How do you solve #y-2x=3# and #2x-3y=21# using substitution?

1 Answer
Mar 10, 2016

Answer:

#x=-7.5#
#y=-12#

Explanation:

Start by writing one of these 2 equations in terms of #x# or #y# so that you have a substitutable form. Note that the substitutable means in the form of "#x=#" or "#y=#"

I'm going to start with getting #y-2x=3# in terms of #y#. I will call this equation 1. All we have to do here is add #2x# to each side,

#y-2x=3#
Adding #2x# to each side:
#y=3+2x#

Now that we have the equation in terms of #y#, we can substitute this term (3+2x) into the other equation (equation 2).

#2x-3y=2x-3(3+2x) =21#

Notice that now we have one equation with one variable. This means we can go ahead and solve for #x#

#2x-3(3+2x)=2x-9-6x=-4x-9=21#
Adding 9 to both sides,
#-4x=21+9=30#
#x=-7.5#

Now that we have a value for #x#, we can substitute back into equation 1 to solve for #y#.

#y-2x=y-2(-7.5)=y+15=3#
Subtracting 15 from both sides and this solving for y,
#y=3-15=-12#

Therefore,
#x=-7.5#
#y=-12#