2.6 Working With Formulas

Terrance Berg

13 2.6 Working With Formulas

In 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.

Example 2.6.1

Evaluate $p(q + 6)$ when $p = 3$ and $q = 5.$

$\begin{array}{rl} (3)((5)+(6))&\text{Replace }p\text{ with 3 and }q\text{ with 5 and evaluate parentheses} \\ (3)(11)&\text{Multiply} \\ 33&\text{Solution} \end{array}$

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 won’t make a difference, but it is a good habit to always use them to prevent problems later.

Example 2.6.2

Evaluate $x + zx(3 - z)\left(\dfrac{x}{3}\right)$ when $x = -6$ and $z = -2.$

$\begin{array}{rl} (-6)+(-2)(-6)\left[(3)-(-2)\right]\left(\dfrac{-6}{3}\right)&\text{Evaluate parentheses} \\ \\ -6+(-2)(-6)(5)(-2)&\text{Multiply left to right} \\ -6+12(5)(-2)&\text{Multiply left to right} \\ -6+60(-2) &\text{Multiply} \\ -6-120 & \text{Subtract} \\ -126& \text{Solution}\end{array}$

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 $A = \pi r^2+ \pi rs$ (the formula for the surface area of a right circular cone) for the variable $s.$ This means isolating the $s$ so the equation has $s$ on one side. So a solution might look like $s = \dfrac{A - \pi r^2}{\pi r}.$ This second equation gives the same information as the first; they are algebraically equivalent. However, one is solved for the area $A,$ while the other is solved for the slant height of the cone $s.$

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 for $x.$

Example 2.6.3

Isolate the variable $x$ in the following equations.

$\begin{array}{ll} \begin{array}{rrl} 3x&=&12 \\ \\ \dfrac{3x}{3}&=&\dfrac{12}{3} \\ \\ x&=&4 \end{array} & \hspace{0.5in} \begin{array}{rrl} wx&=&z \\ \\ \dfrac{wx}{w}&=&\dfrac{z}{w} \\ \\ x&=&\dfrac{z}{w} \end{array} \end{array}$

The same process is used to isolate $x$ in $3x = 12$ as in $wx = z.$ Because $x$ is being solved for, treat all other variables as numbers. For these two equations, both sides were divided by 3 and $w,$ respectively. A similar idea is seen in the following example.

Example 2.6.4

Isolate 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:

$\begin{array}{rrrrl} m&+&n&=&p \\ -m&&&&\phantom{p}-m \\ \midrule &&n&=&p-m \end{array}$

Since $p$ and $m$ are not like terms, they cannot be combined. For this reason, leave the expression as $p - m.$

Example 2.6.5

Isolate the variable $a$ in the equation $a(x - y) = b.$

This means that $(x-y)$ must be isolated from the variable $a.$

$\dfrac{a(x-y)}{(x-y)}=\dfrac{b}{(x-y)}\hspace{0.25in}\Rightarrow\hspace{0.25in}a=\dfrac{b}{(x-y)}$

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, $x$ is being solved for.

Example 2.6.6

Isolate the variable $x$ in the equation $a(x - y) = b.$

First, distribute $a$ throughout $(x-y)$ :

$\begin{array}{rrrrr} a(x&-&y)&=&b \\ ax&-&ay&=&b \end{array}$

Remove the term $ay$ from both sides:

$\begin{array}{rrrrl} ax&-&ay&=&b \\ &+&ay&&\phantom{b}+ay \\ \midrule &&ax&=&b+ay \end{array}$

$ax$ is then divided by $a$ :

$\dfrac{ax}{a}=\dfrac{b+ay}{a}$

The solution is $x=\dfrac{b+ay}{a},$ which can also be shown as $x=\dfrac{b}{a}+y.$

Be very careful when isolating $x$ not to try and cancel the $a$ 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 $x = \dfrac{b + ay}{a}.$ The next example is another two-step problem.

Example 2.6.7

Isolate the variable $m$ in the equation $y=mx+b.$

First, subtract $b$ from both sides:

$\begin{array}{lrrrr} y&=&mx&+&b \\ \phantom{y}-b&&&-&b \\ \midrule y-b&=&mx&& \end{array}$

Now divide both sides by $x$ :

$\dfrac{y-b}{x}=\dfrac{mx}{x}$

Therefore, the solution is $m=\dfrac{y-b}{x}.$

It is important to note that a problem is complete when the variable being solved for 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.

Example 2.6.8

Isolate the variable $s$ in the equation $A= \pi r^2+\pi rs.$

Subtract $\pi r^2$ from both sides:

$\begin{array}{rrrrr} A\phantom{- \pi r^2}&=&\pi r^2&+&\pi rs \\ \phantom{A}-\pi r^2&&-\pi r^2&& \\ \midrule A- \pi r^2&=&\pi rs&& \end{array}$

Divide both sides by $\pi r$ :

$\dfrac{A-\pi r^2}{\pi r}=\dfrac{\pi rs}{\pi r}$

The solution is:

$s=\dfrac{A-\pi r^2}{\pi r}$

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.

Example 2.6.9

Isolate the variable $m$ in the equation $h=\dfrac{2m}{n}.$