Question

In the following exercises, subtract.

$\frac{5q^2+3q-9}{q^2+6q+8}-\frac{4q^2+9q+7}{q^2+6q+8}$

Short answer

The resultant expression is$\frac{q-8}{q+4}$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{5q^2+3q-9}{q^2+6q+8}-\frac{4q^2+9q+7}{q^2+6q+8}$.

Step 2. Subtract the numerator 

As the denominator is the same for both the terms, subtract the numerator and place it over the common denominator.

$\begin{array}{rcl}\frac{5q^2+3q-9}{q^2+6q+8}-\frac{4q^2+9q+7}{q^2+6q+8}& =& \frac{5q^2+3q-9-4q^2-9q-7}{q^2+6q+8}\\ & =& \frac{q^2-6q-16}{q^2+6q+8}\end{array}$

Step 3. Factor the expression  

• Factor the expression in the numerator and denominator, and then cancel out the common terms from the numerator and denominator.

$\frac{(q+2)(q-8)}{(q+4)(q+2)}=\frac{q-8}{q+4}$

• So, the resultant expression is $\frac{q-8}{q+4}$.