Chapter 9.8, Problem 8P

Solve each system by graphing.

$y=x^2-4$

$y=-x-2$

To determine

To solve the system using elimination

Answer to Problem 8P

The solution of the system are $\left(-2,0\right)\text{ and }\left(1,-3\right)$

Explanation of Solution

Given information:

The equation is

  $\begin{array}{l}y=x^2-4\\ y=-x-2\end{array}$

Calculation: The equation can be obtained as

  $\begin{array}{l}y=x^2-\text{4 }\left[\text{equation }1\right]\\ y=-x-2\left[\text{equation }2\right]\end{array}$

Subtract equation 1 and 2 to eliminate y

  $\begin{array}{l}{x}^2-4-\left(-x-2\right)=0\\ x^2-4+x+2=0\\ x^2+x-2=0\\ x^2+\left(2x-x\right)-2=0\left[\text{write }x=\left(2x-x\right)\right]\\ x^2+2x-x-2=0\\ x\left(x+2\right)-1\left(x+2\right)=0\left[\begin{array}{l}\text{the first and second term has a common factor,}x\text{ and}\\ \text{the third and fourth term has a common factor,1}\end{array}\right]\\ \left(x+2\right)\left(x-1\right)=0\end{array}$

Therefore,

  $\begin{array}{l}\left(x+2\right)=0\text{ or }\left(x-1\right)=0\text{ by zero product property}\\ x=-2\text{ or }x=1\end{array}$

Solving for the corresponding y value using equation 2

  $\begin{array}{l}y=-x-2\\ \text{put the value of }x=-2\\ y=-\left(-2\right)-2\\ y=2-2\\ y=0\\ \text{again, put the value of }x=1\\ y=-\left(1\right)-2\\ y=-3\end{array}$

Hence,

The solution of the system are $\left(-2,0\right)\text{ and }\left(1,-3\right)$