How do you solve #3x + y = 4# and #2x + y = 4#?

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Answer 1 · 2015-11-27T12:28:34

The solution for the system of equations is
#color(blue)(x=0,y=4#

#3x+color(blue)(y)=4#......equation #(1)#
#2x+color(blue)(y)=4#......equation #(2)#

Solving by elimination

Subtracting equation #1# from #2# would result in elimination of #color(blue)(y#

#3x+cancelcolor(blue)(y)=4#
#-2x-cancelcolor(blue)(y)=-4#

#color(blue)(x=0#

Substituting #x# in equation #1# to find #y#:

#3x+y=4#

#3 * 0 +y=4#
# 0 +y=4#
# color(blue)(y=4#