterm 2 Preparation Midterm 2: Version AMidterm 2: Version BMidterm 2: Version CMidterm 2: Version DMidterm 2: Version E Chapter 8: Rational Expressions 8.1 Reducing Rational Expressions8.2 Multiplication and Division of Rational Expressions8.3 Least Common Denominators8.4 Addition and Subtraction of Rational Expressions8.5 Reducing Complex Fractions8.6 Solving Complex Fractions8.7 Solving Rational Equations8.8 Rate Word Problems: Speed, Distance and Time Chapter 9: Radicals 9.1 Reducing Square Roots9.2 Reducing Higher Power Roots9.3 Adding and Subtracting Radicals9.4 Multiplication and Division of Radicals9.5 Rationalizing Denominators9.6 Radicals and Rational Exponents9.7 Rational Exponents (Increased Difficulty)9.8 Radicals of Mixed Index9.9 Complex Numbers (Optional)9.10 Rate Word Problems: Work and Time9.11 Radical Pattern Puzzle Chapter 10: Quadratics 10.1 Solving Radical Equations10.2 Solving Exponential Equations10.3 Completing the Square10.4 The Quadratic Equation10.5 Solving Quadratic Equations Using Substitution10.6 Graphing Quadratic EquationsâVertex and Intercept Method10.7 Quadratic Word Problems: Age and Numbers10.8 Construct a Quadratic Equation from its Roots Midterm 3 Preparation and Sample Questions Midterm 3: Version AMidterm 3: Version BMidterm 3: Version CMidterm 3: Version DMidterm 3: Version E Chapter 11: Functions 11.1 Function Notation11.2 Operations on Functions11.3 Inverse Functions11.4 Exponential Functions11.5 Logarithmic Functions11.6 Compound Interest11.7 Trigonometric Functions11.8 Sine and Cosine Laws11.9 Your Next Yearâs Valentine? Final Exam Preparation Final Exam: Version AFinal Exam: Version BReference SectionAnswer Key 1.1Answer Key 1.2Answer Key 1.3Answer Key 1.4Answer Key 1.6Answer Key 1.7Answer Key 2.1Answer Key 2.2Answer Key 2.3Answer Key 2.4Answer Key 2.5Answer Key 2.6Answer Key 2.7Answer Key 2.8Answer Key 3.1Answer Key 3.2Answer Key 3.3Answer Key 3.4Answer Key 3.5Answer Key 3.6Answer Key 3.7Answer Key 4.1Answer Key 4.2Answer Key 4.3Answer Key 4.4Answer Key 4.5Answer Key 4.6Mid Term 1: Review Questions Answer KeyMidterm 1: Version A Answer KeyMidterm 1: Version B Answer KeyMidterm 1: Version C Answer KeyMidterm 1: Version D Answer KeyMidterm 1: Version E Answer KeyAnswer Key 5.1Answer Key 5.2Answer Key 5.3Answer Key 5.4Answer Key 5.5Answer Key 5.6Answer Key 5.7Answer Key 6.1Answer Key 6.2Answer Key 6.3Answer Key 6.4Answer Key 6.5Answer Key 6.6Answer Key 6.7Answer Key 6.8Answer Key 6.9Answer Key 7.1Answer Key 7.2Answer Key 7.3Answer Key 7.4Answer Key 7.5Answer Key 7.6Answer Key 7.7Answer Key 7.8Answer Key 7.9Answer Key 7.10Midterm 2: Prep Answer KeyMidterm 2: Version A Answer KeyMidterm 2: Version B Answer KeyMidterm 2: Version C Answer KeyMidterm 2: Version D Answer KeyMidterm 2: Version E Answer KeyAnswer Key 8.1Answer Key 8.2Answer Key 8.3Answer Key 8.4Answer Key 8.5Answer Key 8.6Answer Key 8.7Answer Key 8.8Answer Key 9.1Answer Key 9.2Answer Key 9.3Answer Key 9.4Answer Key 9.5Answer Key 9.6Answer Key 9.7Answer Key 9.8Answer Key 9.9Answer Key 9.10Answer Key 9.11Answer Key 10.1Answer Key 10.2Answer Key 10.3Answer Key 10.4Answer Key 10.5Answer Key 10.6Answer Key 10.7Answer Key 10.8Midterm 3 Prep Answer KeyMidterm 3: Version A Answer KeyMidterm 3: Version B Answer KeyMidterm 3: Version C Answer KeyMidterm 3: Version D Answer KeyMidterm 3: Version E Answer KeyAnswer Key 11.1Answer Key 11.2Answer Key 11.3Answer Key 11.4Answer Key 11.5Answer Key 11.6Answer Key 11.7Answer Key 11.8Answer Key 11.9Final Exam: Version A Answer KeyFinal Exam: Version B Answer KeyVersioning History1Accessibility StatementKPU supports the creation of free, open, and accessible educational resources. We are actively committed to increasing the accessibility and usability of the textbooks we produce.The web version of this resources has been designed to meet Web Content Accessibility Guidelines 2.0, level AA, as well as follow most of the guidelines in Appendix A: Checklist for Accessibility of the Accessibility Toolkit â 2nd Edition. However, the Math equations in this text were created with Latex and rendered as images with alt text that may not be accessible for people using screen readers. Please see the heading âKnown Accessibility Issues and Areas for Improvementâ for more information.Ways to Read This TextbookThis textbook is available in the following formats:Online web book. You can read this textbook online on a computer or mobile device in one of the following browsers: Chrome, Firefox, Edge, and Safari.PDF. You can download this book as a PDF to read on a computer (Digital PDF) or print it out (Print PDF).Mobile. 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If using the online web book, you can find the full table of contents on the bookâs homepage or by selecting âContentsâ from the top menu when you are in a chapter.Annotate the textbook.If you like to highlight or write on your textbooks, you can do that by getting a print copy, using the Digital PDF in Adobe Reader, or using the highlighting tools in eReader apps.Known Accessibility Issues and Areas for ImprovementThe equations in this textbook were created with Latex and rendered as images with alternative text. The alternative text for these equations is the Latex code, which may or may not be accessible for students using screen readers depending on their knowledge of Latex. For example, the fraction Âœ will be read as âdfrac 1 2.âWe have plans to edit this text so it is compatible with the MathJax plugin, which will make the equations available in MathML and thereby fully screen reader compatible.Let Us Know if You are Having Problems Accessing This BookWe are always looking for ways to make our textbooks more accessible. If you have problems accessing this textbook, please contact us to let us know so we can fix the issue.Please include the following information:The name of the textbookThe location of the problem by providing a web address or page description.A description of the problemThe computer, software, browser, and any assistive technology you are using that can help us diagnose and solve your issue (e.g., Windows 10, Google Chrome (Version 65.0.3325.181), NVDA screen reader)You can contact us at https://www.kpu.ca/open-education-questionsÂ This statement was last updated on May 1, 2020.2PrefaceMathematics can be best described as a language, and when one learns the foundations of mathematics, one starts the process of becoming literate. The mathematics covered in this textbook are at an intermediate algebra level, building upon literacies covered in Mathematical Fundamental and Elementary Algebra.This textbook can trace its origins to three distinct sources:First, it is adapted from an original work by Wallace: Elementary and Introductory Algebra [PDF].Second, it has been modified after many years of observing student preferences in how they learn. My former KPU students all have fingerprints throughout this book.Third, it is the work of the world, in that mathematics is universal and global, having history in all ages and cultures.This being said, this textbook is intended to never be sold for profit. Rather, it is meant to be freely used and adapted by anyone who wishes to teach or learn intermediate algebra.Please feel free to contact me at KPU for insights and additions that can add richness to this document, as this work is intended to be a living document that can grow and help to increase our understanding of this complex world that we live in.Best regardsTerrance Berg, Ph.D. (terry.berg@kpu.ca)IChapter 1: Algebra ReviewLearning ObjectivesThis chapter covers:IntegersFractionsOrder of OperationsProperties of AlgebraTerms & DefinitionsWord Problems11.1 IntegersThe ability to work comfortably with negative numbers is essential to success in algebra. For this reason, a quick review of adding, subtracting, multiplying, and dividing integers is necessary.2 Integers4 are all the positive whole numbers, all the negative whole numbers, and zero. As this is intended to be a review of integers, descriptions and examples will not be as detailed as in a normal lesson.When adding integers, there are two cases to consider. The first is when the signs matchâthat is, the two integers are both positive or both negative.If the signs match, add the numbers together and retain the sign.Â Example 1.1.1Add Â Â Example 1.1.2Add -7+(-5).Â Â Â The second case is when the signs donât match, and there is one positive and one negative number. Subtract the numbers (as if they were all positive), then use the sign from the number with the greatest absolute value. This means that, if the number with the greater absolute value is positive, the answer is positive. If it is negative, the answer is negative.Â Example 1.1.3Add Example 1.1.4Add Example 1.1.5Add 4+(-3).\begin{array}{rl} 4+(-3)&\text{Different signs. Subtract }4-3\text{. Positive number has greater absolute value.} \\ 1&\text{Solution}\end{array}Example 1.1.6Add 7+(-10).\begin{array}{rl} 7+(-10)&\text{Different signs. Subtract }10-7.\text{ Negative number has greater absolute value.} \\ -3&\text{Solution} \end{array}Â For subtraction of negatives, change the problem to an addition problem, which is then solved using the above methods. The way to change a subtraction problem to an addition problem is by adding the opposite of the number after the subtraction sign to the number before the subtraction sign. Often, this method is referred to as âadding the opposite.âÂ Example 1.1.7Subtract Example 1.1.8Subtract Example 1.1.9Subtract Example 1.1.10Subtract -6-(-2).\begin{array}{rl} -6-(-2)&\text{Add the opposite of }-2\text{ to }-6. \\ -6+2&\text{Different signs. Subtract }6-2.\text{ Negative number has greater absolute value.} \\ -4&\text{Solution} \end{array}Â Multiplication and division of integers both work in a very similar pattern. The short description of the process is to multiply and divide like normal. If the signs match (numbers are both positive or both negative), the answer is positive. If the signs donât match (one positive and one negative), then the answer is negative.Â Example 1.1.11Multiply Example 1.1.12Divide -36 \div -9.\begin{array}{rl} -36 \div -9 & \text{Signs match, so the answer is positive.} \\ 4&\text{Solution} \end{array}Example 1.1.13Multiply Example 1.1.14Divide Key Takeaways:Â A few things to be careful of when working with integers.Be sure not to confuse a problem likeÂ â3Â â 8 with â3(â8).The â3Â â 8 problem is subtraction because the subtraction sign separates the â3 from what comes after it.The â3(â8) is a multiplication problem because there is nothing between the â3 and the parenthesis. If there is no operation written in between the parts, then you assume that you are multiplying.Be careful not to mix the pattern for adding and subtracting integers with the pattern for multiplying and dividing integers. They can look very similar. For example:If the two numbers in an addition problem are negative, then keep the negative sign, such as in â3 + (â7)Â = â10.If the signs of the two numbers in a multiplication problem match, the answer is positive, such as in (â3)(â7) Â = 21.QuestionsFor questions 1 to 30, find the sum and/or difference.Â 1- 34 - (-1)(-6)-(-8)(-6) + 8(-3) - 3(-8) - (-3)3 - (-5)7 - 7(-7) - (-5)(-4) + (-1)3 - (-1)(-1) + (-6)6 - 3(-8) + (-1)(-5) + 3(-1) - 82 - 35 - 7(-8) - (-5)(-5) + 7(-2) + (-5)1 + (-1)5 - (-6)8 - (-1)(-6) + 3(-3) + (-1)4 - 77 - 3(-7) + 7(-3) + (-5)For questions 31 to 44, find each product.(4)(-1)(7)(-5)(10)(-8)(-7)(-2)(-4)(-2)(-6)(-1)(-7)(8)(6)(-1)(9)(-4)(-9)(-7)(-5)(2)(-2)(-2)(-5)(4)(-3)(-9)For questions 45 to 58, find each quotient.30 \div -10-49 \div -7-12 \div -4-2 \div -130 \div 620 \div 1027 \div 3-35 \div -580 \div -88 \div -250 \div 5-16 \div 248 \div 860 \div -10Answer Key 1.1Read aboutÂ The History of Integers.The word "integer" is derived from the Latin word integer, which means "whole." Integers are written without using a fractional component. Examples are 2, 3, 1042, 28, 0, â42, â2. Numbers that are fractionalâsuch as \frac{1}{4}, 0.33, and 1.42âare not integers.21.2 Fractions (Review)Working with fractions is a very important foundational skill in algebra. This section will briefly review reducing, multiplying, dividing, adding, and subtracting fractions. As this is a review, concepts will not be explained in as much detail as they are in other lessons. Final answers of questions working with fractions tend to always be reduced. Reducing fractions is simply done by dividing both the numerator and denominator by the same number.