How do you solve 3x+6y=93x+6y=9 and 2x+8y=102x+8y=10 using substitution?

1 Answer
NGBRO
Mar 30, 2017

x=1; y=1x=1;y=1

Explanation:

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

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

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

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

Hence, we can find yy by "[Eqn 4]"-"[Eqn 3]"[Eqn 4]−[Eqn 3]:
6x-6x+24y-12y=30-186x−6x+24y−12y=30−18
12y=1212y=12
y=1y=1

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

3x+6(1)=93x+6(1)=9
3x=9-6=33x=9−6=3
x=1x=1

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

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