Question

How do you solve the system of equations `3x - \frac { 4} { 3} y = - 8` and `- 4y = x + 16`?

Answer 1

See a solution process below:

Explanation:

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

`-4y = x + 16`

`-4y - color(red)(16) = x + 16 - color(red)(16)`

`-4y - 16 = x + 0`

`-4y - 16 = x`

`x = -4y - 16`

Step 2) Substitute `(-4y - 16)` for `x` in the first equation and solve for `y`:

`3x - 4/3y = -8` becomes:

`3(-4y - 16) - 4/3y = -8`

`(3 * -4y) - (3 * 16) - 4/3y = -8`

`-12y - 48 - 4/3y = -8`

`-12y - 4/3y - 48 = -8`

`(3/3 * -12)y - 4/3y - 48 = -8`

`-36/3y - 4/3y - 48 = -8`

`-40/3y - 48 = -8`

`-40/3y - 48 + color(red)(48) = -8 + color(red)(48)`

`-40/3y - 0 = 40`

`-40/3y = 40`

`-color(red)(3)/color(blue)(40) * -(40)/3y = -color(red)(3)/color(blue)(40) * 40`

`-cancel(color(red)(3))/cancel(color(blue)(40)) * -color(blue)(cancel(color(black)(40)))/color(red)(cancel(color(black)(3)))y = -color(red)(3)/cancel(color(blue)(40)) * color(blue)(cancel(color(black)(40)))`

`- -y = -3`

`y = -3`

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

`x = -4y - 16` becomes:

`x = (-4 * -3) - 16`

`x = 12 - 16`

`x = -4`

The Solution Is: `x = -4` and `y = -3` or `(-4, -3)`