# How do you solve the system 2x + 3y = 7 and 2x - 3y = 13 by graphing?

Feb 4, 2018

See a solution process below:

#### Explanation:

To solve for a system of equations by graphing. Solve for two points we each equation, plot the points, draw a line through the points, determine where the two lines intersect:

Equation 1: $2 x + 3 y = 7$

First Point: For $x = 2$

$\left(2 \cdot 2\right) + 3 y = 7$

$4 + 3 y = 7$

$4 - \textcolor{red}{4} + 3 y = 7 - \textcolor{red}{4}$

$0 + 3 y = 3$

$3 y = 3$

$\frac{3 y}{\textcolor{red}{3}} = \frac{3}{\textcolor{red}{3}}$

$y = 1$ or $\left(2 , 1\right)$

Second Point: For $y = 3$

$2 x + \left(3 \times 3\right) = 7$

$2 x + 9 = 7$

$2 x + 9 - \textcolor{red}{9} = 7 - \textcolor{red}{9}$

$2 x + 0 = - 2$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{2}{\textcolor{red}{2}}$

$x = - 1$ or $\left(- 1 , 3\right)$

We can next plot the two points on the coordinate plane and draw a line through the two points:

graph{((x-2)^2+(y-1)^2-0.025)((x+1)^2+(y-3)^2-0.025)(2x+3y-7)=0 [-10, 10, -5, 5]}

Equation 2: $2 x - 3 y = 13$

First Point: For $x = 2$

$\left(2 \cdot 2\right) - 3 y = 13$

$4 - 3 y = 13$

$4 - \textcolor{red}{4} - 3 y = 13 - \textcolor{red}{4}$

$0 - 3 y = 9$

$- 3 y = 9$

$\frac{- 3 y}{\textcolor{red}{- 3}} = \frac{9}{\textcolor{red}{- 3}}$

$y = - 3$ or $\left(2 , - 3\right)$

Second Point: For $y = - 1$

$2 x - \left(3 \times - 1\right) = 13$

$2 x + 3 = 13$

$2 x + 3 - \textcolor{red}{3} = 13 - \textcolor{red}{3}$

$2 x + 0 = 10$

$\frac{2 x}{\textcolor{red}{2}} = \frac{10}{\textcolor{red}{2}}$

$x = 5$ or $\left(5 , - 1\right)$

We can next plot the two points on the coordinate plane and draw a line through the two points for the second equation:

graph{((x-2)^2+(y+3)^2-0.025)((x-5)^2+(y+1)^2-0.025)(2x+3y-7)(2x-3y-13)=0 [-10, 10, -5, 5]}

We can see the lines cross and $\left(5 , - 1\right)$

graph{((x-5)^2+(y+1)^2-0.05)(2x+3y-7)(2x-3y-13)=0 [-10, 10, -5, 5]}