Suggested languages for you:

Americas

Europe

545 - Review Exercises

Expert-verified
Found in: Page 1025

### Elementary Algebra

Book edition 2nd
Author(s) Lynn Marecek, MaryAnne Anthony-Smith
Pages 1240 pages
ISBN 9780998625713

# Subtract Rational Expressions with a Common DenominatorIn the following exercises, subtract.$\frac{4{q}^{2}-q+3}{{q}^{2}+6q+5}-\frac{3{q}^{2}+q+6}{{q}^{2}+6q+5}.$

The answer is $\frac{q-3}{q+5}.$

See the step by step solution

## Step 1. Given Information.

Given:

$\frac{4{q}^{2}-q+3}{{q}^{2}+6q+5}-\frac{3{q}^{2}+q+6}{{q}^{2}+6q+5}$ having same denominator.

## Step 2. Solving the given expression.

$Asthedenominatorissamesowecandirectlysubtractthenumerators:\phantom{\rule{0ex}{0ex}}\frac{4{q}^{2}-q+3}{{q}^{2}+6q+5}-\frac{3{q}^{2}+q+6}{{q}^{2}+6q+5}\phantom{\rule{0ex}{0ex}}=\frac{4{q}^{2}-q+3-\left(3{q}^{2}+q+6\right)}{{q}^{2}+6q+5}\phantom{\rule{0ex}{0ex}}=\frac{4{q}^{2}-q+3-3{q}^{2}-q-6}{{q}^{2}+6q+5}\phantom{\rule{0ex}{0ex}}=\frac{{q}^{2}-2q-3}{{q}^{2}+6q+5}\phantom{\rule{0ex}{0ex}}=\frac{\left(q+1\right)\left(q-3\right)}{\left(q+1\right)\left(q+5\right)}=\frac{q-3}{q+5}.$