Question

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

Answer 1

See a solution process below:

Explanation:

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

`3x - y = 2`

`-color(red)(3x) + 3x - y = -color(red)(3x) + 2`

`0 - y = -3x + 2`

`-y = -3x + 2`

`color(red)(-1) xx -y = color(red)(-1)(-3x + 2)`

`y = 3x - 2`

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

`3x + 2y = -4` becomes:

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

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

`3x + 6x - 4 = -4`

`(3 + 6)x - 4 = -4`

`9x - 4 = -4`

`9x - 4 + color(red)(4) = -4 + color(red)(4)`

`9x - 0 = 0`

`9x = 0`

`9x/color(red)(9) = 0/color(red)(9)`

`color(red)(cancel(color(black)(9)))x/cancel(color(red)(9)) = 0`

`x = 0`

Step 3) Substitute `0` for `x` in the solution to the first equation at the end of Step 1 and calculate `y`:

`y = 3x - 2` becomes:

`y = (3 xx 0) - 2`

`y = 0 - 2`

`y = -2`

The Solution Is: `x = 0` and `y = -2` or `(0, -2)`