Chapter 9.8, Problem 2OE

To determine

To calculate: To find the solutions of system of equations

Answer to Problem 2OE

Solution of system of equations are $\left(2,-4\right),\left(-4,2\right)$

Explanation of Solution

Given information: System of equations is $x^2+y^2=20;x+y+2=0$

Calculation:

Given system of equations as follows:

  $\begin{array}{l}{x}^2+y^2=20----(1)\\ x+y+2=0----(2)\end{array}$

From equation (2),

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

Substituting the value in equation (1)

  $\begin{array}{l}{x}^2+{\left(-2-x\right)}^2=20\\ x^2+4+x^2+4x=20\\ 2x^2+4x-16=0\\ x^2+2x-8=0\\ \left(x+4\right)\left(x-2\right)=0\\ x=2,-4\end{array}$

When $x=2$

  $\begin{array}{l}y=-2-2\\ y=-4\end{array}$

When $x=-4$

  $\begin{array}{l}y=-2+4\\ y=2\end{array}$

Conclusion:

Hence, solution of system of equations are $\left(2,-4\right),\left(-4,2\right)$