Question

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

Answer 1

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: `2x +3y = 7`

First Point: For `x = 2`

`(2 * 2) + 3y = 7`

`4 + 3y = 7`

`4 -color(red)(4) + 3y = 7 - color(red)(4)`

`0 + 3y = 3`

`3y = 3`

`(3y)/color(red)(3) = 3/color(red)(3)`

`y = 1` or `(2, 1)`

Second Point: For `y = 3`

`2x + (3 xx 3) = 7`

`2x + 9 = 7`

`2x + 9 - color(red)(9) = 7 - color(red)(9)`

`2x + 0 = -2`

`(2x)/color(red)(2) = -2/color(red)(2)`

`x = -1` or `(-1, 3)`

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: `2x -3y = 13`

First Point: For `x = 2`

`(2 * 2) - 3y = 13`

`4 - 3y = 13`

`4 -color(red)(4) - 3y = 13 - color(red)(4)`

`0 - 3y = 9`

`-3y = 9`

`(-3y)/color(red)(-3) = 9/color(red)(-3)`

`y = -3` or `(2, -3)`

Second Point: For `y = -1`

`2x - (3 xx -1) = 13`

`2x + 3 = 13`

`2x + 3 - color(red)(3) = 13 - color(red)(3)`

`2x + 0 = 10`

`(2x)/color(red)(2) = 10/color(red)(2)`

`x = 5` or `(5, -1)`

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 `(5, -1)`

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