How do you solve #x+y = 4# and #5x-2y = -1#?

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Answer 1 · 2015-06-01T20:58:34

Starting with #x+y=4#, subtract #x# from both sides to get:

#y = 4-x#

Then substitute this in the second equation as follows:

#-1 = 5x-2y#

#= 5x-2(4-x) = 5x-8+2x = 7x-8#

Add #8# to both ends to get:

#7x = 7#

Divide both sides by #7# to get

#x = 1#

Substitute that value for #x# into #y = 4-x# to get

#y=4-x=4-1=3#