Q. 40 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} y^2-x^2+4=0 \\ 2x^2+3y^2=6 \end{gathered}$$

Step-by-Step Solution

Answer

The point of intersection is not defined as the values are imaginary.

Step 1. Given information

$$\begin{gathered} y^2-x^2+4=0 \\ 2x^2+3y^2=6 \end{gathered}$$

Step 2. Graphing each equation

$y^2-x^2+4=0$

$2x^2+3y^2=6$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ {y}^2-x^2+4=0 ........\left(1\right) \\ 2x^2+3y^2=6 ..........\left(2\right) \\ subtracting eq. 2 from eq.1\times 2 \\ we get \\ 5y^2+14=0 \\ y=\pm \sqrt{\frac{14}{5}} \\ Since the values are imaginary therefore no solution exist. \end{gathered}$$