Question

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

$\left\{\begin{array}{l}2y^2-3xy+6y+2x+4=0\\ 2x-3y+4=0\end{array}\right.$

Short answer

The point of intersection is $\left(7,6\right)$.

Step-by-step solution

Step 1. Given information

$\left\{\begin{array}{l}2y^2-3xy+6y+2x+4=0\\ 2x-3y+4=0\end{array}\right.$

Step 2. Graphing each equation

$2y^2-3xy+6y+2x+4=0$

$2x-3y+4=0$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 2y^2-3xy+6y+2x+4=0.....\left(1\right) \\ 2x-3y+4=0.....\left(2\right) \\ subtractingeq.2fromeq.1 \\ weget \\ 2y^2-3xy+9y=0 \\ 2y-3x+9=0....\left(3\right) \\ Nowmultiplyeq.2with3andeq.3with2. \\ 6x-9y+12=0 \\ 4y-6x+18=0 \\ Noweliminatextogety \\ y=6 \\ substitutey=6ineq.2togetx \\ 2x-3\left(6\right)+4=0 \\ x=7 \end{gathered}$$