Â ExampleÂ 1.2.1Reduce Â The previous example could have been done in one step by dividing both the numerator and the denominator by 12. Another solution could have been to divide by 2 twice and then by 3 once (in any order). It is not important which method is used as long as the fraction is reduced as much as possible.The easiest operation to complete with fractions is multiplication. Fractions can be multiplied straight across, meaning all numerators and all denominators are multiplied together.Â Example 1.2.2Multiply Â Before multiplying, fractions can be reduced. It is possible to reduce vertically within a single fraction, or diagonally within several fractions, as long as one number from the numerator and one number from the denominator are used.Â Example 1.2.3Multiply \dfrac{25}{24} \cdot \dfrac{32}{55}.\begin{array}{rl} \dfrac{\cancel{25}\text{ }5}{\cancel{24}\text{ }3}\cdot \dfrac{\cancel{32}\text{ }4}{\cancel{55}\text{ }11} & \text{Reduce 25 and 55 by dividing by 5, and reduce 32 and 24 by dividing by 8.} \\ \\ \dfrac{5\cdot 4}{3\cdot 11}&\text{Multiply numerators and denominators across.} \\ \\ \dfrac{20}{33}&\text{Solution} \end{array}Â Dividing fractions is very similar to multiplying, with one extra step. Dividing fractions necessitates first taking the reciprocal of the second fraction. Once this is done, multiply the fractions together. This multiplication problem solves just like the previous problem.Â Example 1.2.4Divide Â To add and subtract fractions, it is necessary to first find the least common denominator (LCD). There are several ways to find the LCD. One way is to break the denominators into primes, write out the primes that make up the first denominator, and only add primes that are needed to make the other denominators.Â Example 1.2.5Find the LCD of 8 and 12.Break 8 and 12 into primes:Â Â The LCD will contain all the primes needed to make each number above.Â Â Â Adding and subtracting fractions is identical in process. If both fractions already have a common denominator, simply add or subtract the numerators and keep the denominator.Â Example 1.2.6Add Â While can be written as the mixed number , algebra almost never uses mixed numbers. For this reason, always use the improper fraction, not the mixed number.Â Example 1.2.7Subtract Â If the denominators do not match, it is necessary to first identify the LCD and build up each fraction by multiplying the numerator and denominator by the same number so each denominator is built up to the LCD.Â Example 1.2.8Add \dfrac{5}{6}+\dfrac{4}{9}.\begin{array}{rl} \rlap{$\overbrace{\phantom{2\times 3}}^6$}2\times \underbrace{3\times 3}_9& \text{LCD is }18. \\ \\ \dfrac{3\cdot 5}{3\cdot 6}+\dfrac{4\cdot 2}{9\cdot 2} & \text{Multiply the first fraction by 3 and the second by 2.} \\ \\ \dfrac{15}{18}+\dfrac{8}{18} & \text{Same denominator, so add }15 + 8. \\ \\ \dfrac{23}{18}&\text{Solution} \end{array}Example 1.2.9Subtract QuestionsFor questions 1 to 18, simplify each fraction. Leave your answer as an improper fraction.\dfrac{42}{12}\dfrac{25}{20}\dfrac{35}{25}\dfrac{24}{8}\dfrac{54}{36}\dfrac{30}{24}\dfrac{45}{36}\dfrac{36}{27}\dfrac{27}{18}\dfrac{48}{18}\dfrac{40}{16}\dfrac{48}{42}\dfrac{63}{18}\dfrac{16}{12}\dfrac{80}{60}\dfrac{72}{48}\dfrac{72}{60}\dfrac{126}{108}For questions 19 to 36, find each product. Leave your answer as an improper fraction.(9)\left(\dfrac{8}{9}\right)(-2)\left(-\dfrac{5}{6}\right)(2)\left(-\dfrac{2}{9}\right)(-2)\left(\dfrac{1}{3}\right)(-2)\left(\dfrac{13}{8}\right)\left(\dfrac{3}{2}\right) \left(\dfrac{1}{2}\right)\left(-\dfrac{6}{5}\right)\left(-\dfrac{11}{8}\right)\left(-\dfrac{3}{7}\right)\left(-\dfrac{11}{8}\right)(8)\left(\dfrac{1}{2}\right)(-2)\left(-\dfrac{9}{7}\right)\left(\dfrac{2}{3}\right)\left(\dfrac{3}{4}\right)\left(-\dfrac{17}{9}\right)\left(-\dfrac{3}{5}\right)(2)\left(\dfrac{3}{2}\right)\left(\dfrac{17}{9}\right)\left(-\dfrac{3}{5}\right)\left(\dfrac{1}{2}\right)\left (-\dfrac{7}{5}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{5}{7}\right)\left(\dfrac{5}{2}\right)\left(-\dfrac{0}{5}\right)\left(\dfrac{6}{0}\right)\left(\dfrac{6}{7}\right)For questions 37 to 52, find each quotient. Leave your answer as an improper fraction.-2 \div \dfrac {7}{4}-\dfrac{12}{7} \div -\dfrac{9}{5}-\dfrac{1}{9} \div -\dfrac{1}{2}-2 \div -\dfrac{3}{2}-\dfrac{3}{2} \div \dfrac{13}{7}\dfrac{5}{3} \div \dfrac{7}{5}-1 \div \dfrac{2}{3}\dfrac{10}{9} \div -6\dfrac{8}{9} \div \dfrac{1}{5}\dfrac{1}{6} \div -\dfrac{5}{3}-\dfrac{9}{7} \div \dfrac{1}{5}-\dfrac{13}{8} \div -\dfrac{15}{8}-\dfrac{2}{9} \div -\dfrac{3}{2}-\dfrac{4}{5} \div -\dfrac{13}{8}\dfrac{1}{10} \div \dfrac{3}{2}\dfrac{5}{3} \div \dfrac{5}{3}For questions 53 to 70, evaluate each expression. Leave your answer as an improper fraction.\dfrac{1}{3} + \left(-\dfrac{4}{3}\right)\dfrac{1}{7} + \left(-\dfrac{11}{7}\right)\dfrac{3}{7} - \dfrac{1}{7}\dfrac{1}{3} + \dfrac{5}{3}\dfrac{11}{6} + \dfrac{7}{6}(-2)+ \left(-\dfrac{15}{8}\right)\dfrac{3}{5}+ \dfrac{5}{4}(-1)-\dfrac{2}{3}\dfrac{2}{5}+ \dfrac{5}{4}\dfrac{12}{7}- \dfrac{9}{7}\dfrac{9}{8}+ \left(-\dfrac{2}{7}\right)(-2)+ \dfrac{5}{6}1+ \left(-\dfrac{1}{3}\right)\dfrac{1}{2}- \dfrac{11}{6}\left(-\dfrac{1}{2}\right)+ \dfrac{3}{2}\dfrac{11}{8}- \dfrac{1}{22}\dfrac{1}{5}+ \dfrac{3}{4}\dfrac{6}{5}- \dfrac{8}{5}Answer Key 1.231.3 Order of Operations (Review)When simplifying expressions, it is important to do so in the correct order. Consider the problem 2 + 5Â â
3 done two different ways:Method 1: Add firstMethod 2: Multiply firstAdd: 2 + 5Â â
3Multiply: 2 + 5Â â
3Multiply: 7Â â
3Add: 2 + 15Solution: 21Solution: 17Â The previous example illustrates that if the same problem is done two different ways, it will result in two different solutions. However, only one method can be correct. It turns out the second method is the correct one. The order of operations ends with the most basic of operations, addition (or subtraction). Before addition is completed, do all repeated addition, also known as multiplication (or division). Before multiplication is completed, do all repeated multiplication, also known as exponents. When something is supposed to be done out of order, to make it come first, put it in parentheses (or grouping symbols). This list, then, is the order of operations used to simplify expressions.Â Key Takeaways: Order of Operations1st Brackets (Grouping)2nd Exponents3rd Multiplication and Division (Left to Right)4th Addition and Subtraction (Left to Right)Â Multiplication and division are on the same level because they are the same operation (division is just multiplying by the reciprocal). This means multiplication and division must be performed from left to right. Therefore, division will come first in some problems, and multiplication will come first in others. The same is true for adding and subtracting (subtracting is just adding the opposite).Often, students use the word BEMDAS to remember the order of operations, as the first letter of each operation creates the word (written as B E MD AS). Remember BEMDAS to ensure that multiplication and division are done from left to right (same with addition and subtraction).Â Example 1.3.1Evaluate using the order of operations.Â Â Â It is very important to remember to multiply and divide from left to right!Â Example 1.3.2Evaluate using the order of operations.Â Â Â If there are several sets of parentheses in a problem, start with the innermost set and work outward. Inside each set of parentheses, simplify using the order of operations. To make it easier to know which left parenthesis goes with which right parenthesis, different types of grouping symbols will be used, such as braces { }, brackets [ ], and parentheses ( ). These all do the same thing: they are grouping symbols and must be evaluated first.Â Example 1.3.3Evaluate using the order of operations.Â As Example 1.3.3 illustrates, it can take several steps to complete a problem. The key to successfully solving order of operations problems is to take the time to show your work and do one step at a time. This will reduce the chance of making a mistake along the way.There are several types of grouping symbols that can be used besides parentheses, brackets, and braces. One such symbol is a fraction bar. The entire numerator and the entire denominator of a fraction must be evaluated before reducing. Once the fraction is reduced, the numerator and denominator can be simplified at the same time.Â Example 1.3.4Evaluate using the order of operations.Â Another type of grouping symbol is the absolute value. Everything inside a set of absolute value brackets must be evaluated, just as if it were a normal set of parentheses. Then, once the inside is completed, take the absolute valueâor distance from zeroâto make the number positive.Â Example 1.3.5Evaluate using the order of operations.Key Takeaways: ExponentsThe above example also illustrates an important point about exponents:Exponents are only considered to be on the number they are attached to.This means that, in the expression â42, only the 4 is squared, giving us â(42) or â16.But when the negative is in parentheses, such as in (â5)2, the negative is part of the number and is also squared, giving a positive solution of 25.QuestionsFor questions 1 to 24, reduce and solve the following expressions.-6 \cdot 4(-1)(-6 \div 6)^33 + (8) \div | 4 |5(-5 + 6) \cdot 6^28 \div 4 \cdot 27 - 5 + 6[-9 - (2 - 5)] \div (-6)Â (-2 \cdot 2^3 \cdot 2) \div (-4)-6 + (-3 - 3)^2 \div | 3 |Â (-7 - 5) \div [-2 - 2 - (-6)]4 - 2 | 3^2 - 16 |(-10 - 6) \div (-2)^2 - 5[ -1 - (-5)] | 3 + 2 |-3 - \{3 - [ -3(2 + 4) - (-2)]\}[2 + 4 |7 + 2^2|] \div [4 \cdot 2 + 5 \cdot 3]-4 - [2 + 4(-6) - 4 - 22 - 5 \cdot 2][6 \cdot 2 + 2 - (-6)] (-5 + | (-18 \div 6) |)2 \cdot (-3) + 3 - 6[ -2 - (-1 - 3)]\dfrac{-13-2}{2-(-1)^3+(-6)-[-1-(-3)]}\dfrac{5^2+(-5)^2}{| 4^2-2^5| -2 \cdot 3}\dfrac{6 \cdot -8 - 4 + (-4) - [ -4 -(-3)]}{(4^2 + 3^2) \div 5}\dfrac{-9 \cdot 2 - (3 - 6)}{1-(-2+1)-(-3)}\dfrac{2^3 + 4}{-18 - 6 + (-4) - [ -5(-1)(-5)]}\dfrac{13 + (-3)^2 + 4(-3) + 1 - [ -10 - (-6)]}{\{[4 + 5] \div [4^2 - 3^2(4 - 3) - 8]\} + 12}Answer Key 1.341.4 Properties of Algebra (Review)When doing algebra, it is common not to know the value of the variables. In this case, simplify where possible and leave any unknown variables in the final solution. One way to simplify expressions is to combine like terms.Like terms are terms whose variables match exactly, exponents included. Examples of like terms would be and and or â3 and 5. To combine like terms, add (or subtract) the numbers in front of the variables and keep the variables the same.Â Example 1.4.1Simplify Â Â Example 1.4.2Simplify Â Â Â When combining like terms, subtraction signs must be interpreted as part of the terms they precede. This means that the term following a subtraction sign should be treated like a negative term. The sign always stays with the term.Another method to simplify is known as distributing. Sometimes, when working with problems, there will be a set of parentheses that makes solving a problem difficult, if not impossible. To get rid of these unwanted parentheses, use the distributive property and multiply the number in front of the parentheses by each term inside.Â Â Several examples of using the distributive property are given below.Â Example 1.4.3Simplify Â Â Example 1.4.4Simplify Â Â Â In the previous example, it is necessary to again use the fact that the sign goes with the number. This means â6 is treated as a negative number, which gives (â7)(â6) = 42, a positive number. The most common error in distributing is a sign error. Be very careful with signs! It is possible to distribute just a negative throughout parentheses. If there is a negative sign in front of parentheses, think of it like a â1 in front and distribute it throughout.Â Example 1.4.5Simplify Â Â Â Distributing throughout parentheses and combining like terms can be combined into one problem. Order of operations says to multiply (distribute) first, then add or subtract (combine like terms). Thus, do each problem in two steps: distribute, then combine.Â Example 1.4.6Simplify Â Â Example 1.4.7Simplify Â Â Â In Example 1.4.6, â2 is distributed, not just 2. This is because a number being subtracted must always be treated like it has a negative sign attached to it. This makes a big difference, for in that example, when the â5 inside the parentheses is multiplied by â2, the result is a positive number. More involved examples of distributing and combining like terms follow.