Q. 33 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

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

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

$$\begin{gathered} Consider, \\ \begin{array}{l}{x}^2-4y^2+7=0......\left(1\right)\\ 3x^2+y^2=31.....\left(2\right)\end{array} \\ Multiply 4 in eq. 2 and add to eq.1 \\ we get \\ 13x^2=117 \\ x=\pm \\ substitute the value of x in eq.1 to gey y \\ x=3 \\ {3}^2-4y^2+7=0 \\ y=\pm 2 \\ x=-3 \\ y=\pm 2 \end{gathered}$$