Chapter 3, Problem 10CT

To determine

To calculate: The solution of the system of equations or inequalities.

Answer to Problem 10CT

The required solutions are $\left(-6,-2\right)$ and $\left(2,6\right)$ .

Explanation of Solution

Given information:

The system of equations: $0=x^2+y^2-40$

  $y=x+4$

Calculation:

The given system of equations is,

  $0=x^2+y^2-40$............................ (1)

  $y=x+4$............................ (2)

Substitute $x+4$ for $y$ in the equation (1).

  $\begin{array}{c}{x}^2+{\left(x+4\right)}^2-40=0\\ x^2+x^2+8x+16-40=0\\ \frac{2x^2}{2}+\frac{8x}{2}-\frac{24}{2}=0\\ x^2+4x-12=0\end{array}$

To solve the above quadratic equation, use the factorization method.

  $\begin{array}{c}{x}^2+4x-12=0\\ x^2+6x-2x-12=0\\ x\left(x+6\right)-2\left(x+6\right)=0\\ \left(x+6\right)\left(x-2\right)=0\end{array}$

Simplify the above factors for x .

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

And,

  $\begin{array}{c}x-2=0\\ x=2\end{array}$

Substitute −6 for x in the equation (2) and solve for y .

  $\begin{array}{c}y=-6+4\\ y=-2\end{array}$

Also, Substitute 2 for x in the equation (2) and solve for y .

  $\begin{array}{c}y=2+4\\ y=6\end{array}$

Thus, the solutions are $x=-6$ , $y=-2$ and $x=2$ and $y=6$ .

Hence, the required solutions are $\left(-6,-2\right)$ and $\left(2,6\right)$ .