Chapter EP, Problem 10.7.5EP

To determine

To Find: The solution of the system of equations.

Answer to Problem 10.7.5EP

Thus, the solution of the system of equation are $\left(\frac{-3+\sqrt{23}}{2},\frac{3+\sqrt{23}}{2}\right),\left(\frac{-3-\sqrt{23}}{2},\frac{3-\sqrt{23}}{2}\right)$ .

Explanation of Solution

Given:

The given equation are,

$\begin{array}{l}{x}^2+y^2=16\\ x+y=-3\end{array}$

Calculation:

Consider the given equation are,

$\begin{array}{l}{x}^2+y^2=16\\ x+y=-3\end{array}$

Consider the second equation as,

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

Substitute above value in the first equation as,

$\begin{array}{l}{x}^2+{\left(x+3\right)}^2=16\\ x^2+x^2+6x+9=16\\ 2x^2+6x-7=0\end{array}$

The roots of the above equation are,

$\begin{array}{l}2x^2+6x-7=0\\ x=\frac{-6\pm \sqrt{36-4\left(2\right)\left(-7\right)}}{4}\\ x=\frac{-3+\sqrt{23}}{2},\frac{-3-\sqrt{23}}{2}\end{array}$

The value of $y$ is obtained as,

$\begin{array}{c}y=\frac{-3+\sqrt{23}}{2}+3,\frac{-3-\sqrt{23}}{2}+3\\ =\frac{3+\sqrt{23}}{2},\frac{3-\sqrt{23}}{2}\end{array}$

Thus, the solution of the equations are $\left(\frac{-3+\sqrt{23}}{2},\frac{3+\sqrt{23}}{2}\right),\left(\frac{-3-\sqrt{23}}{2},\frac{3-\sqrt{23}}{2}\right)$ .