Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

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 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

$$\begin{gathered} Consider, \\ {x}^2-3y^2+1=0 ........\left(1\right) \\ 2x^2-7y^2+5=0 ........\left(2\right) \\ subtracting eq. 2 from eq.1\times 2 \\ we get \\ {y}^2-3=0 \\ y=\pm \sqrt{3} \\ substitute y=\pm \sqrt{3} in eq. 1 to get x \\ y=\sqrt{3} \\ x=\pm 2\sqrt{2} \\ y=-\sqrt{3} \\ x=\pm 2\sqrt{2} \end{gathered}$$