How do you solve #x - 2y = 10# and #1/2x - 2y = 4# using substitution?

1 Answer
Alan P.
May 11, 2016

#(x,y)=color(blue)(12,1)#

Explanation:

Given
[1]#color(white)("XXX")x-2y=10#
[2]#color(white)("XXX")1/2x-2y=4#

We can rearrange [1] to give us an equation in terms of just #x#
[3]#color(white)("XXX")x=2y+10#

This allows us to substitute #(2y+10)# for #x# in [2]
[4]#color(white)("XXX")1/2(2y+10)-2y=4#

Simplifying
[5]#color(white)("XXX")y+5-2y=4#

[6]#color(white)("XXX")-y=-1#

[7]#color(white)("XXX")y=1#

Now we can substitute #1# for #y# back in [1]
[8]#color(white)("XXX")x-2(1)=10#

Simplifying
[9]#color(white)("XXX")x=12#

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