Question

Graph each equation of the system. Then solve the system to find the points of intersection.

$$\begin{gathered} 3x^2-2y^2+5=0 \\ 2x^2-y^2+2=0 \end{gathered}$$

Short answer

The point of intersection is $(1,2),(1,-2),(-1,2),(-1,-2)$.

Step-by-step solution

Step 1. Given information

$$\begin{gathered} 3x^2-2y^2+5=0 \\ 2x^2-y^2+2=0 \end{gathered}$$

Step 2. Graphing each equation

$3x^2-2y^2+5=0$

$2x^2-y^2+2=0$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 3x^2-2y^2+5=0....\left(1\right) \\ 2x^2-y^2+2=0.......\left(2\right) \\ Multiply2witheq.2andsubtractfromeq.1 \\ weget \\ -x^2+1=0 \\ x=\pm 1 \\ substitutevalueofxineq.1togety \\ x=1 \\ 3-2y^2+5=0 \\ y=\pm 2 \\ x=-1 \\ y=\pm 2 \end{gathered}$$