How do you solve the following system?: `-x -4y =31, x -y = -42`

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

`x - y = -42`

`x - y + color(red)(y) = -42 + color(red)(y)`

`x - 0 = -42 + y`

`x = -42 + y`

Step 2) Substitute `-42 + y` for `x` in the first equation and solve for `y`:

`-x - 4y = 31` becomes:

`-(-42 + y) - 4y = 31`

`42 - y - 4y = 31`

`42 - 5y = 31`

`-color(red)(42) + 42 - 5y = -color(red)(42) + 31`

`0 - 5y = -11`

`-5y = -11`

`(-5y)/color(red)(-5) = -11/color(red)(-5)`

`(color(red)(cancel(color(black)(-5)))y)/cancel(color(red)(-5)) = 11/5`

`y = 11/5`

Step 3) Substitute `11/5` for `y` into the solution for the second equation at the end of Step 1 and calculate `x`:

`x = -42 + y` becomes:

`x = -42 + 11/5`

`x = (5/5 xx -42) + 11/5`

`x = -210/5 + 11/5`

`x = -199/5`

The solution is: `x = -199/5` and `y = 11/5` or `(-199/5, 11/5)`

`-x -4y =31 " "and " " x -y = -42`

This system of equations is a perfect scenario for eliminating
the `x`-terms because they are additive inverses.
Additive inverses give 0 when added together.

`(-x) + (+x) = 0`

`color(white)(........)-x -4y =31...................................A`
`color(white)(..............)x -y = -42...............................B`

`A+B:color(white)(..)-5y = -11`

`color(white)(.....)y = (-11)/-5 = 11/5 = 2 1/5`

Substitute `2 1/5` for `y` in `B`

`color(white)(..............)x -2 1/5 = -42`

`color(white)(..................)x = -42 +2 1/5`

`color(white)(..................)x = -39 4/5`

NOte that in this case the solutions are easier to work with as mixed numbers rather than as improper fractions which tend to involve large numbers.