How do you solve by substitution #x+y=2# and #3x+2y=5#?

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Answer 1 · 2018-07-23T11:18:17

#x=1# & #y=1#

Given equations

#x+y=2\ ........(1)#

#3x+2y=5\ ........(2)#

Substituting #y=2-x# from (1) in (2) as follows

#3x+2(2-x)=5#

#3x+4-2x=5#

#x+4=5#

#x=1#

substituting #x=1# in (1), we get

#1+y=2#

#y=1#

Answer 2 · 2018-07-23T11:21:59

Here

If #x+y=2#
#x=2-y#

Put this into the second equation:

#3times(2-y) + 2y =5#

#6 - 3y + 2y =5#

#6-5 = y#

#y=1#

Since #x+y=2#

#x=2-y = 2-1 =1#

Your solution:

#x=1# and #y=1#

Or #x=y=1#