Â Example 1.4.8Simplify Example 1.4.9Simplify QuestionsFor questions 1 to 28, reduce and combine like terms.r - 9 + 10-4x + 2 - 4n + n4b + 6 + 1 + 7b8v + 7v-x + 8x-7x - 2x-7a - 6 + 5k - 2 + 7-8p + 5pÂ x - 10 - 6x + 11 - 10n - 10m - 2m1 - r - 6-8(x - 4)3(8v + 9)8n(n + 9)-(-5 + 9a)7k(-k + 6)10x(1 + 2x)-6(1 + 6x)-2(n + 1)8m(5 - m)-2p(9p - 1)-9x(4 - x)4(8n - 2)-9b(b - 10)-4(1 + 7r)For questions 29 to 58, simplify each expression.9(b + 10) + 5b4v - 7(1 - 8v)-3x(1 - 4x) - 4x^2-8x + 9(-9x + 9)-4k^2 - 8k(8k + 1)-9 - 10(1 + 9a)1 - 7(5 + 7p)-10(x - 2) - 3-10 - 4(n - 5)-6(5 - m) + 3m4(x + 7) + 8(x + 4)-2r(1 + 4r) + 8r(-r + 4)-8(n + 6) - 8n(n + 8)9(6b + 5) - 4b(b + 3)7(7 + 3v) + 10(3 - 10v)-7(4x - 6) + 2(10x - 10)2n(- 10n + 5) - 7(6 - 10n)-3(4 + a) + 6a(9a + 10)5(1 - 6k) + 10(k - 8)-7(4x + 3) - 10(10x + 10)(8n^2 - 3n) - (5 + 4n^2)(7x^2 - 3) - (5x^2 + 6x)(5p - 6) + (1 - p)(3x^2 - x) - (7 - 8x)(2 - 4v^2) + (3v^2 + 2v)(2b - 8) + (b - 7b^2)(4 - 2k^2) + (8 - 2k^2)(7a^2 + 7a) - (6a^2 + 4a)(x^2 - 8) + (2x^2 - 7)(3 - 7n^2) + (6n^2 + 3)Answer Key 1.451.5 Terms and DefinitionsDigits can be defined as the alphabet of the HinduâArabic numeral system that is in common usage today. This alphabet is: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Written in set-builder notation, digits are expressed as:Â Â Natural numbers are often called counting numbers and are usually denoted by These numbers start at 1 and carry on to infinity, which is denoted by the symbol â. Writing the set of natural numbers in set-builder notation gives:Â Â Whole numbers include the set of natural numbers and zero. Whole numbers are generally designated by In set-builder notation, the set of whole numbers is denoted by:Â Â Integers include the set of all whole numbers and their negatives. This means the set of integers is composed of positive whole numbers, negative whole numbers, and zero (fractions and decimals are not integers). Common symbols used to represent integers are and For this textbook, the symbol will be used to represent integers.Rational numbers include all integersÂ and all fractions, terminating decimals, and repeating decimals. Every rational number can be written as a fraction where and are integers. Rational numbers are denoted by the symbol In set-builder notation, the set of rational numbers can be informally written as:Irrational numbers include any number that cannot be defined by the fraction where and are integers. These are numbers that are non-repeating or non-terminating. Classic examples of irrational numbers are pi Â and the square roots of 2 and 3. The symbol for irrational numbers is commonly given as or For this textbook, the symbol will be used. In set-builder notation, the set of irrational numbers can be informally written as:Real numbers include the set of all rational numbers and irrational numbers. The symbol for real numbers is commonly given as In set-builder notation, the set of real numbers can be informally written as:Â Â Numbers that may not yet have been encountered are imaginary numbers (commonly sometimes ) and complex numbers These numbers will be properly defined later in the textbook.Imaginary numbers include any real number multiplied by the square root of â1.Complex numbers are combinations of any real number, imaginary number, or a sum and difference of them.Consecutive integers are integers that follow each other sequentially. Examples are:Â Â Consecutive even or odd integers are numbers that skip the odd/even sequence to just show odd, odd, odd, or even, even, even. Examples are:Â Â Prime numbers are numbers that cannot be divided by any integer other than 1 and itself. The following is a list of all the prime numbers that are found between 0 and 1000. (Note: 1 is not considered prime.)Squares are numbers multiplied by themselves. A number that is being squared is shown as having a superscript 2 attached to it. For example, 5 squared is written as 52, which equals 5 Ă 5 or 25.Perfect squares are squares of whole numbers, such as 1, 4, 9, 16, 25, 36, and 49. They are found by squaring natural numbers. The following is the list of perfect squares using numbers up to 20:Â Â Cubes are numbers multiplied by themselves three times. A number that is being cubed is shown as having a superscript 3 attached to it. For example, 5 cubed is written as 53, which equals 5Â Ă 5Â Ă 5 or 125.Perfect cubes are cubes of whole numbers, such as 1, 8, 27, 64, 125, 216, and 343. They are found by cubing natural numbers. The following is the list of perfect cubesÂ using numbers up to 20:Â Â Percentage means parts per hundred. A percentage can be thought of as a fraction where the numerator, is the number to the left of the % sign, and the denominator, is 100. For example: 42\% = \dfrac{42}{100} = 0.42.Absolute values. The absolute value of an expression denoted is the distance from zero of the number or operation that occurs between the absolute value signs. For example:Â Â Examples of absolute values of simple operations are:Â Â Set-builder notation follows standard patterns and is as follows:Begin the set with a left brace {A vertical bar | means âsuch thatâEnd the set with a right brace }So to say is an integer, write this as:This means âthe set of such that is an integer.âAnother way of writing this is to use the symbols that mean âelement ofâ and ânot an element of.ââElement ofâ is shown by the symbol â, and ânot an element ofâ is shown by the element symbol with a line drawn through it, â.In simplest terms, if something is an element of something else, it means that it belongs to or is part of it. For example, a set of numbers called can only be made up of any natural number like 4, 6, 9, and 15. This can be stated as which reads as âthe set of such that is an element of the natural number system.ââNot an element ofâ can be used to state that the set cannot contain excluded values. For example, say there is a set of all numbers except counting numbers This can be written as:Â Â This can be read as âthe set of such that is an element of the set of all real numbers, excluding those numbers that are natural numbers.âSets of numbers giving excluded values can be seen throughout this textbook. The standard example is to exclude values that would result in a denominator of zero. This exclusion avoids division by zero and getting an undefinable answer.The empty set. Sometimes, a set contains no elements. This set is termed the âempty setâ or the ânull set.â To represent this, write either { } or Ă.Â Â Â Â 61.6 Unit Conversion Word ProblemsOne application of rational expressions deals with converting units. Units of measure can be converted by multiplying several fractions together in a process known as dimensional analysis.The trick is to decide what fractions to multiply. If an expression is multiplied by 1, its value does not change. The number 1 can be written as a fraction in many different ways, so long as the numerator and denominator are identical in value. Note that the numerator and denominator need not be identical in appearance, but rather only identical in value. Below are several fractions, each equal to 1, where the numerator and the denominator are identical in value. This is why, when doing dimensional analysis, it is very important to use units in the setup of the problem, so as to ensure that theÂ conversion factor is set up correctly.Â Example 1.6.1Â If 1 pound = 16 ounces, how many pounds are in 435 ounces?Â The same process can be used to convert problems with several units in them. Consider the following example.Â Example 1.6.2A student averaged 45 miles per hour on a trip. What was the studentâs speed in feet per second?Example 1.6.3Convert 8 ft3 to yd3.Example 1.6.4A room is 10 ft by 12 ft. How many square yards are in the room? The area of the room is 120 ft2 (area = lengthÂ Ă width).Converting the area yields:Â The process of dimensional analysis can be used to convert other types of units as well. Once relationships that represent the same value have been identified, a conversion factor can be determined.Â Example 1.6.5A child is prescribed a dosage of 12 mg of a certain drug per day and is allowed to refill his prescription twice. If there are 60 tablets in a prescription, and each tablet has 4 mg, how many doses are in the 3 prescriptions (original + 2 refills)?Metric and Imperial (U.S.) ConversionsDistanceÂ Â Imperial to metric conversions:Â Â AreaÂ Â Imperial to metric conversions:Â Â VolumeÂ Â Imperial to metric conversions:Â Â MassÂ Â Imperial to metric conversions:Â Â TemperatureFahrenheit to Celsius conversions:Â Â Fahrenheit to Celsius conversion scale. Long description available.Celsius to Fahrenheit conversion scale. [Long Description]QuestionsFor questions 1 to 18, use dimensional analysis to perform the indicated conversions.7 miles to yards234 oz to tons11.2 mg to grams1.35 km to centimetres9,800,000 mm to miles4.5 ft2 to square yards435,000 m2 to square kilometres8 km2 to square feet0.0065 km3 to cubic metres14.62 in3 to square centimetres5500 cm3 to cubic yards3.5 mph (miles per hour) to feet per second185 yd per min. to miles per hour153 ft/s (feet per second) to miles per hour248 mph to metres per second186,000 mph to kilometres per year7.50 tons/yd2 to pounds per square inch16 ft/s2 to kilometres per hour squaredFor questions 19 to 27, solve each conversion word problem.On a recent trip, Jan travelled 260 miles using 8 gallons of gas. What was the carâs miles per gallon for this trip? Kilometres per litre?A certain laser printer can print 12 pages per minute. Determine this printerâs output in pages per day.An average human heart beats 60 times per minute. If the average person lives to the age of 86, how many times does the average heart beat in a lifetime?Blood sugar levels are measured in milligrams of glucose per decilitre of blood volume. If a personâs blood sugar level measured 128 mg/dL, what is this in grams per litre?You are buying carpet to cover a room that measures 38 ft by 40 ft. The carpet cost $18 per square yard. How much will the carpet cost?A cargo container is 50 ft long, 10 ft wide, and 8 ft tall. Find its volume in cubic yards and cubic metres.A local zoning ordinance says that a houseâs âfootprintâ (area of its ground floor) cannot occupy more than ÂŒ of the lot it is built on. Suppose you own a \frac{1}{3}-acre lot (1 acre = 43,560 ft2). What is the maximum allowed footprint for your house in square feet? In square metres?A car travels 23 km in 15 minutes. How fast is it going in kilometres per hour? In metres per second?The largest single rough diamond ever found, the Cullinan Diamond, weighed 3106 carats. One carat is equivalent to the mass of 0.20 grams. What is the mass of this diamond in milligrams? Weight in pounds?Answer Key 1.6Long DescriptionsCelsius to Fahrenheit conversion scale long description: Scale showing conversions between Celsius and Fahrenheit. The following table summarizes the data:CelsiusFahrenheitâ40Â°Câ40Â°Fâ30Â°Câ22Â°Fâ20Â°Câ4Â°Fâ10Â°C14Â°F0Â°C32Â°F10Â°C50Â°F20Â°C68Â°F30Â°C86Â°F40Â°C104Â°F50Â°C122Â°F60Â°C140Â°F70Â°C158Â°F80Â°C176Â°F90Â°C194Â°F100Â°C212Â°F[Return to Celsius to Fahrenheit conversion scale]71.7 Puzzles for HomeworkExercise 1.7.1There are four known solutions to the following math puzzle, in which you can move just one line to fix the equation. How many solutions can you find?