## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 362: 53

x = 20

#### Work Step by Step

$\frac{3x}{4}$ - 3 = $\frac{x}{2}$ + 2 We can start by adding 3 to both sides of the equation, getting: $\frac{3x}{4}$ - 3 + 3 = $\frac{x}{2}$ + 2 + 3 Performing the arithmetic, we get: $\frac{3x}{4}$ = $\frac{x}{2}$ + 5 Subtract $\frac{x}{2}$ from both sides of the equation. $\frac{3x}{4}$ - $\frac{x}{2}$ = $\frac{x}{2}$ - $\frac{x}{2}$ + 5 $\frac{3x}{4}$ - $\frac{x}{2}$ = 5 Now, rewrite each fraction using the common denominator of 4. $\frac{3x}{4}$ - $\frac{2x}{4}$ = 5 Next, complete the operation of subtraction on the left side of the equation and write 5 as a fraction by writing it over a denominator of 1. : $\frac{x}{4}$ = $\frac{5}{1}$ WE can then cross multiply and get x = 20. To check our answer, we can substitute 20 back into the original equation for x. $\frac{(3)(20)}{4}$ - 3 = $\frac{20}{2}$ + 2 Then complete the arithmetic getting: $\frac{60}{4}$ - 3 = $\frac{20}{2}$ + 2. Continue with the arithmetic: 15 - 3 = 10 + 2 Last, add/subtract on each side of the equation. 12 = 12 Our answers match, so our solution of x = 20 is correct.

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