Question

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

$$\begin{gathered} x^2-3y^2+1=0 \\ 2x^2-7y^2+5=0 \end{gathered}$$

Short answer

the point of intersection is $\left(2\sqrt{2},\sqrt{3}\right),\left(-2\sqrt{2},\sqrt{3}\right),\left(2\sqrt{2},-\sqrt{3}\right),\left(-2\sqrt{2},-\sqrt{3}\right)$

Step-by-step solution

Step 1. Given 

$$\begin{gathered} x^2-3y^2+1=0 \\ 2x^2-7y^2+5=0 \end{gathered}$$

Step 2. Graphing each equation

$x^2-3y^2+1=0$

$2x^2-7y^2+5=0$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ {x}^2-3y^2+1=0........\left(1\right) \\ 2x^2-7y^2+5=0........\left(2\right) \\ subtractingeq.2fromeq.1\times 2 \\ weget \\ {y}^2-3=0 \\ y=\pm \sqrt{3} \\ substitutey=\pm \sqrt{3}ineq.1togetx \\ y=\sqrt{3} \\ x=\pm 2\sqrt{2} \\ y=-\sqrt{3} \\ x=\pm 2\sqrt{2} \end{gathered}$$