How do you solve #1/2x+1/3y=-1/6# and #1/2x+2y=13/2# using substitution?

1 Answer
Alan P.
May 9, 2016

#(x,y)=color(blue)(""(-3,4))#

Explanation:

To simplify this process multiply each equation by appropriate constants to clear the fractions:
[1a]#color(white)("XXX") 1/2x+1/3y=-1/6#
#color(white)("XXXXXXXXXXXXX")rarr #[1b]#color(white)("XXX") 3x+2y=-1#

[2a]#color(white)("XXX") 1/2x+2y=13/2#
#color(white)("XXXXXXXXXXXXX") rarr #[2b]#color(white)("XXX") x+4y=13#

[2b} appears to be the easiest one to convert into a form for substitution:
[2c]#color(white)("XXXXXXXXXXX") x=13-4y#

Substituting this back into [1b]
#color(white)("XXXXX") 3(13-4y)+2y=-1#

#color(white)("XXXXX") 39-10y=-1#

#color(white)("XXXXX") 10y=40#

#color(white)("XXXXX") y=4#

We can now substitute #4# for #y# back in [2c] to get
#color(white)("XXXXX") x=-3#

[as always, you should substitute these values back into the original equations to verify the results... assume that I've done this.]

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