Chapter 7.1, Problem 14E

To determine

To find:The solution of system of equations $\{\begin{array}{c}-2x+y=-5\\ x^2+y^2=25\end{array}$ by method of substitution and check solution graphically.

Answer to Problem 14E

Thesolution of the system of equations $\{\begin{array}{c}-2x+y=-5\\ x^2+y^2=25\end{array}$ are $\left(0,-5\right)$ and $\left(4,3\right)$ .

Explanation of Solution

Given information:

The given system of equationsis $\{\begin{array}{c}-2x+y=-5\\ x^2+y^2=25\end{array}$ .

Calculation:

The equation $-2x+y=-5$ can be written as $y=2x-5$ .

Substitute $2x-5$ for $y$ in the equation $x^2+y^2=25$ .

  $\begin{array}{c}{x}^2+{\left(2x-5\right)}^2=25\\ x^2+4x^2-20x+25=25\\ 5x^2-20x=0\\ 5x\left(x-4\right)=0\end{array}$

Solve the above equation as follows:

  $\begin{array}{c}5x=0\\ x=0\\ x-4=0\\ x=4\end{array}$

Substitute 0 for $x$ in equation $-2x+y=-5$ to find $y$ .

  $\begin{array}{c}-2\left(0\right)+y=-5\\ y=-5\end{array}$

Substitute 4 for $x$ in equation $-2x+y=-5$ to find $y$ .

  $\begin{array}{c}-2\left(4\right)+y=-5\\ -8+y=-5\\ y=-5+8\\ =3\end{array}$

So, the solution of the system of equationsare $\left(0,-5\right)$ and $\left(4,3\right)$ .

Verify the solution using graphical utility:

Draw the graph for the first equation $-2x+y=-5$ and the second equation $x^2+y^2=25$ using graphical utility.

  

From the above graph, the coordinates of the intersection pointsare $\left(0,-5\right)$ and $\left(4,3\right)$ . It satisfied the result calculated algebraically.

Therefore, the solution of the system of equations $\{\begin{array}{c}-2x+y=-5\\ x^2+y^2=25\end{array}$ are $\left(0,-5\right)$ and $\left(4,3\right)$ .