How do you solve #3x+6y=9# and #2x+8y=10# using substitution?

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Answer 1 · 2017-03-30T05:41:05

#x=1; y=1#

#["Eqn 1"] 3x+6y=9; ["Eqn 2"] 2x+8y=10#

Multiply [Eqn 1] by 2; [Eqn 2] by 3, so #x# can be in the same multiple:

#["Eqn 3"] 6x+12y=18#

#["Eqn 4"] 6x+24y=30#

Hence, we can find #y# by #"[Eqn 4]"-"[Eqn 3]"#:
#6x-6x+24y-12y=30-18#
#12y=12#
#y=1#

Now, just substitute #y# into either [Eqn 1] or [2]:
For example, I will subst. into [Eqn 1].

#3x+6(1)=9#
#3x=9-6=3#
#x=1#

Hence, the solution:
#x=1; y=1#