Â Exercise 1.7.2Are the following statements true?Letters A, B, C and D do not appear anywhere in the spellings of 1 to 99Letter D appears for the first time in âhundredâLetters A, B and C do not appear anywhere in the spellings of 1 to 999Letter A appears for the first time in âthousandâLetters B and C do not appear anywhere in the spellings of 1 to 999,999,999Letter B appears for the first time in âbillionâLetter C does not appear anywhere in any word used to count in EnglishIIChapter 2: Linear EquationsThe study of linear equations is the study of the foundations of algebra that lead into multiple applications and more advanced mathematics, physics, and engineering fields. Linear equations are used quite frequently in these fields in part because the solutions to complex, non-linear systems can often be well approximated using a linear equation.Linear equations are equations that can define multiple dimensions. Consider, for instance, the following two equations:Â Â Â Here they are in graphed form:Â Linear equations can also be shown in three dimensions, but that would require time-spaced snapshots to show how a three-dimensional line would change if the fourth dimension of time were to be added.Fundamental to all linear equations is that the variables being worked with have no powers attached to them. This means that a four-dimensional space-time linear equation could look like but it could not carry any powers, like in the equation Remember:Â Â 82.1 Elementary Linear EquationsSolving linear equations is an important and fundamental skill in algebra. In algebra, there are often problems in which the answer is known, but the variable part of the problem is missing. To find this missing variable, it is necessary to follow a series of steps that result in the variable equalling some solution.Addition and Subtraction ProblemsTo solve equations, the general rule is to do the opposite of the order of operations. Consider the following.Example 2.1.1Solve for x-7=5\phantom{1}\begin{array}{rrrrr} x&-&7&=&-5\\ &+&7&&+7\\ \midrule &&x&=&2 \end{array}4+x=8\phantom{1}\begin{array}{rrrrr} 4&+&x&=&8\\ -4&&&&-4\\ \midrule &&x&=&4 \end{array}7=x-9\phantom{1}\begin{array}{rrrrr} 7&=&x&-&9\\ +9&&&+&9\\ \midrule 16&=&x&& \end{array}5=8+x\phantom{1}\begin{array}{rrrrr} 5&=&8&+&x\\ -8&&-8&&\\ \midrule -3&=&x&& \end{array}Multiplication ProblemsIn a multiplication problem, get rid of the coefficient in front of the variable by dividing both sides of the equation by that number. Consider the following examples.Example 2.1.2Solve for \begin{array}{rrl} \\ \\ \\ \\ \\ 4x&=&20\\ \\ \dfrac {4x}{4}&=&\dfrac{20}{4}\\ \\ x&=&5 \end{array}\begin{array}{rrl} \\ \\ \\ \\ \\ 8x&=&-24\\ \\ \dfrac {8x}{8}&=&\dfrac{-24}{8}\\ \\ x&=&-3 \end{array}\begin{array}{rrl} \\ \\ \\ \\ \\ -4x&=&-20\\ \\ \dfrac {-4x}{-4}&=&\dfrac{-20}{-4}\\ \\ x&=&5 \end{array}Division ProblemsIn division problems, remove the denominator by multiplying both sides by it. Consider the following examples.Example 2.1.3Solve for \phantom{1}\begin{array}{rrl}\\ \dfrac{x}{-7}&=&-2\\ \\ -7\left(\dfrac{x}{-7}\right)&=&(-2)-7 \\ \\ x&=&14\end{array}\phantom{1}\begin{array}{rrl}\\ \dfrac{x}{8}&=&5\\ \\ 8\left(\dfrac{x}{8}\right)&=&(5)8\\ \\ x&=&40\end{array}\phantom{1}\begin{array}{rrl}\\ \dfrac{x}{-4}&=&9\\ \\ -4\left(\dfrac{x}{-4}\right)&=&(9) -4\\ \\ x&=&-36\end{array}QuestionsFor questions 1 to 28, solve each linear equation.v + 9 = 1614 = b + 3x - 11 = -16-14 = x - 1830 = a + 20-1 + k = 5x - 7 = -26-13 + p = -1913 = n - 522 = 16 + m340 = -17x4r = -28{-9} = \dfrac{n}{12}27 = 9b20v = -160-20x = -80340 = 20n12 = 8a16x = 3208k = -16-16 + n = -13-21 = x - 5p-8 = -21m - 4 = -13\dfrac{r}{14} = \dfrac{5}{14}\dfrac{n}{8} = {40}20b = -200-\dfrac{1}{3} = \dfrac{x}{12}Answer Key 2.1Extra Reading and Instructional VideosArticle to read: New theory finds âtraffic jamsâ in jet stream cause abnormal weather patterns.The abstract reads:A study offers an explanation for a mysterious and sometimes deadly weather pattern in which the jet stream, the global air currents that circle the Earth, stalls out over a region. Much like highways, the jet stream has a capacity, researchers said, and when itâs exceeded, blockages form that are remarkably similar to traffic jams â and climate forecasters can use the same math to model them both.92.2 Solving Linear EquationsWhen working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable.Example 2.2.1Solve for Â In solving the above equation, notice that the general pattern followed was to do the opposite of the equation. was added to 16, so 16 was then subtracted from both sides. The variable was multiplied by 4, so both sides were divided by 4.Â Example 2.2.2\begin{array}{rrrrr} \\ \\ \\ \\ \\ 5x &+& 7 &=& 7 \\ &-&7&&-7 \\ \midrule &&\dfrac{5x}{5} &=& \dfrac{0}{5}\\ \\ &&x& =& 0 \end{array}\begin{array}{rrrrr} \\ \\ \\ \\ \\ 4 &- &2x& =& 10 \\ -4&&&&-4\\ \midrule &&\dfrac{-2x}{-2}& = &\dfrac{6}{-2}\\ \\ &&x& =& -3 \end{array}\begin{array}{rrrrr} \\ \\ \\ \\ \\ -3x& -& 7& =& 8 \\ &+&7& = & + 7 \\ \midrule &&\dfrac{-3x}{-3}& =& \dfrac{15}{-3} \\ \\ &&x &= &-5\\ \end{array}Â QuestionsFor questions 1 to 20, solve each linear equation.5 + \dfrac{n}{4} = 4-2 = -2m + 12102 = -7r + 427 = 21 - 3x-8n + 3 = -77-4 - b = 80 = -6v-2 + \dfrac{x}{2} = 4-8 = \dfrac{x}{5} - 6-5 = \dfrac{a}{4} - 1Â 0 = -7 + \dfrac{k}{2}-6 = 15 + 3p-12 + 3x = 0-5m + 2 = 27\dfrac{b}{3} + 7 = 10\dfrac{x}{1} - 8 = -8152 = 8n + 64-11 = -8 + \dfrac{v}{2}-16 = 8a + 64-2x - 3 = -29Answer Key 2.2102.3 Intermediate Linear EquationsWhen working with linear equations with parentheses, the first objective is to isolate the parentheses. Once isolated, the parentheses can be removed and then the variable solved.Example 2.3.1Solve for in the equation Example 2.3.2Solve for in the equation Â For some problems, it is too difficult to isolate the parentheses. In these problems, it is necessary to multiply or divide throughout the parentheses by whatever coefficient is in front of it.Â Example 2.3.3Solve for in the equation QuestionsFor questions 1 to 26, solve each linear equation.2 - (-3a - 8) = 12(-3n + 8) = -20-5(-4 + 2v) = -502 - 8(-4 + 3x) = 3466 = 6(6 + 5x)32 = 2 - 5(-4n + 6)-2 + 2(8x -9) = -16-(3 - 5n) = 12-1 - 7m = -8m + 756p - 48 = 6p +21 - 12r = 29 - 8r4 + 3x = -12x + 420 - 7b = -12b + 30-16n + 12 = 39 - 7n-2 - 5(2 - 4m) = 33 + 5m-25 - 7x = 6(2x - 1)-4n + 11 = 2(1 - 8n) + 3n-7(1 + b) = -5 - 5b-6v-29 = -4v - 5(v+1)-8(8r - 2) = 3r + 162(4x - 4) = -20 - 4x-8n - 19 = -2(8n - 3) + 3n-2(m - 2) + 7(m - 8) = -677 = 4(n - 7) + 5(7n + 7)50 = 8(7 + 7r) - (4r + 6)-8(6 + 6x) + 4(-3 + 6x) = -12Answer Key 2.3112.4 Fractional Linear EquationsWhen working with fractions built into linear equations, it is often easiest to remove the fraction in the very first step. This generally means finding the LCD of the fraction and then multiplying every term in the entire equation by the LCD.Example 2.4.1Solve for in the equation For this equation, theÂ LCD isÂ 12, so every term in this equation will be multiplied byÂ 12.Â Â Cancelling out the denominator yields:Â Â Multiplying results in:Â Â Example 2.4.2Solve for x in the equation \dfrac{3\left(\dfrac{5}{9}x+\dfrac{4}{27}\right)}{2}=3.First, remove the outside denominatorÂ 2 by multiplying both sides byÂ 2:Â Â Â Â Now divide both sides byÂ 3, which leaves:Â Â To remove theÂ 9Â andÂ 27, multiply both sides by theÂ LCD, 27:Â Â This leaves:Â Â QuestionsFor questions 1 to 18, solve each linear equation.\dfrac{3}{5}\left(1 + p\right) = \dfrac{21}{20}-\dfrac{1}{2} = \dfrac{3k}{2} + \dfrac{3}{2}0 = -\dfrac{5}{4}\left(x-\dfrac{6}{5}\right)\dfrac{3}{2}n - 8 = -\dfrac{29}{12}\dfrac{3}{4} - \dfrac{5}{4}m = \dfrac{108}{24}\dfrac{11}{4} + \dfrac{3}{4}r = \dfrac{160}{32}2b + \dfrac{9}{5} = -\dfrac{11}{5}\dfrac{3}{2} - \dfrac{7}{4}v = -\dfrac{9}{8}\dfrac{3}{2}\left(\dfrac{7}{3}n+1\right) = \dfrac{3}{2}\dfrac{41}{9} = \dfrac{5}{2}\left(x+\dfrac{2}{3}\right) - \dfrac{1}{3}x-a - \dfrac{5}{4}\left(-\dfrac{8}{3}a+ 1\right) = -\dfrac{19}{4}\dfrac{1}{3}\left(-\dfrac{7}{4}k + 1\right) - \dfrac{10}{3}k = -\dfrac{13}{8}\dfrac{55}{6} = -\dfrac{5}{2}\left(\dfrac{3}{2}p-\dfrac{5}{3}\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x-\dfrac{3}{4}\right)-\dfrac{7}{2}x=-\dfrac{83}{24}-\dfrac{5}{8}=\dfrac{5}{4}\left(r-\dfrac{3}{2}\right)\dfrac{1}{12}=\dfrac{4}{3}x+\dfrac{5}{3}\left(x-\dfrac{7}{4}\right)-\dfrac{11}{3}+\dfrac{3}{2}b=\dfrac{5}{2}\left(b-\dfrac{5}{3}\right)\dfrac{7}{6}-\dfrac{4}{3}n=-\dfrac{3}{2}n+2\left(n+\dfrac{3}{2}\right)Answer Key 2.4122.5 Absolute Value EquationsWhen solving equations with absolute values, there can be more than one possible answer. This is because the variable whose absolute value is being taken can be either negative or positive, and both possibilities must be accounted for when solving equations.Â Example 2.5.1Solve |x| = 7.Â Â Â When there are absolute values in a problem, it is important to first isolate the absolute value, then remove the absolute value by considering both the positive and negative solutions. Notice that, in the next two examples, all the numbers outside of the absolute value are moved to one side first before the absolute value bars are removed and both positive and negative solutions are considered.Â Example 2.5.2Solve 5 + | x | = 8.Â Â Example 2.5.3Solve -4 | x | = -20.Â Â Â Note: the objective in solving for absolute values is to isolate the absolute value to yield a solution.Â Example 2.5.4Solve 5|x| -4 = 26.Â Â Â Often, the absolute value of more than just a variable is being taken. In these cases, it is necessary to solve the resulting equations before considering positive and negative possibilities. This is shown in the next example.Â Example 2.5.5Solve | 2x - 1 | = 7.Since absolute value can be positive or negative, this means that there are two equations to solve.Â Â Â Remember: the absolute value must be isolated first before solving.Â Example 2.5.6Solve 2 - 4 | 2x + 3 | = -18.Â Â Now, solve two equations to get the positive and negative solutions:Â Â Â There exist two other possible results from solving an absolute value besides what has been shown in the above six examples.Â Example 2.5.7Consider the equation | 2x - 1 | = 7 from Example 2.5.5.What happens if, instead, the equation to solve is |2x-1|=0 or |2x-1|=-5?For |2x-1|=0, there is no \pm 0, so there will be just one solution instead of two. Solving this equation yields:Â Â For |2x-1|=-5, the result will be âno solutionâ or \emptyset, since an absolute value is never negative.Â One final type of absolute value problem covered in this chapter is when two absolute values are equal to each other. There will still be both a positive and a negative resultâthe difference is that a negative must be distributed into the second absolute value for the negative possibility.Â Example 2.5.8Solve | 2x - 7 | = | 4x + 6 |.Solving this means solving:Â Â Removing the absolute value signs leaves:\begin{array}{ll} \begin{array}{rrrrrrr} 2x&-&7&=&4x&+&6 \\ -4x&+&7&&-4x&+&7 \\ \midrule &&\dfrac{-2x}{-2}&=&\dfrac{13}{-2}&& \\ \\ &&x&=&-\dfrac{13}{2}&& \end{array} & \hspace{0.25in}\text{ and }\hspace{0.25in} \begin{array}{rrrrrrr} 2x&-&7&=&-4x&-&6 \\ +4x&+&7&&+4x&+&7 \\ \midrule &&\dfrac{6x}{6}&=&\dfrac{1}{6}&& \\ \\ &&x&=&\dfrac{1}{6}&& \end{array} \end{array}QuestionsFor questions 1 to 24, solve each absolute value equation.| x | = 8| n | = 7| b | = 1| x | = 2| 5 + 8a | = 53| 9n + 8 | = 46| 3k + 8 | = 2| 3 - x | = 6-7 | -3 - 3r | = -21| 2 + 2b | + 1 = 37 | -7x - 3 | = 21| -4 - 3n | = 28 | 5p + 8 | - 5 = 113 - | 6n + 7 | = -405 | 3 + 7m | + 1 = 514 | r + 7 | + 3 = 59-7 + 8 | -7x - 3 | = 738 | 3 - 3n | - 5 = 91| 5x + 3 | = | 2x - 1 || 2 + 3x | = | 4 - 2x || 3x - 4 | = | 2x + 3 || 2x - 5 | = | 3x + 4 || 4x - 2 | = | 6x + 3 ||3x+2|=|2x-3|Answer Key 2.5132.6 Working With FormulasIn algebra, expressions often need to be simplified to make them easier to use. There are three basic forms of simplifying, which will be reviewed here. The first form of simplifying expressions is used when the value of each variable in an expression is known. In this case, each variable can be replaced with the equivalent number, and the rest of the expression can be simplified using the order of operations.Â Example 2.6.1Evaluate p(q + 6) when p = 3 and q = 5.\begin{array}{rl} (3)((5)+(6))&\text{Replace }p\text{ with 3 and }q\text{ with 5 and evaluate parentheses} \\ (3)(11)&\text{Multiply} \\ 33&\text{Solution} \end{array}Â Whenever a variable is replaced with something, the new number is written inside a set of parentheses. Notice the values of 3 and 5 in the previous example are in parentheses. This is to preserve operations that are sometimes lost in a simple substitution. Sometimes, the parentheses wonât make a difference, but it is a good habit to always use them to prevent problems later.Â Example 2.6.2Evaluate x + zx(3 - z)\left(\dfrac{x}{3}\right) when x = -6 and z = -2.Â Â Â Isolating variables in formulas is similar to solving general linear equations. The only difference is, with a formula, there will be several variables in the problem, and the goal is to solve for one specific variable. For example, consider solving a formula such as (the formula for the surface area of a right circular cone) forÂ the variable This means isolating the so the equation has on one side. So a solution might look like This second equation gives the same information as the first; they are algebraically equivalent. However, one is solved for the area while the other is solvedÂ forÂ the slant height of the cone When solving a formula for a variable, focus on the one variable that is being solved for; all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown: the first is a normal one-step equation, and the second is a formula that you are solving forÂ Â Example 2.6.3Isolate the variable x in the following equations.Â Â Â The same process is used to isolate in as in Because is being solved for, treat all other variables as numbers.Â For these two equations, both sides were divided by 3 and respectively. A similar idea is seen in the following example.Â Example 2.6.4Isolate the variable n in the equation m+n=p.To isolate n, the variable m must be removed, which is done by subtracting m from both sides:Â Â Â Since and are not like terms, they cannot be combined.Â ForÂ this reason, leave the expression as Â Example 2.6.5Isolate the variable a in the equation a(x - y) = b.This means that (x-y) must be isolated from the variable a.