Question

How do you solve the system of equations `0=2y+4-2x` and `-3x-12=-12y` by graphing?

Answer 1

See a solution process below:

Explanation:

To solve this system of equations by graphing we need to find and plot two points from each equation, draw a line through the two points and then identify where the lines cross.

-Equation 1:
First Point `x = 0`

`0 = 2y + 4 - (2 * 0)`
`0 = 2y + 4 - 0`
`0 = 2y + 4`
`0 - color(red)(4) = 2y + 4 - color(red)(4)`
`-4 = 2y`
`-4/color(red)(2) = (2y)/color(red)(2)`
`-2 = y`
`(0, -2)`

Second Point `x = 2`

`0 = 2y + 4 - (2 * 2)`
`0 = 2y + 4 - 4`
`0 = 2y + 0`
`0 = 2y`
`0/color(red)(2) = (2y)/color(red)(2)`
`0 = y`
`(2, 0)`

Plot Points and Draw Line Through Both Points
graph{(2y+4-2x)(x^2+(y+2)^2-0.025)((x-2)^2+y^2-0.025)=0 [-10. 10, -5, 5]}

-Equation 2:
First Point `x = 0`

`(-3 xx 0) - 12 = -12y`
`0 - 12 = -12y`
`-12 = -12y`
`-12/color(red)(-12) = (-12y)/color(red)(-12)`
`1 = y`
`(0, 1)`

Second Point `x = -4`

`(-3 xx -4) - 12 = -12y`
`12 - 12 = -12y`
`0 = -12y`
`0/color(red)(-12) = (-12y)/color(red)(-12)`
`0 = y`
`(-4, 0)`

Plot Points and Draw Line Through Both Points
graph{(12y-3x-12)(2y+4-2x)(x^2+(y-1)^2-0.025)((x+4)^2+y^2-0.025)=0 [-10. 10, -5, 5]}

We can see the lines cross at the solution which is: `(4, 2)`

graph{(12y-3x-12)(2y+4-2x)((x-4)^2+(y-2)^2-0.025)=0 [0, 10, 0, 5]}