Question

How do you solve the system `-2x+3y=0` and `3x+3y=-15` by graphing?

Answer 1

See a solution process below:

Explanation:

First, find two point on the line for the first equation.

Set `x = 0 : (-2 * 0) + 3y = 0`

`0 + 3y = 0`

`3y = 0`

`(3y)/color(red)(3) = 0/color(red)(3)`

`(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 0`

`y = 0`

`(0, 0)`

Set `x = 3: (-2 * 3) + 3y = 0`

`-6 + 3y = 0`

`color(red)(6) - 6 + 3y = color(red)(6) + 0`

`0 + 3y = 6`

`3y = 6`

`(3y)/color(red)(3) = 6/color(red)(3)`

`(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 2`

`y = 2`

`(3, 2)`

Now plot the two points and draw a line through the points:

graph{(x^2 + y^2 - 0.125)((x-3)^2+(y - 2)^2 - 0.125)(-2x + 3y)=0 [-20, 20, -10, 10]}

Find two points on the line for the second equation.

Set `x = 0: (3 * 0) + 3y = -15`

`0 + 3y = -15`

`3y = -15`

`(3y)/color(red)(3) = -15/color(red)(3)`

`(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = -5`

`y = -5`

`(0, -5)`

Set `y = 0: 3x + (3 * 0) = -15`

`3x + 0 = -15`

`3x = -15`

`(3x)/color(red)(3) = -15/color(red)(3)`

`(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -5`

`x = -5`

`(-5, 0)`

Now, plot these two points and draw a line through them.

graph{((x+5)^2+y^2 - 0.125)(3x + 3y + 15)(x^2+(y+5)^2 - 0.125)(x^2 + y^2 - 0.125)((x-3)^2+(y - 2)^2 - 0.125)(-2x + 3y)=0 [-15, 15, -7.5, 7.5]}

The two lines intersect at: `(-3, -2)`