Question

In the following exercises, solve the system of equations by using substitution.

$$\begin{gathered} x^2+y^2=4 \\ x-y=4 \end{gathered}$$

Short answer

Solution for the system of equations

$$\begin{gathered} \begin{array}{cc}x=2+\sqrt{2}i,\hspace{0.17em}& y=-2+\sqrt{2}i\end{array} \\ \begin{array}{cc}x=2-\sqrt{2}i,\hspace{0.17em}& y=-2-\sqrt{2}i\end{array} \end{gathered}$$

Step-by-step solution

Given Information 

System of equations are

$$\begin{gathered} x^2+y^2=4 \\ x-y=4 \end{gathered}$$

use substitution method 

$$\begin{gathered} x^2+y^2=4 \\ x-y=4 \\ x-4=y \\ Substitutex-4foryinfirstequation \\ {x}^2+{(x-4)}^2=4x^2+x^2-8x+16=4 \\ 2x^2-8x+16-4=0 \\ 2x^2-8x+12=0 \\ Divideby2 \\ {x}^2-4x+6=0 \\ applyquadraticformula \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\left(1\right)\left(6\right)}}{2\left(1\right)} \\ x=\frac{4\pm \sqrt{-8}}{2} \\ x=\frac{4\pm 2i\sqrt{2}}{2} \\ x=2\pm i\sqrt{2} \\ x=2-i\sqrt{2} \\ x=2+i\sqrt{2} \end{gathered}$$

Solve for y 

substitute the x value and solve for y

$$\begin{gathered} y=x-4 \\ x=2-i\sqrt{2} \\ y=(2-i\sqrt{2})-4 \\ y=-2-i\sqrt{2} \\ x=2+i\sqrt{2} \\ y=(2+i\sqrt{2})-4 \\ y=-2+i\sqrt{2} \end{gathered}$$

solution 

solution for the given system of equations

$$\begin{gathered} \begin{array}{cc}x=2+\sqrt{2}i,\hspace{0.17em}& y=-2+\sqrt{2}i\end{array} \\ \begin{array}{cc}x=2-\sqrt{2}i,\hspace{0.17em}& y=-2-\sqrt{2}i\end{array} \end{gathered}$$