Question

Perform the indicated operation and simplify.

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

Short answer

The solution 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. Simplify the expression.

$$\begin{gathered} \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{10x^2+16x-7}{8x-3}-\frac{2x^2+3x-1}{8x-3} \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{10x^2-2x^2+16x-3x-7+1}{8x-3} \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{8x^2+13x-6}{8x-3} \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{8x^2+16x-3x-6}{8x-3} \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{8x\left(x+2\right)-3\left(x+2\right)}{\left(8x-3\right)} \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=\frac{\left(x+2\right)\left(8x-3\right)}{\left(8x-3\right)}(Removethecommonfactor) \\ \frac{10x^2+16x-7}{8x-3}+\frac{2x^2+3x-1}{3-8x}=x+2 \end{gathered}$$

Step 3. Simplified information.

The simplified expression is$x+2$.