How do you solve the following system?: `x +12y = 13 , 5x - 2y = -4`

Because this question is in the section for using the Substitution method of solving systems of equations, that is the method I am going to use.

So, we start off with the equations:

`x+12y=13`
`5x-2y=-4`

1) Set an equation equal to a variable

Since the first equation has a lone variable, on without a coefficient, I'm going to use that one, but you can use whichever one you want.

`x+12y=13`
`color(red)(12-)x+12y=13color(red)(-12)`
`cancel(color(red)(12y-))x cancel(+12y)=13color(red)(-12y)`
`color(blue)(x=13-2y`

2) Substitute the new equation into the other one

`5x-2y=-4`
`color(blue)(x=13-2y`
`5(color(blue)(13-2y))-2y=-4`
`65-10y-2y=-4`
`65-12y=-4`
`color(red)(65-)65-12y=-4color(red)(-65)`
`cancel(color(red)(65-)65)-12y=-4color(red)(-65)`
`-12y=-69`
`(-12y=-69)/color(red)(-12)`
`y=color(green)5.75` or `y=color(green)(23/4`

3) Plug answer back into original equation

`x=13-2y`
`x=13-2(color(green)(5.75))`
`x=13-11.5`
`x=color(orange)1.5` or `y=color(orange)(3/2`

4) Write the solution as a coordinate (x,y)

The solution is: `(color(orange)(3/2),color(green)(23/4))`