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

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Answer 1 · 2015-08-04T05:21:19

The solution for the system of equations is :
#color(blue)(x=0,y=3#

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

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

Solving by elimination, as adding the two equations would result in elimination of #color(blue)(y#

#x+cancelcolor(blue)(y)=3#
# 2x−cancelcolor(blue)(y)=−3#

#3x=0#
#color(blue)(x=0#

Substituting #x# in equation #1# to find #y#
#x + y=3#

# y=3-x#
#y=3-0#
#color(blue)(y=3#