Question

In the following exercises, add or subtract.

$\frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}$

Short answer

The sum of the expressions is$x+2$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}$.

Step 2. Multiply Numerator and denominator of the second fraction by -1

Multiply the numerator and denominator of the second term by $-1$.

$\frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{10x^2+16x-7}{8x-3}+\frac{-2x^2-3x+1}{8x-3}$

Step 3. Add the Numerator

• As the denominator is same for both the terms, add the numerator and place over the common denominator.
• Add the like terms of the numerator.

$\begin{array}{rcl}\frac{10x^2+16x-7}{8x-3}+\frac{-2x^2-3x+1}{8x-3}& =& \frac{10x^2+16x-7-2x^2-3x+1}{8x-3}\\ & =& \frac{8x^2+13x-6}{8x-3}\end{array}$

Step 4. Simplify

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

$\begin{array}{rcl}\frac{8x^2+13x-6}{8x-3}& =& \frac{\cancel{\left(8x-3\right)}\left(x+2\right)}{\cancel{8x-3}}\\ & =& x+2\end{array}$

• So, the sum is$x+2$.