Â Â Â If no individual term inside parentheses is being solved for, keep the terms inside them together and divide by them as a unit. However, if an individual term inside parentheses is being solved for, it is necessary to distribute. The following example is the same formula as in Example 2.6.5, but this time, is being solved for.Â Example 2.6.6Isolate the variable x in the equation a(x - y) = b.First, distribute a throughout (x-y):Â Â Remove the term ay from both sides:Â Â ax is then divided by a:Â Â The solution is x=\dfrac{b+ay}{a}, which can also be shown as x=\dfrac{b}{a}+y.Â Be very careful when isolating not to try and cancel the on the top and the bottom of the fraction. This is not allowed if there is any adding or subtracting in the fraction. There is no reducing possible in this problem, so the final reduced answer remains The next example is another two-step problem.Â Example 2.6.7Isolate the variable m in the equation y=mx+b.First, subtract b from both sides:Â Â Now divide both sides by x:Â Â Therefore, the solution is m=\dfrac{y-b}{x}.Â It is important to note that a problem is complete when the variable being solved forÂ is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation.The next example is also a two-step equation. It is a problem from earlier in the lesson.Â Example 2.6.8Isolate the variable s in the equation A= \pi r^2+\pi rs.Subtract \pi r^2 from both sides:Â Â Divide both sides by \pi r:Â Â The solution is:Â Â Â Formulas often have fractions in them and can be solved in much the same way as any fraction. First, identify the LCD, and then multiply each term by the LCD. After reducing, there will be no more fractions in the problem.Â Example 2.6.9Isolate the variable m in the equation h=\dfrac{2m}{n}.To clear the fraction, multiply both sides by n:Â Â This leaves:Â Â Divide both sides by 2:Â Â Which reduces to:Â Â Example 2.6.10Isolate the variable b in the equation A=\dfrac{a}{2-b}.To clear the fraction, multiply both sides by (2-b):Â Â Which reduces to:Â Â Distribute A throughout (2-b), then isolate:Â Â Finally, divide both sides by -A:Â Â Solution:Â Â QuestionsFor questions 1 to 10, evaluate each expression using the values given.p + 1 + q - m\text{ (}m = 1, p = 3, q = 4)y^2+y-z\text{ (}y=5, z=1)p- \left[pq \div 6\right]\text{ (}p=6, q=5)\left[6+z-y\right]\div 3\text{ (}y=1, z=4)c^2-(a-1)\text{ (}a=3, c=5)x+6z-4y\text{ (}x=6, y=4, z=4)5j+kh\div 2\text{ (}h=5, j=4, k=2)5(b+a)+1+c\text{ (}a=2, b=6, c=5)\left[4-(p-m)\right]\div 2+q\text{ (}m=4, p=6, q=6)z+x-(1^2)^3\text{ (}x=5, z=4)For questions 11 to 34, isolate the indicated variable from the equation.b\text{ in }ab=ch\text{ in }g=\dfrac{h}{i}x\text{ in }\left(\dfrac{f}{g}\right)x=by\text{ in }p=\dfrac{3y}{q}x\text{ in }3x=\dfrac{a}{b}y\text{ in }\dfrac{ym}{b}=\dfrac{c}{d}\pi\text{ in }V=\dfrac{4}{3}\pi r^3m\text{ in }E=mv^2y\text{ in }c=\dfrac{4y}{m+n}r\text{ in }\dfrac{rs}{a-3}=kD\text{ in }V=\dfrac{\pi Dn}{12}R\text{ in }F=k(R-L)c\text{ in }P=n(p-c)L\text{ in }S=L+2BD\text{ in }T=\dfrac{D-d}{L}E_a\text{ in }I=\dfrac{E_a-E_q}{R}L_o\text{ in }L=L_o(1+at)m\text{ in }2m+p=4m+qk\text{ in }\dfrac{k-m}{r}=qT\text{ in }R=aT+bQ_2\text{ in }Q_1=P(Q_2-Q_1)r_1\text{ in }L=\pi(r_1+r_2)+2dT_1\text{ in }R=\dfrac{kA(T+T_1)}{d}V_2\text{ in }P=\dfrac{V_1(V_2-V_1)}{g}Answer Key 2.6142.7 Variation Word ProblemsDirect Variation ProblemsThere are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:A historical example of direct variation can be found in the changing measurement of pi, whichÂ has been symbolized using the Greek letter Ï since the mid 18th century. Variations of historical Ï calculations are Babylonian Egyptian and Indian In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of Ï to seven decimal places (3.1415926), representing the most accurate value of Ï for over 1000 years.Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716âŠ (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate Ï to 1013 decimals.Â Â All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:x=kyx varies directly as yx varies as yx varies directly proportional to yx is proportional to yx=ky^2x varies directly as the square of yx varies as y squaredx is proportional to the square of yx=ky^3x varies directly as the cube of yx varies as y cubedx is proportional to the cube of yx=ky^{\frac{1}{2}}x varies directly as the square root of yx varies as the root of yx is proportional to the square root of yÂ Example 2.7.1Find the variation equation described as follows:The surface area of a square surface is directly proportional to the square of either side Solution:Â Â Example 2.7.2When looking at two buildings at the same time, the length of the buildingsâ shadows varies directly as their height If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?The equation that describes this variation is:Â Â Breaking the data up into the first and second parts gives:Inverse Variation ProblemsInverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure and the volume of a gas, called Boyleâs Law (1662). This law is written as:Â Â Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:Â Â Another example is the historically famous inverse square laws. Examples of this are the force of gravity electrostatic force and the intensity of light In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.In equation form, these look like:Â Â These equations would be verbalized as:The force of gravity (F_{\text{g}}) varies inversely as the square of the distance.Electrostatic force (F_{\text{el}}) varies inversely as the square of the distance.The intensity of a light source (I) varies inversely as the square of the distance.All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:x=\dfrac{k}{y}x varies inversely as yx varies as the inverse of yx varies inversely proportional to yx is inversely proportional to yx=\dfrac{k}{y^2}x varies inversely as the square of yx varies inversely as y squaredx is inversely proportional to the square of yx=\dfrac{k}{y^3}x varies inversely as the cube of yx varies inversely as y cubedx is inversely proportional to the cube of yx=\dfrac{k}{y^{\frac{1}{2}}}x varies inversely as the square root of yx varies as the inverse root of yx is inversely proportional to the square root of yÂ Example 2.7.3Find the variation equation described as follows:The force experienced by a magnetic field is inversely proportional to the square of the distance from the source Solution:Â Â Example 2.7.4The time it takes to travel from North Vancouver to Hope varies inversely as the speed at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.The equation that describes this variation is:Â Â Breaking the data up into the first and second parts gives:Joint or Combined Variation ProblemsIn real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:Â Â where:F_{\text{g}} stands for the gravitational force of attractionG is Newtonâs constant, which would be represented by k in a standard variation problemm_1 and m_2 are the masses of the two bodiesd^2 is the distance between the centres of both bodiesTo write this out as a variation problem, first state that the force of gravitational attraction between two bodies is directly proportional to the product of the two masses and inversely proportional to the square of the distance separating the two masses.Â From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.Â Example 2.7.5Find the variation equation described as follows:The force of electrical attraction between two statically charged bodies is directly proportional to the product of the charges on each of the two objects and inversely proportional to the square of the distance separating these two charged bodies.Solution:Â Â Â Solving these combined or joint variation problems is the same as solving simpler variation problems.First, decide what equation the variation represents. Second, break up the data into the first data givenâwhich is used to find âand then the second data, which is used to solve the problem given. Consider the following joint variation problem.Â Example 2.7.6 varies jointly with and and inversely with the square of If when and find the constant then use to find when and The equation that describes this variation is:Â Â Breaking the data up into the first and second parts gives:QuestionsFor questions 1 to 12, write the formula defining the variation, including the constant of variation x varies directly as yx is jointly proportional to y and zx varies inversely as yx varies directly as the square of yx varies jointly as z and yx is inversely proportional to the cube of yx is jointly proportional with the square of y and the square root of zx is inversely proportional to y to the sixth powerx is jointly proportional with the cube of y and inversely to the square root of zx is inversely proportional with the square of y and the square root of zx varies jointly as z and y and is inversely proportional to the cube of px is inversely proportional to the cube of y and square of zFor questions 13 to 22, find the formula defining the variation and the constant of variation If A varies directly as B, find k when A=15 and B=5.If P is jointly proportional to Q and R, find k when P=12, Q=8 and R=3.If A varies inversely as B, find k when A=7 and B=4.If A varies directly as the square of B, find k when A=6 and B=3.If C varies jointly as A and B, find k when C=24, A=3, and B=2.If Y is inversely proportional to the cube of X, find k when Y=54 and X=3.If X is directly proportional to Y, find k when X=12 and Y=8.If A is jointly proportional with the square of B and the square root of C, find k when A=25, B=5 and C=9.If y varies jointly with m and the square of n and inversely with d, find k when y=10, m=4, n=5, and d=6.If P varies directly as T and inversely as V, find k when P=10, T=250, and V=400.For questions 23 to 37, solve each variation word problem.The electrical current I (in amperes, A) varies directly as the voltage (V) in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?The current I in an electrical conductor varies inversely as the resistance R (in ohms, Î©) of the conductor. If the current is 12 A when the resistance is 240 Î©, what is the current when the resistance is 540 Î©?Hookeâs law states that the distance (d_s) that a spring is stretched supporting a suspended object varies directly as the mass of the object (m). If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?The volume (V) of an ideal gas at a constant temperature varies inversely as the pressure (P) exerted on it. If the volume of a gas is 200 cm3 under a pressure of 32 kg/cm2, what will be its volume under a pressure of 40 kg/cm2?The number of aluminum cans (c) used each year varies directly as the number of people (p) using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?The time (t) required to do a masonry job varies inversely as the number of bricklayers (b). If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?The wavelength of a radio signal (Î») varies inversely as its frequency (f). A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?The number of kilograms of water (w) in a human body is proportional to the mass of the body (m). If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?The time (t) required to drive a fixed distance (d) varies inversely as the speed (v). If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?The volume (V) of a cone varies jointly as its height (h) and the square of its radius (r). If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm3, what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?The centripetal force (F_{\text{c}}) acting on an object varies as the square of the speed (v) and inversely to the radius (r) of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the objectâs speed increases to 25 m/s and the path is now 1.0 m?The maximum load (L_{\text{max}}) that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter (d) and inversely as the square of the height (h). If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?The volume (V) of gas varies directly as the temperature (T) and inversely as the pressure (P). If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm2, what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm2?The electrical resistance (R) of a wire varies directly as its length (l) and inversely as the square of its diameter (d). A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Î©. Find the electrical resistance in a 10.0 m long wire having twice the diameter.The volume of wood in a tree (V) varies directly as the height (h) and the diameter (d). If the volume of a tree is 377 m3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m3 and a diameter of 1.75 m?Answer Key 2.7152.8 The Mystery X PuzzleThe centre number of each square is found by using the order of operations on the numbers that surround it. The challenge is to solve for the variable Â Â Can you solve for ? Can you find any other possible solution?Answer Key 2.8IIIChapter 3: GraphingLearning ObjectivesThis chapter covers:Points & CoordinatesMidpoint & Distance Between Two PointsSlopes & Their GraphsGraphing Linear EquationsConstructing Linear EquationsPerpendicular & Parallel LinesNumeric Word Problems163.1 Points and CoordinatesOften, to get an idea of the behaviour of an equation or some function, a visual representation that displays the solutions to the equation or function in the form of a graph will be made. Before exploring this, it is necessary to review the foundations of a graph. The following is an example of what is called the coordinate plane of a graph.