Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

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

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

$$\begin{gathered} Consider, \\ 3x^2-2y^2+5=0 ....\left(1\right) \\ 2x^2-y^2+2=0 .......\left(2\right) \\ Multiply 2 with eq.2 and subtract from eq.1 \\ we get \\ -x^2+1=0 \\ x=\pm 1 \\ substitute value of x in eq. 1 to get y \\ x=1 \\ 3-2y^2+5=0 \\ y=\pm 2 \\ x=-1 \\ y=\pm 2 \end{gathered}$$