## College Algebra (10th Edition)

Published by Pearson

# Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 45

#### Answer

$\dfrac{4-x}{x-2}; x \ne 2$

#### Work Step by Step

The given expression is equivalent to: $=\dfrac{4}{x-2}+\dfrac{x}{-x+2} \\=\dfrac{4}{x-2} + \dfrac{x}{-1(x-2)} \\=\dfrac{4}{x-2} + \dfrac{-x}{x-2}$ The expressions are similar so add the numerators and copy the denominator to obtain: $=\dfrac{4+(-x)}{x-2} \\=\dfrac{4-x}{x-2}; x \ne 2$ ($x$ cannot be $2$ as it makes the expression undefined.)

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