How do you solve the system #y=x-1# and #x+y=3# using substitution?

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Answer 1 · 2017-01-20T18:34:33

#x=2# and #y=1#

#y=x-1#
#x+y=3#

In the second equation, substitute #y# with its value from the first equation, #color(red)((x-1))#.

#x+y=3#

#x+color(red)((x-1))=3#

Open the brackets and simplify.

#x+color(red)(x-1)=3#

#2x-1=3#

Add #1# to each side.

#2x=4#

Divide both sides by #2#.

#x=2#

In the first equation, substitute #x# with #color(blue)2#.

#y=x-1#

#y=color(blue)2-1#

#y=1#