Â Â The plane is divided into four sections by a horizontal number line (-axis) and a vertical number line (-axis). Where the two lines meet in the centre is called the origin. This centre origin is where and and is represented by the ordered pair .Â x-axisÂ For the -axis, moving to the right from the centre 0, the numbers count up, and To the left of the centre 0, the numbers count down, and Â y-axisÂ Similarly, for the -axis, moving up from the centre 0, the numbers count up, and Moving down from the centre 0, the numbers count down, and When identifying points on a graph, a dot is generally used with a set of parentheses following that gives the -value followed by the -value. This will look like or and is given the formal name of an ordered pair.This coordinate system is universally used, with the simplest example being the kind of treasure map that is usually encountered in childhood, or the longitude and latitude system used to identify any position on the Earth. For this system, the -axis (which represents latitude) is the equator and the -axis (which represents longitude) or the prime meridian is the line that passes though Greenwich, England. The origin of the Earthâs latitude and longitude (0Â°, 0Â°) is a fictional island called âNull Island.âÂ Representations of the globe that demonstrate latitude and longitude. Long description available.Latitude and longitude. [Long Description]Â A hand-drawn treasure map of an island.The treasure map of Robert Louis Stevenson made popular by his work Treasure Island. From Cordingly, David (1995). Under the Black Flag: The romance and the reality of life among the pirates. Times Warner, 1996.Â Example 3.1.1Identify the coordinates of the following data points.Â Â For the -coordinate, move 4 to the right from the origin. For the -coordinate, move 4 up. This gives the final coordinates of (4, 4). For the -coordinate, stay at the origin. For the -coordinate, move 2 up. This gives the final coordinates of (0, 2). For the -coordinate, move 3 to the left from the origin. For the -coordinate, move 2 up. This gives the final coordinates of (â3, 2). For the -coordinate, move 2 to the left from the origin. For the -coordinate, move 4 down. This gives the final coordinates of (â2, â4). For the -coordinate, move 3 to the right from the origin. For the -coordinate, move 2 down. This gives the final coordinate of (3, â2).Example 3.1.2Graph the points A(3, 2), B(â2, 1), C(3, â4), and D(â2, â3).The first point, A, is at (3, 2). This means (3 to the right) and (up 2). Following these instructions, starting from the origin, results in the correct point.The second point, B(â2, 1), is left 2 for the -coordinate and up 1 for the -coordinate.The third point, C(3 ,â4), is right 3, down 4.The fourth point, D(â2, â3), is left 2, down 3.Â QuestionsWhat are the coordinates of each point on the graph below? graph with points a-h in placePlot and label the following points on the graph. Graph with points a-h in place \begin{array}{lll} \text{A }(-5,5)&\text{B }(1,0)&\text{C }(-3,4) \\ \text{D }(-3,0)&\text{E }(-4, 2)&\text{F }(4,-2) \\ \text{G }(-2,-2)&\text{H }(3,-2)&\text{I }(0,3) \end{array}Answer Key 3.1Long DescriptionsLatitude and longitude long description: Two views of the globe that show the landmark points of the latitude and longitude system.The first globe demonstrates the lines of latitude. The centre line of latitude is called the equator and represents 0Â° latitude. It wraps around the centre of the Earth from west to east. The globe shows North and South America, and the equator runs through the northern part of South America. The North Pole is at 90Â° latitude and the South Pole is at â90Â° latitude. Positive latitude is above the equator, and negative latitude is below it.The second globe demonstrates the lines of longitude. The centre line of longitude is called the prime meridian and represents 0Â° longitude. It wraps around the centre of the Earth from north to south. It passes through Greenwich, England, by convention, as well as parts of France, Spain, and western Africa. Positive longitude is east of the prime meridian, and negative longitude is west of it. [Return to Latitude and longitude]173.2 Midpoint and Distance Between PointsFinding the Distance Between Two PointsThe logic used to find the distance between two data points on a graph involves the construction of a right triangle using the two data points and the Pythagorean theorem to find the distance.To do this for the two data points and , the distance between these two points will be found using and Using the Pythagorean theorem, this will end up looking like:Â Â or, in expanded form:Â Â Â On graph paper, this looks like the following. For this illustration, both and are 7 units long, making the distance or .Â Â The square root of 98 is approximately 9.899 units long.Â Example 3.2.1Find the distance between the points and .Start by identifying which are the two data points and . Let be and be .Now: or and or .This means thatorwhich reduces toorTaking the square root, the result is .Finding the Midway Between Two Points (Midpoint)The logic used to find the midpoint between two data points and on a graph involves finding the average values of the data points and the of the data points . The averages are found by adding both data points together and dividing them by .In an equation, this looks like: and Â Example 3.2.2Find the midpoint between theÂ points and .Â Â We start by adding the two data points and then dividing this result by 2.orThe midpointâs -coordinate is found by adding the two data points and then dividing this result by 2.orThe midpoint between the points and is at the data point .QuestionsFor questions 1 to 8, find the distance between the points.Â (â6, â1) and (6, 4)(1, â4) and (5, â1)(â5, â1) and (3, 5)(6, â4) and (12, 4)(â8, â2) and (4, 3)(3, â2) and (7, 1)(â10, â6) and (â2, 0)(8, â2) and (14, 6)For questions 9 to 16, find the midpoint between the points.(â6, â1) and (6, 5)(1, â4) and (5, â2)(â5, â1) and (3, 5)(6, â4) and (12, 4)(â8, â1) and (6, 7)(1, â6) and (3, â2)(â7, â1) and (3, 9)(2, â2) and (12, 4)Answer Key 3.2183.3 Slopes and Their GraphsAnother important property of any line or linear function is slope. Slope is a measure of steepness and indicates in some situations how fast something is changingâspecifically, its rate of change. A line with a large slope, such as 10, is very steep. A line with a small slope, such as is very flat or nearly level. Lines that rise from left to right are called positive slopes and lines that sink are called negative slopes. Slope can also be used to describe the direction of a line. A line that goes up as it moves from from left to right is described as having a positive slope whereas a line that goes downward has a negative slope. Slope, therefore, will define a line as rising or falling.Slopes in real life have significance. For instance, roads with slopes that are potentially dangerous often carry warning signs. For steep slopes that are rising, extra slow moving lanes are generally provided for large trucks. For roads that have steep down slopes, runaway lanes are often provided for vehicles that lose their ability to brake.Â Caution signs for steep upward and downward slopes.Â When quantifying slope, use the measure of the rise of the line divided by its run. The symbol that represents slope is the letter which has unknown origins. Its first recorded usage is in an 1844 text by Matthew OâBrian, âA Treatise on Plane Co-Ordinate Geometry,â4 which was quickly followed by George Salmonâs âA Treatise on Conic Sectionsâ (1848), in which he used in the equation Since the rise of a line is shown by the change in the -value and the run is shown by the change in the -value, this equation is shortened to:This equation is often expanded to:Â Â Example 3.3.1Find the slope of the following line.Â Â First, choose two points on the line on this graph. Any points can be chosen, but they should fall on one of the corner grids. These are labelled and To find the slope of this line, consider the rise, or vertical change, and the run, or horizontal change. Observe in this example that the -value (the rise) goes from 4 to â2.Therefore, , orÂ (4Â â â2), which equals (4 + 2), or 6.Â Â The -value (the run) goes fromÂ â2 to 4.Therefore, , orÂ (â2 â 4), which equals (â2 + â4), or â6.This means the slope of this line is , or , or â1.Â Â Example 3.3.2Find the slope of the following line.Â Â First, choose two points on the line on this graph. Any points can be chosen, but to fall on a corner grid, they should be on opposite sides of the graph. These are and To find the slope of this line, consider the rise, or vertical change, and the run, or horizontal change. Observe in this example that the -value (the rise) goes from â4 to 1.Therefore, , orÂ (1Â â â4), which equals 5.Â Â The -value (the run) goes from â6 to 6.Therefore, or (6Â â â6), which equals 12.This means the slope of this line is , or , which cannot be further simplified.Â Â There are two special lines that have unique slopes that must be noted: lines with slopes equal to zero and slopes that are undefined.Undefined slopes arise when the line on the graph is vertical, going straight up and down. In this case, which means that zero is divided by while calculating the slope, which makes it undefined.Zero slopes are flat, horizontal lines that do not rise or fall; therefore, In this case, the slope is simply 0.Â Â Â Most often, the slope of the line must be found using data points rather than graphs. In this case, two data points are generally given, and the slope is found by dividing by This is usually done using the expanded slope equation of:Â Example 3.3.3Find the slope of a line that would connect the data points and .Choose Point 1 to be and Point 2 to be .Â Â Example 3.3.4Find the slope of a line that would connect the data points and .Choose Point 1 to be and Point 2 to be .Â Â This is an example of a flat, horizontal line.Example 3.3.5Find the slope of a line that would connect the data points and .Choose Point 1 to be and Point 2 to be .Â Â This is a vertical line.Example 3.3.6Find the slope of a line that would connect the data points and .Choose Point 1 to be and Point 2 to be .Â Â QuestionsFor questions 1 to 6, find the slope of each line shown on the graph.Â Â For questions 7 to 26, find the slope of the line that would connect each pair of points.(2, 10), (â2, 15)(1, 2), (â6, â12)(â5, 10), (0, 0)(2, â2), (7, 8)(4, 6), (â8, â10)(â3, 6), (9, â6)(â2 â4), (10, â4)(3, 5), (2, 0)(â4, 4), (â6, 8)(9, â6), (â7, â7)(2, â9), (6, 4)(â6, 2), (5, 0)(â5, 0), (â5, 0)(8, 11), (â3, â13)(â7, 9), (1, â7)(1, â2), (1, 7)(7, â4), (â8, â9)(â8, â5), (4, â3)(â5, 7), (â8, 4)(9, 5), (5, 1)Answer Key 3.3Derivation of Slope: https://services.math.duke.edu//education/webfeats/Slope/Slopederiv.html193.4 Graphing Linear EquationsThere are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.If the equation is given in the form , then gives the rise over run value and the value gives the point where the line crosses the -axis, also known as the -intercept.Â Example 3.4.1Given the following equations, identify the slope and the -intercept.\begin{array}{lll} y = 2x - 3\hspace{0.14in} & \text{Slope }(m)=2\hspace{0.1in}&y\text{-intercept } (b)=-3 \end{array}\begin{array}{lll} y = \dfrac{1}{2}x - 1\hspace{0.08in} & \text{Slope }(m)=\dfrac{1}{2}\hspace{0.1in}&y\text{-intercept } (b)=-1 \end{array}\begin{array}{lll} y = -3x + 4 & \text{Slope }(m)=-3 &y\text{-intercept } (b)=4 \end{array}\begin{array}{lll} y = \dfrac{2}{3}x\hspace{0.34in} & \text{Slope }(m)=\dfrac{2}{3}\hspace{0.1in} &y\text{-intercept } (b)=0 \end{array}Â When graphing a linear equation using the slope-intercept method, start by using the value given for the -intercept. After this point is marked, then identify other points using the slope.This is shown in the following example.