Question

How do you solve this system of equations: `-8x - 6y = - 12; 8x + 8y = 0`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `x`:

`8x + 8y = 0`

`8x + 8y - color(red)(8y) = 0 - color(red)(8y)`

`8x + 0 = -8y`

`8x = -8y`

`(8x)/color(red)(8) = -8y/color(red)(8)`

`(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = -color(red)(cancel(color(black)(8)))y/cancel(color(red)(8))`

`x = -y`

Step 2) Substitute `(-y)` for `x` in the first equation and solve for `y`

`-8x - 6y = -12` becomes:

`-8(-y) - 6y = -12`

`8y - 6y = -12`

`(8 - 6)y = -12`

`2y = -12`

`(2y)/color(red)(2) = -12/color(red)(2)`

`(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -6`

`y = -6`

Step 3)

Substitute `-6` for `y` in the solution to the second equation at the end of Step 1 and calculate `x`:

`x = -y` becomes:

`x = -(-6)`

`x = 6`

The Solution Is:

`x = 6` and `y = -6`

Or

`(6, -6)`

Answer 2

`x = 6 and y =-6`

Explanation:

This is the BEST scenario you could hope for!.

Notice that the `x` terms are additive inverses. `-8x+8x =0`

Simply add the two equations together and the `x` terms will be eliminated.

`color(white)(xxxxx)-8x-6y =-12" "A`
`color(white)(xxxxx)+8x+8y =" "0" "B`

`A+B:rarr" "2y = -12`
`color(white)(xxxxxxxxxxx)y=-6" "`

Substitute `-6` for `y` in either equation. Let's use `B`

`color(white)(xxxxx)8x+8y" " =0" "B`
`color(white)(xxxxx)8x+8(-6) =0`

`color(white)(xxxxx)8x-48 =" "0`
`color(white)(xxxxxxxx)8x=48`
`color(white)(xxxxxxxx)x=6`

Check in equation `A`

`color(white)(xxxxx)-8(6)-6(-6) =-12" "A`
`color(white)(xxxxxx.xxx)-48+36 =-12`
`color(white)(xxxxxxx.xx.xxx)-12 =-12" "larr` correct