Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}7x^2-3y^2+5=0\\ 3x^2+5y^2=12\end{array}\right.$

Step-by-Step Solution

Answer

The point of intersection is  $\left(\frac{1}{2},\frac{3}{2}\right),\left(-\frac{1}{2},\frac{3}{2}\right),\left(\frac{1}{2},-\frac{3}{2}\right),\left(-\frac{1}{2},-\frac{3}{2}\right)$.

Step 1. Given information

$\left\{\begin{array}{l}7x^2-3y^2+5=0\\ 3x^2+5y^2=12\end{array}\right.$

Step 2. Graphing each equation

$7x^2-3y^2+5=0$

$3x^2+5y^2=12$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ \begin{array}{l}7x^2-3y^2+5=0 .......\left(1\right)\\ 3x^2+5y^2=12 ..........\left(2\right)\end{array} \\ Multiply 5 with eq.1 and 3 with eq.2 and add both \\ we get \\ 44x^2=11 \\ x=\pm \frac{1}{2} \\ substitute x=\pm \frac{1}{2} in eq. 2 to get y \\ x=\pm \frac{1}{2} \\ 3{\left(\frac{1}{2}\right)}^2+5y^2=12 \\ y=\pm \frac{3}{2} \\ x=-\frac{3}{2} \\ 3\left(-\frac{1}{2}\right)+5y^2=12 \\ y=\pm \frac{3}{2} \end{gathered}$$