How do you solve the linear system #y = -1/2x + 2# and #y = 3x -5#?

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Answer 1 · 2016-01-09T05:16:05

#x=2# and #y=1#

the given #y=-x/2+2# and #y=3x-5#

just equate the two given y values:

#y=y#

therefore

#-x/2+2=3x-5#

get rid of the fraction by multiplying by 2 both sides of the equation

#-x+4=6x-10#

#10+4=6x+x# by transposition

#14=7x# by simplification

#7x=14# by symmetric property

#x=2# after dividing both sides of the equation

using now #y=3x-5# substitute #x=2#

#y=3(2)-5#

#y=1#

so that the required values are #x=2# and #y=1#