Question

How do you solve the system of equations `-3x+4y=12` and `2x+y=-8`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `y`

`2x + y = -8`

`-color(red)(2x) + 2x + y = -color(red)(2x) - 8`

`0 + y = -2x - 8`

`y = -2x - 8`

Step 2) Substitute `(-2x - 8)` for `y` in the first equation and solve for `x`:

`-3x + 4y = 12` becomes:

`-3x + 4(-2x - 8) = 12`

`-3x + (4 xx -2x) + (4 xx -8) = 12`

`-3x + (-8x) + (-32) = 12`

`-3x - 8x - 32 = 12`

`(-3 - 8)x - 32 = 12`

`-11x - 32 = 12`

`-11x - 32 + color(red)(32) = 12 + color(red)(32)`

`-11x - 0 = 44`

`-11x = 44`

`(-11x)/color(red)(-11) = 44/color(red)(-11)`

`(color(red)(cancel(color(black)(-11)))x)/cancel(color(red)(-11)) = -4`

`x = -4`

Step 3) Substitute `-4` for `x` in the solution to the second equation at the end of Step 1 and calculate `y`:

`y = -2x - 8` becomes:

`y = (-2 xx -4) - 8`

`y = 8 - 8`

`y = 0`

The Solution Is:

`x = -4` and `y = 0`

Or

`(-4, 0)`

Answer 2

`x = -4 and y=0`

Explanation:

You can use the elimination method by making the coefficients of one of the variables the same,

Notice that the two `x` terms have opposite signs. Lets work with those.

`color(white)(xxxxxxx)color(blue)(-3x) +" "4y = 12" ".........A`
`color(white)(xxxxxxx)color(blue)(+2x) +" " y = -8" ".......B`

`A xx 2rarr:color(blue)(-6x) +" "8y = +24" ".......C`
`B xx 3rarr:color(blue)(+6x) +" "3y = -24" ".......D`

The `x` terms are now additive inverses. Their sum will be `0`

`C+D rarr:color(white)(xxxxxxx)11y = 0`

`color(white)(xxxxxxxxxxxxxxxxx)y = 0`

Substitute `0` for `y` in any of the equations: I will use `B`

`color(white)(xxxxxxxxx)color(blue)(+2x) +0 = -8" ".......B`
`color(white)(xxxxxxxxxxxx)color(blue)(+2x) = -8" ".......B`
`color(white)(xxxxxxxxxxxxxx)color(blue)(x) = -4" ".......B`

Check in the other equation using `x= -4 and y=0`

`color(white)(xxxxxxxxxx)2(-4) + 0 = -8" ".......B`

`color(white)(xxxxxxxxxx)-8 + 0 = -8`

`color(white)(xxxxxxxxxxxxx)-8 = -8`

The values are correct.