Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Test - Page 499: 13

Answer

$ 1-x$.

Work Step by Step

Both numerator and denominator of the given expression contain fractions. In order to simplify the given expression we will multiply both numerator and denominator of the given expression by the Least Common Denominator (LCD) of all the fractions inside them. Factor $x^2+2x$. Factor out $x$. $\Rightarrow x(x+2)$. Back substitute factor into the fraction. $\Rightarrow \frac{\frac{1}{x}-\frac{3}{x+2}}{\frac{2}{x^2+2x}}=\frac{\frac{1}{x}-\frac{3}{x+2}}{\frac{2}{x(x+2)}}$ The LCD of all fractions in the numerator and the denominator is $x(x+2)$. Multiply the numerator and the denominator by the LCD. $\Rightarrow \frac{x(x+2)\left (\frac{1}{x}-\frac{3}{x+2}\right )}{x(x+2)\left (\frac{2}{x(x+2)}\right )}$ Use the distributive property. $\Rightarrow \frac{x(x+2)\left (\frac{1}{x}\right )-x(x+2)\left (\frac{3}{x+2}\right )}{x(x+2)\left (\frac{2}{x(x+2)}\right )}$ Cancel common factors. $\Rightarrow \frac{x+2-3x }{2}$ Simplify. $\Rightarrow \frac{2-2x }{2}$ Factor out $2$. $\Rightarrow \frac{2(1-x) }{2}$ Cancel common factors. $\Rightarrow 1-x$.
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