Question

In the following exercises, add or subtract.

$\frac{2b^2+30b-13}{b^2-49}-\frac{2b^2-5b-8}{49-b^2}$

Short answer

The difference of the expressions is$\frac{4b-3}{b-7}$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{2b^2+30b-13}{b^2-49}-\frac{2b^2-5b-8}{49-b^2}$.

Step 2. Equalise the denominators

• Multiply the numerator and denominator of the second term by $-1$ to make both the numerators equal.

$\frac{2b^2+30b-13}{b^2-49}-\frac{2b^2-5b-8}{49-b^2}=\frac{2b^2+30b-13}{b^2-49}-\frac{-2b^2+5b+8}{b^2-49}$

Step 3. Subtract the numerators 

• As the denominators of the rational expressions is equal, subtract the numerators and place them over a common denominator.

$\begin{array}{rcl}\frac{2b^2+30b-13}{b^2-49}-\frac{-2b^2+5b+8}{b^2-49}& =& \frac{2b^2+30b-13-\left(-2b^2+5b+8\right)}{b^2-49}\\ & =& \frac{2b^2+30b-13+2b^2-5b-8}{b^2-49}\\ & =& \frac{4b^2+25b-21}{b^2-49}\end{array}$

Step 4. Factor the numerator and denominator 

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

$\begin{array}{rcl}\frac{4b^2+25b-21}{b^2-49}& =& \frac{\left(4b-3\right)\cancel{\left(b+7\right)}}{\left(b-7\right)\cancel{\left(b+7\right)}}\\ & =& \frac{4b-3}{b-7}\end{array}$