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))#