How do you solve #3x+3y=15# and #2x+3y=-5#?

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Answer 1 · 2015-08-03T08:35:17

The solution for the system of equations is
#color(blue)(x=20,y=-15#

#3x+color(blue)(3y)=15 #.....equation #(1)#

#2x+color(blue)(3y)=−5#....equation #(2)#

Solving by elimination
Subtracting equation #2# form #1# results in elimination of #color(blue)(3y#:

#3x+cancelcolor(blue)(3y)=15 #

#-2x-cancelcolor(blue)(3y)=+5#

#x=color(blue)(20#

Substituting #x# in equation #1# to find #y#
#3y=15-3x #
#3y=15-3*(20) #

#3y=15-60 #

#y=-45/3 #

#color(blue)(y=-15#