Question

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

$$\begin{gathered} 9x^2+4y^2=36 \\ y-x=5 \end{gathered}$$

Short answer

Solution

$$\begin{gathered} \begin{array}{cc}x=-\frac{20}{13}+i\frac{12\sqrt{3}}{13},\hspace{0.17em}& y=\frac{45}{13}+i\frac{12\sqrt{3}}{13}\end{array} \\ \begin{array}{cc}x=-\frac{20}{13}-i\frac{12\sqrt{3}}{13},\hspace{0.17em}& y=\frac{45}{13}-i\frac{12\sqrt{3}}{13}\end{array} \end{gathered}$$

Step-by-step solution

Given Information 

System of equations

$$\begin{gathered} 9x^2+4y^2=36 \\ y-x=5 \end{gathered}$$

Substitute y equation 

$$\begin{gathered} 9x^2+4y^2=36 \\ y-x=5 \\ y=5+x \\ subsituteitinfirstequation \\ 9x^2+4{(5+x)}^2=369x^2+4(x^2+10x+25)=36 \\ 9x^2+4x^2+40x+100=36 \\ 13x^2+40x+64=0 \\ Applyquadraticformula \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-40\pm \sqrt{{40}^2-4\left(13\right)\left(64\right)}}{2\left(13\right)} \\ x=\frac{-40\pm \hspace{0.17em}24\sqrt{3}i}{2·\hspace{0.17em}13} \\ x=-\frac{20}{13}-i\frac{12\sqrt{3}}{13} \\ x=-\frac{20}{13}+i\frac{12\sqrt{3}}{13} \end{gathered}$$

Substitute x  and find out y 

$$\begin{gathered} y=x+5 \\ x=-\frac{20}{13}+i\frac{12\sqrt{3}}{13} \\ y=-\frac{20}{13}+i\frac{12\sqrt{3}}{13}+5 \\ y=y=\frac{45}{13}+i\frac{12\sqrt{3}}{13} \\ x=-\frac{20}{13}-i\frac{12\sqrt{3}}{13} \\ y=-\frac{20}{13}-i\frac{12\sqrt{3}}{13}+5 \\ y=\frac{45}{13}-i\frac{12\sqrt{3}}{13} \end{gathered}$$

Solution 

solution for the system of equations

$$\begin{gathered} \begin{array}{cc}x=-\frac{20}{13}+i\frac{12\sqrt{3}}{13},\hspace{0.17em}& y=\frac{45}{13}+i\frac{12\sqrt{3}}{13}\end{array} \\ \begin{array}{cc}x=-\frac{20}{13}-i\frac{12\sqrt{3}}{13},\hspace{0.17em}& y=\frac{45}{13}-i\frac{12\sqrt{3}}{13}\end{array} \end{gathered}$$