How do you solve #x+y=3# and #x-y=1#?

1 Answer
George C.
Aug 1, 2015

Add the two equations together to get #2x=4# hence #x=2#
Subtract the two equations to get #2y=2# hence #y=1#

Explanation:

Add the two equations to find:

#2x = (x + color(red)(cancel(color(black)(y))))+(x-color(red)(cancel(color(black)(y)))) = 3 + 1 = 4#

Divide both ends by #2# to get #x=2#

Subtract the two equations to find:

#2y = (color(red)(cancel(color(black)(x))) + y) - (color(red)(cancel(color(black)(x))) - y) = 3 - 1 = 2#

Divide both ends by #2# to get #y = 1#

Related questions
See all questions in Linear Systems with Multiplication
Impact of this question
16148 views around the world
You can reuse this answer
Creative Commons License