Question

How do you solve the following system: `x-4y=2, 3x+4y=-10`?

Answer 1

See a solution process below:

Explanation:

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

`x - 4y + color(red)(4y) = 2 + color(red)(4y)`

`x - 0 = 2 + 4y`

`x = 2 + 4y`

Step 2) Substitute `(2 + 4y)` for `x` in the second equation and solve for `y`:

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

`3(2 + 4y) + 4y = -10`

`(3 xx 2) + (3 xx 4y) + 4y = -10`

`6 + 12y + 4y = -10`

`6 + (12 + 4)y = -10`

`6 + 16y = -10`

`6 - color(red)(6) + 16y = -10 - color(red)(6)`

`0 + 16y = -16`

`16y = -16`

`(16y)/color(red)(16) = -16/color(red)(16)`

`(color(red)(cancel(color(black)(16)))y)/cancel(color(red)(16)) = -1`

`y = -1`

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

`x = 2 + 4y` becomes:

`x = 2 + (4 xx -1)`

`x = 2 + (-4)`

`x = 2 - 4`

`x = -2`

The Solution Is:

`x = -2` and `y = -1`

Or

`(-2, -1)`