Chapter 7.1, Problem 9E

To determine

ToSolveThe system of equations $x-y=-4$ and $x^2-y=-2$ by method of substitution and check solution(s) graphically.

Answer to Problem 9E

The solutions of the given system of equationsare $\left(x,y\right)=\left(2,6\right)$ and $\left(x,y\right)=\left(-1,3\right)$

Explanation of Solution

Given:

The system of equations $x-y=-4$ and $x^2-y=-2$

Concept Used:

For solving the system of two equations by the method of substitution.

The first step is the take out the value of say $x$ (or $y$ ) from any of the equation and then substitute that value into the other equation to find the value of other unknown $y$ .

And then use that value of $y$ to find the value of $x$ .

Calculation:

For the given system of equations $x-y=-4$ and $x^2-y=-2$

Let $x-y=-4------\left(1\right)$

  $x^2-y=-2------\left(2\right)$

From equation 1, we have

  $x-y=-4$

i.e., $y=x+4------\left(3\right)$

Substituting in equation 2, we have

i.e., $x^2-y=-2$

i.e., $\begin{array}{c}{x}^2-\left(x+4\right)=-2\\ x^2-x-4=-2\\ x^2-x-4+2=0\\ x^2-x-2=0\end{array}$

Further simplify as shown:

  $\begin{array}{c}{x}^2-2x+x-2=0\\ x\left(x-2\right)+1\left(x-2\right)=0\\ \left(x-2\right)\left(x+1\right)=0\end{array}$

i.e., $x-2=0$

and

  $x+1=0$

i.e., $x=2$

and

  $x=-1$

Substituting $x=2$ in equation 3, we have

i.e., $\begin{array}{c}y=x+4\\ y=2+4\\ y=6\end{array}$

Substituting $x=-1$ in equation 3, we have

i.e., $\begin{array}{c}y=x+4\\ y=-1+4\\ y=3\end{array}$

Therefore, the solutions of the given system of equationsare $\left(x,y\right)=\left(2,6\right)$ and $\left(x,y\right)=\left(-1,3\right)$

From the given graph of system of equation

  

The solutions of the given system of equationsare $\left(x,y\right)=\left(2,6\right)$ and $\left(x,y\right)=\left(-1,3\right)$ .

Conclusion:

The solutions of the given system of equationsare $\left(x,y\right)=\left(2,6\right)$ and $\left(x,y\right)=\left(-1,3\right)$ .