11 2.4 Fractional Linear Equations
When 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 2.4.1
Solve for \(x\) in the equation \(\frac{3}{4}x - \frac{7}{2} = \frac{5}{6}\).
The LCD is 12, so multiply every term by 12
$$
\frac{3}{4}x \cdot 12 - \frac{7}{2} \cdot 12 = \frac{5}{6} \cdot 12
$$
Cancel the denominators
$$
3x \cdot 3 - 7 \cdot 6 = 5 \cdot 2
$$
Multiply each term
$$
9x - 42 = 10
$$
Add 42 to both sides
$$
9x - 42 + 42 = 10 + 42
$$
Divide both sides by 9
$$
\frac{9x}{9} = \frac{52}{9}
$$
Solution
$$
x = \frac{52}{9}
$$
Example 2.4.2
Solve for \(x\) in the equation \(\dfrac{3\left(\frac{5}{9}x+\frac{4}{27}\right)}{2} = 3\).
First, remove the outside denominator 2 by multiplying both sides by 2
$$
2 \cdot \frac{3\left(\frac{5}{9}x+\frac{4}{27}\right)}{2} = 3 \cdot 2
$$
Simplify the left-hand side
$$
3\left(\frac{5}{9}x + \frac{4}{27}\right) = 6
$$
Divide both sides by 3
$$
\frac{5}{9}x + \frac{4}{27} = 2
$$
Multiply both sides by the LCD 27 to eliminate denominators
$$
\frac{5}{9}x \cdot 27 + \frac{4}{27} \cdot 27 = 2 \cdot 27
$$
Simplify the multiplication
$$
5x \cdot 3 + 4 = 54
$$
Subtract 4 from both sides
$$
5x \cdot 3 + 4 - 4 = 54 - 4
$$
Combine terms
$$
15x = 50
$$
Divide both sides by 15
$$
x = \frac{50}{15} \text{ or } \frac{10}{3}
$$
Solution
$$
x = \frac{10}{3}
$$
Questions
For questions 1 to 18, solve each linear equation.
- \(\dfrac{3}{5}\left(1 + p\right) = \dfrac{21}{20}\)
- \(-\dfrac{1}{2} = \dfrac{3k}{2} + \dfrac{3}{2}\)
- \(0 = -\dfrac{5}{4}\left(x-\dfrac{6}{5}\right)\)
- \(\dfrac{3}{2}n - 8 = -\dfrac{29}{12}\)
- \(\dfrac{3}{4} - \dfrac{5}{4}m = \dfrac{108}{24}\)
- \(\dfrac{11}{4} + \dfrac{3}{4}r = \dfrac{160}{32}\)
- \(2b + \dfrac{9}{5} = -\dfrac{11}{5}\)
- \(\dfrac{3}{2} - \dfrac{7}{4}v = -\dfrac{9}{8}\)
- \(\dfrac{3}{2}\left(\dfrac{7}{3}n+1\right) = \dfrac{3}{2}\)
- \(\dfrac{41}{9} = \dfrac{5}{2}\left(x+\dfrac{2}{3}\right) - \dfrac{1}{3}x\)
- \(-a - \dfrac{5}{4}\left(-\dfrac{8}{3}a+ 1\right) = -\dfrac{19}{4}\)
- \(\dfrac{1}{3}\left(-\dfrac{7}{4}k + 1\right) - \dfrac{10}{3}k = -\dfrac{13}{8}\)
- \(\dfrac{55}{6} = -\dfrac{5}{2}\left(\dfrac{3}{2}p-\dfrac{5}{3}\right)\)
- \(-\dfrac{1}{2}\left(\dfrac{2}{3}x-\dfrac{3}{4}\right)-\dfrac{7}{2}x=-\dfrac{83}{24}\)
- \(-\dfrac{5}{8}=\dfrac{5}{4}\left(r-\dfrac{3}{2}\right)\)
- \(\dfrac{1}{12}=\dfrac{4}{3}x+\dfrac{5}{3}\left(x-\dfrac{7}{4}\right)\)
- \(-\dfrac{11}{3}+\dfrac{3}{2}b=\dfrac{5}{2}\left(b-\dfrac{5}{3}\right)\)
- \(\dfrac{7}{6}-\dfrac{4}{3}n=-\dfrac{3}{2}n+2\left(n+\dfrac{3}{2}\right)\)