Â Example 3.4.2Graph the equation .First, place a dot on the -intercept, , which is placed on the coordinate Â Â Now, place the next dot using the slope of 2.A slope of 2 means that the line rises 2 for every 1 across.Simply, is the same as , where and .Placing these points on the graph becomes a simple counting exercise, which is done as follows:Â Â Once several dots have been drawn, draw a line through them, like so:Â Â Note that dots can also be drawn in the reverse of what has been drawn here.Slope is 2 when rise over run is or , which would be drawn as follows:Â Â Example 3.4.3Graph the equation .First, place a dot on the -intercept, .Now, place the dots according to the slope, .Â Â This will generate the following set of dots on the graph. All that remains is to draw a line through the dots.Â Â The second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find when and find when .Â Example 3.4.4Graph the equation .To find the first coordinate, choose .This yields:Â Â Coordinate is .Now choose .This yields:Â Â Coordinate is .Draw these coordinates on the graph and draw a line through them.Â Example 3.4.5Graph the equation .To find the first coordinate, choose .This yields:Â Â Coordinate is .Now choose .This yields:Â Â Coordinate is .Draw these coordinates on the graph and draw a line through them.Â Example 3.4.6Graph the equation .To find the first coordinate, choose .This yields:Â Â Coordinate is .Since the intercept is , finding the other intercept yields the same coordinate. In this case, choose any value of convenience.Choose .This yields:Â Â Coordinate is .Draw these coordinates on the graph and draw a line through them.Â QuestionsFor questions 1 to 10, sketch each linear equation using the slope-intercept method.y = -\dfrac{1}{4}x - 3y = \dfrac{3}{2}x - 1y = -\dfrac{5}{4}x - 4y = -\dfrac{3}{5}x + 1y = -\dfrac{4}{3}x + 2y = \dfrac{5}{3}x + 4y = \dfrac{3}{2}x - 5y = -\dfrac{2}{3}x - 2y = -\dfrac{4}{5}x - 3y = \dfrac{1}{2}xFor questions 11 to 20, sketch each linear equation using the and -intercepts.x + 4y = -42x - y = 22x + y = 43x + 4y = 122x - y = 24x + 3y = -12x + y = -53x + 2y = 6x - y = -24x - y = -4For questions 21 to 28, sketch each linear equation using any method.y = -\dfrac{1}{2}x + 3y = 2x - 1y = -\dfrac{5}{4}xy = -3x + 2y = -\dfrac{3}{2}x + 1y = \dfrac{1}{3}x - 3y = \dfrac{3}{2}x + 2y = 2x - 2For questions 29 to 40, reduce and sketch each linear equation using any method.y + 3 = -\dfrac{4}{5}x + 3y - 4 = \dfrac{1}{2}xx + 5y = -3 + 2y3x - y = 4 + x - 2y4x + 3y = 5 (x + y)3x + 4y = 12 - 2y2x - y = 2 - y \text{ (tricky)}7x + 3y = 2(2x + 2y) + 6x + y = -2x + 33x + 4y = 3y + 62(x + y) = -3(x + y) + 59x - y = 4x + 5Answer Key 3.4203.5 Constructing Linear EquationsQuite often, students are required to find the equation of a line given only a data point and a slope or two data points. The simpler of these problems is to find the equation when given a slope and a data point. To do this, use the equation that defines the slope of a line:This becomes:To illustrate this method, consider the following example.Â Example 3.5.1Find the equation that has slope and passes through the point .First, replace with 2 and with .Next, reduce the resulting equation.First, multiply both sides by the denominator to eliminate the fraction:This leaves:Which simplifies to:This can be written in the -intercept form by isolating the variable :It is also useful to write the equation in the general form of , where , , and are integers and is positive.In general form, becomes:The standard form of a linear equation is written as .In standard form,Â becomes:The three common forms that a linear equation can be written in are:Â Â Example 3.5.2Find the equation having slope that passes though the point Write the solutions in slope-intercept form and in both general and standard forms.First, replace with and with Multiplying both sides by to eliminate the denominators yields:Which simplifies to:Â Writing this solution in all three forms looks like:Â Â Â The more difficult variant of this type of problem is that in which the equation of a line that connects two data points must be found. However, this is simpler than it may seem.The first step is to find the slope of the line that would connect those two points. Use the slope equation, as has been done previously in this textbook. After this is done, use this slope and one of the two data points given at the beginning of the problem. The following example illustrates this.Â Example 3.5.3Find the equation of the line that runs through and .First, find the slope:Â Â Now, treat this as a problem of finding a line with a given slope running through a point .The slope is , and there are two points to choose from: and . Choose the simplest point to work with. For this problem, either point works. For this example, choose .Â Â Eliminate the fraction by multiplying both sides by .This leaves which must be simplified:Â Â This answer is in standard form, but it can easily be converted to the -intercept form or general form if desired.QuestionsFor questions 1 to 12, write the slope-intercept form of each linear equation using the given point and slope.(2, 3) and m = \dfrac{2}{3}(1, 2) and m = 4(2, 2) and m = \dfrac{1}{2}(2, 1) and m = -\dfrac{1}{2}(-1, -5) and m = 9(2, -2) and m = -2(-4 , 1) and m = \dfrac{3}{4}(4, -3) and m = -2(0, -2) and m = -3(-1, 1) and m = 4(0, -5) and m = -\dfrac{1}{4}(0, 2) and m= -\dfrac{5}{4}For questions 13 to 24, write the general form of each linear equation using the given point and slope.(-1, -5) and m = 2(2, -2) and m= -2(5, -1) and m= -\dfrac{3}{5}(-2, -2) and m= -\dfrac{2}{3}(-4, 1) and m= \dfrac{1}{2}(4, -3) and m=-\dfrac{7}{4}(4, -2) and m= -\dfrac{3}{2}(-2, 0) and m= -\dfrac{5}{2}(-5, -3) and m= -\dfrac{2}{5}(3, 3) and m= \dfrac{7}{3}(2, -2) and m= 1(-3, 4) and m= -\dfrac{1}{3}For questions 25 and 32, write the slope-intercept form of each linear equation using the given points.(-4, 3) and (-3, 1)(1, 3) and (-3, -3)(5, 1) and (-3, 0)(-4, 5) and (4, 4)(-4, -2) and (0, 4)(-4, 1) and (4, 4)(3, 5) and (-5, 3)(-1, -4) and (-5, 0)For questions 33 to 40, write the general form of each linear equation using the given points.(3, -3) and (-4, 5)(-1, -5) and (-5, -4)(3, -3) and (-2, 4)(-6, -7) and (-3, -4)(-5,1) and (-1, -2)(-5,-1) and (5, -2)(-5, 5) and (2, -3)(1, -1) and (-5, -4)Answer Key 3.5213.6 Perpendicular and Parallel LinesPerpendicular, parallel, horizontal, and vertical lines are special lines that have properties unique to each type. Parallel lines, for instance, have the same slope, whereas perpendicular lines are the opposite and have negative reciprocal slopes. Vertical lines have a constant -value, and horizontal lines have a constant -value.Two equations govern perpendicular and parallel lines:For parallel lines, the slope of the first line is the same as the slope for the second line. If the slopes of these two lines are called and , then .Perpendicular lines are slightly more difficult to understand. If one line is rising, then the other must be falling, so both lines have slopes going in opposite directions. Thus, the slopes will always be negative to one another. The other feature is that the slope at which one is rising or falling will be exactly flipped for the other one. This means that the slopes will always be negative reciprocals to each other. If the slopes of these two lines are called and , then .Â Example 3.6.1Find the slopes of the lines that are parallel and perpendicular to The parallel line has the identical slope, so its slope is also 3.The perpendicular line has the negative reciprocal to the other slope, so it is Example 3.6.2Find the slopes of the lines that are parallel and perpendicular to The parallel line has the identical slope, so its slope is also The perpendicular line has the negative reciprocal to the other slope, so it is Â Typically, questions that are asked of students in this topic are written in the form of âFind the equation of a line passing through point that is perpendicular/parallel to .â The first step is to identify the slope that is to be used to solve this equation, and the second is to use the described methods to arrive at the solution like previously done. For instance:Â Example 3.6.3Find the equation of the line passing through the point that is parallel to the line The first step is to identify the slope, which here is the same as in the given equation, .Now, simply use the methods from before:Â Â Clearing the fraction by multiplying both sides by leaves:Â Â Now put this equation in one of the three forms. For this example, use the standard form:Â Â Example 3.6.4Find the equation of the line passing through the point that is perpendicular to the line The first step is to identify the slope, which here is the negative reciprocal to the one in the given equation, so Now, simply use the methods from before:Â Â First, clear the fraction by multiplying both sides by . This leaves:Â Â which reduces to:Â Â Now put this equation in one of the three forms. For this example, choose the general form:Â Â For the general form, the coefficient in front of the must be positive. So for this equation, multiply the entire equation by â1 to make positive.Â Questions that are looking for the vertical or horizontal line through a given point are the easiest to do and the most commonly confused.Vertical lines always have a single -value, yielding an equation like Horizontal lines always have a single -value, yielding an equation like Â Example 3.6.5Find the equation of the vertical and horizontal lines through the point The vertical line has the same -value, so the equation is .The horizontal line has the same -value, so the equation is .QuestionsFor questions 1 to 6, find the slope of any line that would be parallel to each given line.y = 2x + 4y = -\dfrac{2}{3}x + 5y = 4x - 5y = -10x - 5x - y = 46x - 5y = 20For questions 7 to 12, find the slope of any line that would be perpendicular to each given line.y = \dfrac{1}{3}xy = -\dfrac{1}{2}x - 1y = -\dfrac{1}{3}xy = \dfrac{4}{5}xx - 3y = -63x - y = -3For questions 13 to 18, write the slope-intercept form of the equation of each line using the given point and line.(1, 4) and parallel to y = \dfrac{2}{5}x + 2(5, 2) and perpendicular to y = \dfrac{1}{3}x + 4(3, 4) and parallel to y = \dfrac{1}{2}x - 5(1, â1) and perpendicular to y = -\dfrac{3}{4}x + 3(2, 3) and parallel to y = -\dfrac{3}{5}x + 4(â1, 3) and perpendicular to y = -3x - 1For questions 19 to 24, write the general form of the equation of each line using the given point and line.(1, â5) and parallel to -x + y = 1(1, â2) and perpendicular to -x + 2y = 2(5, 2) and parallel to 5x + y = -3(1, 3) and perpendicular to -x + y = 1(4, 2) and parallel to -4x + y = 0(3, â5) and perpendicular to 3x + 7y = 0For questions 25 to 36, write the equation of either the horizontal or the vertical line that runs through each point.Horizontal line through (4, â3)Vertical line through (â5, 2)Vertical line through (â3,1)Horizontal line through (â4, 0)Horizontal line through (â4, â1)Vertical line through (2, 3)Vertical line through (â2, â1)Horizontal line through (â5, â4)Horizontal line through (4, 3)Vertical line through (â3, â5)Vertical line through (5, 2)Horizontal line through (5, â1)Answer Key 3.6223.7 Numeric Word ProblemsNumber-based word problems can be very confusing, and it takes practice to convert a word-based sentence into a mathematical equation. The best strategy to solve these problems is to identify keywords that can be pulled out of a sentence and use them to set up an algebraic equation.Variables that are to be solved for are often written as âa number,â âan unknown,â or âa value.ââEqualâ is generally represented by the words âis,â âwas,â âwill be,â or âare.âAddition is often stated as âmore than,â âthe sum of,â âadded to,â âincreased by,â âplus,â âall,â or âtotal.â Addition statements are quite often written backwards. An example of this is âthree more than an unknown number,â which is written as Subtraction is often written as âless than,â âminus,â âdecreased by,â âreduced by,â âsubtracted from,â or âthe difference of.â Subtraction statements are quite often written backwards. An example of this is âthree less than an unknown number,â which is written as Multiplication can be seen in written problems with the words âtimes,â âthe product of,â or âmultiplied by.âDivision is generally found by a statement such as âdivided by,â âthe quotient of,â or âper.