Chapter 7.1, Problem 33E

To determine

To find:The solution of the system of equations $\{\begin{array}{c}{x}^2-2x+y=8\\ x-y=-2\end{array}$ by the method of substitution and verify the result graphically.

Answer to Problem 33E

The solutions of the system of equations $\{\begin{array}{c}{x}^2-2x+y=8\\ x-y=-2\end{array}$ are $\left(3,5\right)$ and $\left(-2,0\right)$ .

Explanation of Solution

Given information:

The given system of equations is $\{\begin{array}{c}{x}^2-2x+y=8\\ x-y=-2\end{array}$ .

Calculation:

Consider the first equation be $x-y=-2$ and the second equation be $x^2-2x+y=8$ . Solve the first equation for $y$ and substitute the value of $y$ in the second equation.

  $\begin{array}{c}x-y=-2\\ y=x+2\end{array}$

Substitute the value of $y$ in the second equation and solve.

  $\begin{array}{c}{x}^2-2x+x+2=8\\ x^2-x+2-8=0\\ x^2-x-6=0\end{array}$

Simplify the above equation by determining the factors.

  $\begin{array}{c}{x}^2-3x+2x-6=0\\ x\left(x-3\right)+2\left(x-3\right)=0\\ \left(x+2\right)\left(x-3\right)=0\\ x=3,-2\end{array}$

Substitute the value of $x$ as 3 in the first equation.

  $\begin{array}{c}3-y=-2\\ y=3+2\\ y=5\end{array}$

Substitute the value of $x$ as $-2$ in the first equation.

  $\begin{array}{c}\left(-2\right)-y=-2\\ y=0\end{array}$

From the above, the ordered pairs are $\left(3,5\right)$ and $\left(-2,0\right)$ .Hence, the solutions of the system of equations $\{\begin{array}{c}{x}^2-2x+y=8\\ x-y=-2\end{array}$ are $\left(3,5\right)$ and $\left(-2,0\right)$ .

Verify the solution using graphical utility:

Draw the graph for the first equation $x-y=-2$ and the second equation $x^2-2x+y=8$ using graphical utility.

  

From the above graph, the coordinate of the intersection points are $\left(3,5\right)$ and $\left(-2,0\right)$ . It satisfied the result calculated algebraically.

Therefore, the solutions of the system of equations $\{\begin{array}{c}{x}^2-2x+y=8\\ x-y=-2\end{array}$ are $\left(3,5\right)$ and $\left(-2,0\right)$ .