Chapter 7.1, Problem 11E

To determine

The solution of the system of equations by substitution method.

The solution is to be verifiedgraphically.

Answer to Problem 11E

  $\{(5,0),(-3,-4)\}$

Explanation of Solution

Given information:

The system of the equation are:

  $\{\begin{array}{l}\frac{-1}{2}x+y=\frac{-5}{2}\\ x^2+y^2=25\end{array}$

Calculation:

The given equations are $\{\begin{array}{l}\frac{-1}{2}x+y=\frac{-5}{2}\\ x^2+y^2=25\end{array}$ .

Multiply the first equation by $2$ and isolate $x$ and write it in terms of $y$ are as shown below: $\begin{array}{l}\{\begin{array}{l}-x+2y=-5\\ x^2+y^2=25\end{array}\\ -x+2y=-5\\ x=2y+5\end{array}$

Substitute $x$ in terms of $y$ in the second equation:

  $\begin{array}{l}{x}^2+y^2=25\\ $ {(2y+5)}^2+y^2=25$4y^2+20y+25+y^2=25\\ 5y^2+20y+25=25\end{array}$

Subtract $25$ from the both side of the equation.

  $\begin{array}{l}5y^2+20y+25-25=25-25\\ 5y^2+20y=0\end{array}$

Divide the equation by $5$ :

  $\begin{array}{l}{y}^2+4y=0\\ y(y+4)=0\end{array}$

Find the value of $y$ from the above equation:

  $\begin{array}{l}y=0\Rightarrow y_1=0\\ y+4=0\Rightarrow y_2=-4\end{array}$

Find the value of $x$ by plug the value of $y_1$ and $y_2$ .

  $\begin{array}{l}\text{for }{y}_1=0\Rightarrow x_1=2y_1+5=2\cdot 0+5=5\\ \text{for }{y}_2=-4\Rightarrow x_2=2y_2+5=2\cdot (-4)+5=-3\end{array}$

The solutions are $\{(5,0),(-3,-4)\}$ .

Now, verify the given equation by graphically are as shown below:

   Thus,the solutions are $\{(5,0),(-3,-4)\}$ .