How do you solve the following system: `3x-y=12, 3x+4y=-10`?

Step 1) Solve the first equation for `y`:

`3x - y = 12`

`3x - y + color(red)(y) - color(blue)(12) = color(red)(y) + 12 - color(blue)(12)`

`3x cancel( - y + color(red)(y)) - 12 = y + cancel(12 - color(blue)(12))`

`3x - 12 = y`

`y = 3x - 12`

Step 2) Substitute `(3x - 12)` for `y` in the second equation and solve for `x`:

`3x + 4y = -10` becomes:

`3x + 4(3x - 12) = -10`

`3x + (4 xx 3x) - (4 xx 12) = -10`

`3x + 12x - 48 = -10`

`(3 + 12)x - 48 = -10`

`15x - 48 = -10`

`15x - 48 + color(red)(48) = -10 + color(red)(48)`

`15x cancel(- 48 + color(red)(48)) = 38`

`15x = 38`

`15x/color(red)(15) = 38/color(red)(15)`

`color(red)(cancel(color(black)(15)))x/cancel(color(red)(15)) = 38/15`

`x = 38/15`

Step 3) Substitute `38/15` for `x` in the solution to the first equation at the end of Step 1 and calculate `y`:

`y = 3x - 12` becomes:

`y = (3 xx 38/15) - 12`

`y = (color(red)(cancel(color(black)(3))) xx 38/(color(red)(cancel(color(black)(15)))5)) - 12`

`y = 38/5 - 12`

`y = 38/5 - (5/5 xx 12)`

`y = 38/5 - 60/5`

`y = -22/5`

The solution is: `x = 38/15` and `y = -22/5` or `(38/15, -22/5)`