Question

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

$$\begin{gathered} x^2-y^2=5 \\ -2x^2-3y^2=-30 \end{gathered}$$

Short answer

Solution

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

Step-by-step solution

Given Information 

Given equations are

$$\begin{gathered} x^2-y^2=5 \\ -2x^2-3y^2=-30 \end{gathered}$$

Eliminate a variable 'x'

$$\begin{gathered} x^2-y^2=5 \\ -2x^2-3y^2=-30 \\ Multiplythefirstequationby2 \\ 2x^2-2y^2=10 \\ -2x^2-3y^2=-30 \\ ------------- \\ -5y^2=-20 \\ {y}^2=4 \\ y=\pm 2 \end{gathered}$$

solve for x

we got y= +2, y=-2

Substitute it in any one of the equation and solve for x

$$\begin{gathered} x^2-y^2=5 \\ y=+2 \\ {x}^2-2^2=5 \\ {x}^2-4=5 \\ {x}^2=9 \\ x=\pm 3 \\ (2,3)and(2,-3) \\ {x}^2-y^2=5 \\ y=-2 \\ {x}^2-{(-2)}^2=5x^2-4=5 \\ {x}^2=9 \\ x=\pm 3 \\ (-2,3)and(-2,-3) \end{gathered}$$

verify the solution 

$$\begin{gathered} (3,2) \\ {x}^2-y^2=5 \\ {\left(3\right)}^2-{\left(2\right)}^2=5,True \\ -2x^2-3y^2=-30 \\ -2{\left(3\right)}^2-3{\left(2\right)}^2=-30,true \\ (-3,2) \\ {x}^2-y^2=5 \\ {(-3)}^2-{\left(2\right)}^2=5,True \\ -2x^2-3y^2=-30 \\ -2{(-3)}^2-3{\left(2\right)}^2=-30,true \\ (3,-2) \\ {x}^2-y^2=5 \\ {\left(3\right)}^2-{(-2)}^2=5,True \\ -2x^2-3y^2=-30 \\ -2{\left(3\right)}^2-3{(-2)}^2=-30,true \\ (-3,-2) \\ {x}^2-y^2=5 \\ {(-3)}^2-{(-2)}^2=5,True \\ -2x^2-3y^2=-30 \\ -2{(-3)}^2-3{(-2)}^2=-30,true \end{gathered}$$