How do you solve the following system: `3x - 2y = 9, 2x + 3y = 12`?

`9x-6y=27`
`4x-6y=24`

when you expand the first equation (3) and the second (2)
Combine the above equations:

`9x+4x=51`
`x=51/13`

Put this value in any equation to get y

`2*51/13 + 3y=12`

`3y=(156-102)/13`

`y=54/(13*3)`

`y=18/13`

Your `x=51/13` and your `y=18/13`

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

`3x - 2y = 9`

`3x - 2y + color(red)(2y) = 9 + color(red)(2y)`

`3x - 0 = 9 + 2y`

`3x = 9 + 2y`

`(3x)/color(red)(3) = (9 + 2y)/color(red)(3)`

`(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 9/3 + (2y)/3`

`x = 3 + 2/3y`

Step 2) Substitute `3 + 2/3y` for `x` in the second equation and solve for `y`:

`2x + 3y = 12` becomes:

`2(3 + 2/3y) + 3y = 12`

`(2 * 3) + (2 * 2/3y) + 3y = 12`

`6 + 4/3y + 3y = 12`

`6 + 4/3y + (3/3 * 3y) = 12`

`6 + 4/3y + 9/3y = 12`

`6 + 13/3y = 12`

`-color(red)(6) + 6 + 13/3y = -color(red)(6) + 12`

`0 + 13/3y = 6`

`13/3y = 6`

`color(red)(3)/color(blue)(13) xx 13/3y = color(red)(3)/color(blue)(13) xx 6`

`cancel(color(red)(3))/cancel(color(blue)(13)) xx color(blue)(cancel(color(black)(13)))/color(red)(cancel(color(black)(3)))y = 18/13`

`y = 18/13`

Step 3) Substitute `18/13` for `y` in the solution to the first equation at the end of Step 1 and calculate `x`:

`x = 3 + 2/3y` becomes:

`x = 3 + (2/3 xx 18/13)`

`x = 3 + (2/color(red)(cancel(color(black)(3))) xx (color(red)(cancel(color(black)(18)))6)/13)`

`x = 3 + 12/13`

`x = (13/13 xx 3) + 12/13`

`x = 39/13 + 12/13`

`x = 51/13`

The solution is: `x = 51/13` and `y = 18/13` or `(51/13, 18/131)`