Question

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

$\left\{\begin{array}{l}{x}^2-4y^2+7=0\\ 3x^2+y^2=31\end{array}\right.$

Short answer

tThe point of intersection is $\left(3,2\right),\left(3,-2\right),\left(-3,2\right),\left(-3,-2\right)$.

Step-by-step solution

Step 1. Given information

$\left\{\begin{array}{l}{x}^2-4y^2+7=0\\ 3x^2+y^2=31\end{array}\right.$

Step 2. Graphing each equation

$x^2-4y^2+7=0$

$3x^2+y^2=31$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ \begin{array}{l}{x}^2-4y^2+7=0......\left(1\right)\\ 3x^2+y^2=31.....\left(2\right)\end{array} \\ Multiply4ineq.2andaddtoeq.1 \\ weget \\ 13x^2=117 \\ x=\pm \\ substitutethevalueofxineq.1togeyy \\ x=3 \\ {3}^2-4y^2+7=0 \\ y=\pm 2 \\ x=-3 \\ y=\pm 2 \end{gathered}$$