Question

In the following exercises, evaluate the rational expression for the given values.

$$\begin{gathered} \frac{y^2+5y+6}{y^2-1} \\ \left(a\right)y=0 \\ \left(b\right)y=2 \\ \left(c\right)y=-2 \end{gathered}$$

Short answer

The required answers are

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)-6 \\ \mathrm{Part}\left(\mathrm{b}\right)\frac{20}{3} \\ \mathrm{Part}\left(\mathrm{c}\right)0 \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information   

The given expression is $\frac{y^2+5y+6}{y^2-1}$

Part (a) Step 2. Calculation   

Substitute 0 for y,

$$\begin{gathered} \frac{y^2+5y+6}{y^2-1}=\frac{{\left(0\right)}^2+5\left(0\right)+6}{{\left(0\right)}^2-1} \\ =\frac{6}{-1} \\ =-6 \end{gathered}$$

Part (b) Step 1. Calculation 

Substitute 2 for y,

$$\begin{gathered} \frac{y^2+5y+6}{y^2-1}=\frac{{\left(2\right)}^2+5\left(2\right)+6}{{\left(2\right)}^2-1} \\ =\frac{4+10+6}{4-1} \\ =\frac{20}{3} \end{gathered}$$

Part (c) Step 1. Calculation    

Substitute $-2$ for y,

$$\begin{gathered} \frac{y^2+5y+6}{y^2-1}=\frac{{(-2)}^2+5(-2)+6}{{(-2)}^2-1} \\ =\frac{4-10+6}{4-1} \\ =\frac{0}{3} \\ =0 \end{gathered}$$