âÂ Example 3.7.128 less than five times a certain number is 232. What is the number?28 less means that it is subtracted from the unknown number (write this as â28)five times an unknown number is written as 5xis 232 means it equals 232 (write this as = 232)Putting these pieces together and solving gives:Â Â Example 3.7.2Fifteen more than three times a number is the same as nine less than six times the number. What is the number?Fifteen more than three times a number is 3x + 15 or 15 + 3xis means =nine less than six times the number is 6x-9Putting these parts together gives:Â Â Â Another type of number problem involves consecutive integers, consecutive odd integers, or consecutive even integers. Consecutive integers are numbers that come one after the other, such as 3, 4, 5, 6, 7. The equation that relates consecutive integers is:Â Â Consecutive odd integers and consecutive even integers both share the same equation, since every second number must be skipped to remain either odd (such as 3, 5, 7, 9) or even (2, 4, 6, 8). The equation that is used to represent consecutive odd or even integers is:Â Â Â Example 3.7.3The sum of three consecutive integers is 93. What are the integers?The relationships described in equation form are as follows:Â Â Which reduces to:Â Â This means that the three consecutive integers are 30, 31, and 32.Example 3.7.4The sum of three consecutive even integers is 246. What are the integers?The relationships described in equation form are as follows:Â Â Which reduces to:Â Â This means that the three consecutive even integers are 80, 82, and 84.QuestionsFor questions 1 to 8, write the formula defining each relationship. Do not solve.Five more than twice an unknown number is 25.Twelve more than 4 times an unknown number is 36.Three times an unknown number decreased by 8 is 22.Six times an unknown number less 8 is 22.When an unknown number is decreased by 8, the difference is half the unknown number.When an unknown number is decreased by 4, the difference is half the unknown number.The sum of three consecutive integers is 21.The sum of the first two of three odd consecutive integers, less the third, is 5.For questions 9 to 16, write and solve the equation describing each relationship.When five is added to three times a certain number, the result is 17. What is the number?If five is subtracted from three times a certain number, the result is 10. What is the number?Sixty more than nine times a number is the same as two less than ten times the number. What is the number?Eleven less than seven times a number is five more than six times the number. Find the number.The sum of three consecutive integers is 108. What are the integers?The sum of three consecutive integers is â126. What are the integers?Find three consecutive integers such that the sum of the first, twice the second, and three times the third is â76.Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 70.Answer Key 3.7233.8 The Newspaper Delivery PuzzleA newspaper delivery driver wants to visit of number of specific stores in a city to drop off bundles of newspapers and return to the starting position by taking the shortest distance. Since there are a number of stores and choices of what path is best to take, what is the shortest distance you can find?One possible path is shown below. Can you find the distance covered in this path if each square represents 1 km?ShapeÂ Â Can you find a shorter path? How many km of driving would you save?Â Â These problems can rapidly become more difficult as the number of stops are increased. An example of this is as follows. What is the shortest distance that you can find for this puzzle?Why UPS Drivers Donât Turn Left and You Probably Shouldnât EitherThe company UPS has analyzed delivery routs thoroughly. Read this article by Graham Kendall2 to find out how this affects the routes their drivers take.Why UPS drivers donât turn left and you probably shouldnât either Kendall, Graham. Why UPS Drivers Don't Turn Right and Why You Probably Shouldn't Either. https://theconversation.com/why-ups-drivers-dont-turn-left-and-you-probably-shouldnt-either-71432IVChapter 4: InequalitiesLearning ObjectivesThis chapter covers:Solving and Graphing Linear EquationsCompound InequalitiesLinear Absolute Value Inequalities2D Inequality & Absolute Value GraphsGeometric Word Problems244.1 Solve and Graph Linear InequalitiesWhen given an equation, such as or there are specific values for the variable. However, with inequalities, there is a range of values for the variable rather than a defined value. To write the inequality, use the following notation and symbols:Â SymbolMeaningRight arrow attached to a left parenthesis.> Greater thanRight arrow attached to a left square bracket.â€Â Greater than or equal toLeft arrow attached to a right parenthesis.< Less thanLeft arrow attached to a right square bracket.â„Â Less than or equal toExample 4.1.1Given a variable such that , this means that can be as close to 4 as possible but always larger. For , can equal 5, 6, 7, 199. Even 4.000000000000001 is true, since is larger than 4, so all of these are solutions to the inequality. The line graph of this inequality is shown below:Â Â Written in interval notation, is shown as .Example 4.1.2Likewise, if , then can be any value less than 3, such as 2, 1, â102, even 2.99999999999. The line graph of this inequality is shown below:Â Â Written in interval notation, is shown as .Example 4.1.3For greater than or equal (â„) and less than or equal (â€), the inequality starts at a defined number and then grows larger or smaller. For can equal 5, 6, 7, 199, or 4. The line graph of this inequality is shown below:Â Â Written in interval notation, is shown as .Example 4.1.4If , then can be any value less than or equal to 3, such as 2, 1, â102, or 3. The line graph of this inequality is shown below:Â Â Written in interval notation, is shown as Â When solving inequalities, the direction of the inequality sign (called the sense) can flip over. The sense will flip under two conditions:First, the sense flips when the inequality is divided or multiplied by a negative. For instance, in reducing , it is necessary to divide both sides by â3. This leaves Second, the sense will flip over if the entire equation is flipped over. For instance, , when flipped over, would look like In both cases, the 2 must be shown to be smaller than the , or the is always greater than 2, no matter which side each term is on.Â Example 4.1.5Solve the inequality and show the solution on both a number line and in interval notation.First, subtract 5 from both sides:Â Â Divide both sides by â2:Â Â Since the inequality is divided by a negative, it is necessary to flip the direction of the sense.This leaves:Â Â In interval notation, the solution is written as .On a number line, the solution looks like:Â Â Inequalities can get as complex as the linear equations previously solved in this textbook. All the same patterns for solving inequalities are used for solving linear equations.Â Example 4.1.6Solve and give interval notation of .Multiply out the parentheses:Â Â Simplify both sides:Â Â Combine like terms:Â Â The last thing to do is to isolate from the â2. This is done by dividing both sides by â2. Because both sides are divided by a negative, the direction of the sense must be flipped.This means:Â Â Will end up looking like:Â Â The solution written on a number line is:Â Â Written in interval notation, is shown as .QuestionsFor questions 1 to 6, draw a graph for each inequality and give its interval notation.n > -5n > 4-2 \le k1 \ge k5 \ge x-5 < xFor questions 7 to 12, write the inequality represented on each number line and give its interval notation.For questions 13 to 38, draw a graph for each inequality and give its interval notation.\dfrac{x}{11}\ge 10-2 \le \dfrac{n}{13}2 + r < 3\dfrac{m}{5} \le -\dfrac{6}{5}8+\dfrac{n}{3}\ge 611 > 8+\dfrac{x}{2}2 > \dfrac{(a-2)}{5}\dfrac{(v-9)}{-4} \le 2-47 \ge 8 -5x\dfrac{(6+x)}{12} \le -1-2(3+k) < -44-7n-10 \ge 6018 < -2(-8+p)5 \ge \dfrac{x}{5} + 124 \ge -6(m - 6)-8(n - 5) \ge 0-r -5(r - 6) < -18-60 \ge -4( -6x - 3)24 + 4b < 4(1 + 6b)-8(2 - 2n) \ge -16 + n-5v - 5 < -5(4v + 1)-36 + 6x > -8(x + 2) + 4x4 + 2(a + 5) < -2( -a - 4)3(n + 3) + 7(8 - 8n) < 5n + 5 + 2-(k - 2) > -k - 20-(4 - 5p) + 3 \ge -2(8 - 5p)Answer Key 4.1254.2 Compound InequalitiesSeveral inequalities can be combined together to form what are called compound inequalities.The first type of compound inequality is the âorâ inequality, which is true when either inequality results in a true statement. When graphing this type of inequality, one useful trick is to graph each individual inequality above the number line before moving them both down together onto the actual number line.When giving interval notation for a solution, if there are two different parts to the graph, put a âȘ (union) symbol between two sets of interval notation, one for each part.Â Example 4.2.1Solve the inequality Graph the solution and write it in interval notation.Isolate the variables from the numbers:Â Â Isolate the variable (remember to flip the sense where necessary):Â Â Solution:Â Â Position the inequalities and graph:Â Â In interval notation, the solution is written as .Â Note: there are several possible results that result from an âorâ statement. The graphs could be pointing different directions, as in the graph above, or pointing in the same direction, as in the graph representing that is shown below.Â Â In interval notation, this solution is written as .It is also possible to have solutions that point in opposite directions but are overlapping, as shown by the solutions and graph below.Â Â In interval notation, this solution is written as , or simply ,Â since the graph is all possible numbers.The second type of compound inequality is the âandâ inequality. âAndâ inequalities require both inequality statements to be true. If one part is false, the whole inequality is false. When graphing these inequalities, follow a similar process as before, sketching both solutions for both inequalities above the number line. However, this time, it is only the overlapping portion that is drawn onto the number line. When a solution for an âandâ compound inequality is given in interval notation, it will be expressed in a manner very similar to single inequalities. The symbol that can be used for âandâ is the intersection symbol, â©.Â Example 4.2.2Solve the compound inequality Graph the solution and express it in interval notation.Move all variables to the right side and all numbers to the left:Â Â Isolate the variable for both (flip the sense for the negative):Â Â Solution:Â Â Â Â In interval notation, this solution is written as Note: there are several different results that could result from an âandâ statement. The graphs could be pointing towards each other as in the graph above, or pointing in the same direction, as in the graph representing (shown below). In this case, the solution must be true for both inequalities, which make a combined graph of:Â Â In interval notation, this solution is written as It is also possible to have solutions that point in opposite directions but do not overlap, as shown by the solutions and graph below. Since there is no overlap, there is no real solution.Â Â In interval notation, this solution is written as no solution, or .The third type of compound inequality is a special type of âandâ inequality. When the variable (or expression containing the variable) is between two numbers, write it as a single math sentence with three parts, such as to show is greater than 5 and less than or equal to 8. To stay balanced when solving this type of inequality, because there are three parts to work with, it is necessary to perform the same operation on all three parts. The graph, then, is of the values between the benchmark numbers with appropriate brackets on the ends.Â Example 4.2.3Solve the inequality Graph the solution and write it in interval notation.Isolate the variable by subtracting 2 from all three parts:Â Â Isolate the variable by dividing all three parts by â4 (remember to flip the sense):Â Â Â Â In interval notation, this is written as QuestionsFor questions 1 to 32, solve each compound inequality, graph its solution, and write it in interval notation.\dfrac{n}{3} < 3 \text{ or } -5n < -106m \ge -24 \text{ or } m - 7 < -12x + 7 \ge 12 \text{ or } 9x < -4510r > 0 \text{ or } r - 5 < -12x - 6 < -13 \text{ or } 6x < -609 + n < 2 \text{ or } 5n > 40\dfrac{v}{8} > -1 \text{ and } v - 2 < 1-9x < 63 \text{ and } \dfrac{x}{4} < 1-8 + b < -3 \text{ and } 4b < 20-6n < 12 \text{ and } \dfrac{n}{3} < 2a + 10 \ge 3 \text{ and } 8a < 48-6 + v \ge 0 \text{ and } 2v > 43 < 9 + x < 70 \ge \dfrac{x}{9} \ge -111 < 8 + k < 12-11 < n - 9 < -5-3 < x - 1 < 1-1 < \dfrac{p}{8} < 0-4 < 8 - 3m < 113 + 7r > 59 \text{ or } -6r - 3 > 33-16 < 2n - 10 < -2-6 - 8x \ge -6 \text{ or } 2 + 10x > 82