Question

Evaluate $\frac{x^2+x-6}{x^2-9}$for each value:

$ⓐx=0ⓑx=-2ⓒx=1$

Short answer

$$\begin{gathered} ⓐ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+x-6}{x^2-9}\mathrm{for}x=0\mathrm{is}\frac{2}{3}. \\ ⓑ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+x-6}{x^2-9}\mathrm{for}x=-2\mathrm{is}\frac{4}{5}. \\ ⓒ\mathrm{The}\mathrm{value}\mathrm{of}\frac{x^2+x-6}{x^2-9}\mathrm{for}x=1\mathrm{is}\frac{1}{2}. \end{gathered}$$

Step-by-step solution

Step 1. Given Information 

Evaluate $\frac{x^2+x-6}{x^2-9}$for each value:

$ⓐx=0ⓑx=-2ⓒx=1$

Part (a) Step 1. Finding the value of x2+x−6x2−9 for x=0

Putting $x=0$on the given expression

role="math" localid="1648380202595" $$\begin{gathered} \frac{x^2+x-6}{x^2-9}=\frac{{\left(0\right)}^2+0-6}{{\left(0\right)}^2-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{0+0-6}{0-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{-6}{-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{2}{3} \end{gathered}$$

Part (b) Step 1. Finding the value of 

Putting $x=-2$on the given expression

role="math" localid="1648380281405" $$\begin{gathered} \frac{x^2+x-6}{x^2-9}=\frac{{(-2)}^2+(-2)-6}{{(-2)}^2-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{4-2-6}{4-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{-4}{-5} \\ \frac{x^2+x-6}{x^2-9}=\frac{4}{5} \end{gathered}$$

Part (c) Step 1. Finding the value of 

Putting $x=1$on the given expression

role="math" localid="1648380355148" $$\begin{gathered} \frac{x^2+x-6}{x^2-9}=\frac{{\left(1\right)}^2+\left(1\right)-6}{{\left(1\right)}^2-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{1+1-6}{1-9} \\ \frac{x^2+x-6}{x^2-9}=\frac{-4}{-8} \\ \frac{x^2+x-6}{x^2-9}=\frac{1}{2} \end{gathered}$$