Q. 38 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} 5xy+13y^2+36=0 \\ xy+7y^2=6 \end{gathered}$$

Step-by-Step Solution

Answer

The point of intersection is  $\left(-5\sqrt{3},\sqrt{3}\right),\left(5\sqrt{3},-\sqrt{3}\right)$.

Step 1. Given information

$$\begin{gathered} 5xy+13y^2+36=0 \\ xy+7y^2=6 \end{gathered}$$

Step 2. Graphing each equation

$5xy+13y^2+36=0$

$xy+7y^2=6$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 5xy+13y^2+36=0 .........\left(1\right) \\ xy+7y^2=6........\left(2\right) \\ subtracting eq. 2\times 5 from eq.1 \\ we get \\ -22y^2+66=0 \\ y=\pm \sqrt{3} \\ substitute y=\pm \sqrt{3} in eq. 2 to get x \\ y=\sqrt{3} \\ x=-5\sqrt{3} \\ y=-\sqrt{3} \\ x=5\sqrt{3} \end{gathered}$$