How do you solve the system #2x-3y=12# and #x=4y+1# by substitution?

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Answer 1 · 2015-05-21T03:35:40

The answer is #x=9# and #y=2#.

Problem: Solve the system #2x-3y=12# and #x=4y+1# .

#x=4y+1# is already solved for #x#. Substitute it into the other equation: #2x-3y=12#.

#2(4y+1)-3y=12#

Distribute the #2#.

#8y+2-3y=12#

Simplify.

#5y+2=12# =

#5y=10#

Divide both sides by #5#.

#y=2#

Substitute the #2# for the #y# into one of the original equations.

#x=4*2+1=9#

#x=9#

Substitute the values for #x# and #y# to check the answers.

#2x-3y=12#

#2*9-3*2=12#

#18-6=12#

#12=12#

#x=4y+1#

#9=4*2+1# =

#9=9#

Both equations check!