Chapter 8.10, Problem 13PPS

To determine

The solution of system equations $x+y=-3,x-y=-3$ is to be calculated by using the graphing method.

Answer to Problem 13PPS

The graphing solution for system equations $x+y=-3,x-y=-3$

  

Explanation of Solution

Given:

Two equations are as:

  $\begin{array}{l}x+y=-3\\ x-y=-3\end{array}$

The equations are:

  $x+y=-3$....... (1)

  $x-y=-3$....... (2)

For eq. (1):

Subtract $x$ from both sides of eq. (1):

  $\begin{array}{l}x+y-x=-3-x\\ y=-x-3\end{array}$

Let, the line crosses the $x$ -axis at $y=0$

  $\begin{array}{l}y=-x-3\\ 0=-x-3\\ x=-3\end{array}$

So, the $x_{\mathrm{int}ercept}=\left(x,y\right)=\left(-3,0\right)$

Let, the line crosses the $y$ -axis at $x=0$

  $\begin{array}{l}y=0-3\\ y=-3\\ y=-3\end{array}$

So, the $y_{\mathrm{int}ercept}=\left(x,y\right)=\left(0,-3\right)$

For eq. (2):

Addition of $y$ on both sides of eq. (2):

  $\begin{array}{l}x-y+y=-3+y\\ x=y-3\end{array}$

Let, the line crosses the $x$ -axis at $y=0$

  $\begin{array}{l}x=y-3\\ x=0-3\\ x=-3\end{array}$

So, the $x_{\mathrm{int}ercept}=\left(x,y\right)=\left(-3,0\right)$

Let, the line crosses the $y$ -axis at $x=0$

  $\begin{array}{l}x=y-3\\ 0=y-3\\ y=3\end{array}$

So, the $y_{\mathrm{int}ercept}=\left(x,y\right)=\left(0,3\right)$

Now, plot these two eqn. on graph which is shown by the following image:

  

The solution $\left(-3,0\right)$ which is the point of intersection for two equations.

Conclusion:

So, the solution of the system of eqns. is $\left(-3,0\right)$