How do you solve #5x-2y=48-4x# and #6x+7y=x+6y+33# using substitution?

1 Answer
Jul 19, 2016

Answer:

Solve one of the equations for one of the variables, then use that to substitute that variable out of the other equation, which you can then solve for the other variable.

Explanation:

Let's pick one of the equations -- it ultimately doesn't matter which -- and solve it for one of the variables in terms of the other.

#6x+7y=x+6y+33#
#7y-6y=x-6x+33#
#y=-5x+33#

Now, let's use this to substitute for #y# in the other equation:

#5x-2y=48-4x#
#5x-2(-5x+33)=48-4x#
#5x+10x-66=48-4x#
#15x-66=48-4x#
#15x+4x=48+66#
#19x=114#
#x=6#

Then, substitute this back into the solved-for-#y# form of the first equation:

#y=-5(6)+33#
#y=-30+33#
#y=3#

Done.