Chapter 9.7, Problem 16PPS

To determine

To find: The solution for the given system of the equations.

Answer to Problem 16PPS

The solution for the given equation is $\left(3,-5\right),\left(2,-3\right)$ .

Explanation of Solution

Given:

The given system of equation is,

  $\begin{array}{l}y=-2x+1\\ x^2+3x=-4y-2\end{array}$

Calculation:

Consider the given equation are,

  $\begin{array}{l}y=-2x+1\\ x^2+3x=-4y-2\end{array}$

Substitute the first equation in the second equation as,

  $\begin{array}{l}{x}^2+3x=-4\left(-2x+1\right)-2\\ x^2+3x=8x-4-2\\ x^2-5x+6=0\\ x^2-3x-2x+6=0\\ \left(x-3\right)\left(x-2\right)=0\\ x=3,2\end{array}$

Substitute $x=3$ in $y=-2x+1$ .

  $\begin{array}{l}y=-2\left(3\right)+1\\ y=-6+1\\ y=-5\end{array}$

Substitute $x=2$ in $y=-2x+1$ .

  $\begin{array}{l}y=-2\left(2\right)+1\\ y=-4+1\\ y=-3\end{array}$

Thus, the solution of the equations are $\left(3,-5\right),\left(2,-3\right)$ .