How do you solve #x+2y=1# and #3x-4y=1#?

1 Answer
Sep 27, 2015

#(x,y) = (2/5,1/5)#

Explanation:

[1]#color(white)("XXX")x+2y=1#
[2]#color(white)("XXX")3x-4y=1#

Multiply [1] by #2#
[3]#color(white)("XXX")2x+4y=2#

Add [2] and [3]
[4]#color(white)("XXX")5x=3#

Divide [4] by #5#
[5]#color(white)("XXX")x=3/5#

Substitute #3/5# for #x# in [1]
[6]#color(white)("XXX")3/5+2y=1#

Subtract #3/5# from both sides of [6]
[7]#color(white)("XXX")2y = 2/5#

Divide [7] by #2#
[8]#color(white)("XXX")y=1/5#