How do you solve by substitution #3y=-1/2x+2# and #y=-x+9#?

1 Answer
Alan P.
Jun 15, 2015

For the given equations, #(x,y) = (10,-1)#

Explanation:

[1]#color(white)("XXXX")##3y = -1/2x+2#
[2]#color(white)("XXXX")##y = -x +9#

While it is not technically necessary, I find it easier to clear the fractions before actually beginning; so multiplying [1] by #2#
[3]#color(white)("XXXX")##6y = -x+4#

Substituting (from [2]) #(-x+9)# for #y# in [3]
[4]#color(white)("XXXX")##6(-x+9) = -x+4#

Simplifying
[5]#color(white)("XXXX")##-6x+54 = -x+4#
[6]#color(white)("XXXX")##5x = 50#
[7]#color(white)("XXXX")##x=10#

Substituting (from [7]) #10# for #x# in [2]
[8]#color(white)("XXXX")##y = -10 +9 = -1#

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