Question

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

$$\begin{gathered} 4x^2+9y^2=36 \\ 3y^2-4x=12 \end{gathered}$$

Short answer

Solution for the system of equation

$$\begin{gathered} \begin{array}{cc}x=0,\hspace{0.17em}& y=2\end{array} \\ (0,2) \\ \begin{array}{cc}x=0,\hspace{0.17em}& y=-2\end{array} \\ (0,-2) \\ \begin{array}{cc}x=-3,\hspace{0.17em}& y=0\end{array} \\ (-3,0) \end{gathered}$$

Step-by-step solution

Given information 

Two given equations are

$$\begin{gathered} 4x^2+9y^2=36 \\ 3y^2-4x=12 \end{gathered}$$

Eliminate x  or y 

$$\begin{gathered} 4x^2+9y^2=36 \\ 3y^2-4x=12 \\ Multiplythesecondequationby-3 \\ 4x^2+9y^2=36 \\ -9y^2+12x=-36 \\ ------------ \\ 4x^2+12x=0 \\ 4x(x+3)=0 \\ 4x=0,x=0 \\ x+3=0 \\ x=-3 \end{gathered}$$

Solve for y 

Substitute the value of x and solve for x

$$\begin{gathered} 3y^2-4x=12 \\ x=0 \\ 3y^2-4\left(0\right)=12 \\ 3y^2=12 \\ {y}^2=4 \\ y=\pm 2 \\ x=-3 \\ 3y^2-4(-3)=12 \\ 3y^2=0 \\ {y}^2=0 \\ y=0 \end{gathered}$$

verify solution 

Solution is

$$\begin{gathered} \begin{array}{cc}x=0,\hspace{0.17em}& y=2\end{array} \\ 4{\left(0\right)}^2+9{\left(2\right)}^2=36,True \\ 3{\left(2\right)}^2-4\left(0\right)=12,true \\ \begin{array}{cc}x=0,\hspace{0.17em}& y=-2\end{array} \\ 4{\left(0\right)}^2+9{(-2)}^2=36,True \\ 3{(-2)}^2-4\left(0\right)=12,true \\ \begin{array}{cc}x=-3,\hspace{0.17em}& y=0\end{array} \\ 4{(-3)}^2+9{\left(0\right)}^2=36,True \\ 3{\left(0\right)}^2-4(-3)=12,true \end{gathered}$$