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

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Answer 1 · 2016-02-28T05:06:16

#x=1#
#y=0#

#x+y=1#--------------------- (1)
#2x-y=2# ------------------(2)

Solving the 1st equation for #y#, we get

#y=1-x#

Substitute #y=1-x# in equation (2)

#2x-(1-x)=2#

Simplify and solve for #x#

#2x-1+x=2#
#3x=2+1#
#x=3/3=1#
#x=1#

Substitute the value #x=1#in equation (1) and solve for #y#

#1+y=1#
#y=1-1#
#y=0#