How do you solve #x + y =1# and #2x-y=-2# using substitution?

1 Answer
Jul 10, 2016

Answer:

By eliminating #x# in the expression #x+y=1#, we get

#x = 1-y#, which we can plug in the second expression:

#2(1-y)-y=-2#

#2-2y-y=-2#

#2-3y=-2#

#-3y+2=-2#

#-3y=-4#

#y=4/3#

Knowing #y#, we can now find #x# by plugging the #y#-value into any of the above equations. For example, by plugging #y=4# into the first equation, #x+y=1#, we get

#x+4/3=1#

#x=-1/3#

Explanation:

We can check whether these values satisfy the equations by plugging them back in:

First: #x+y=1#

#-1/3 + 4/3 = 1 -> 4/3-1/3 = 1 -> (12-3)/(9) = 1 -> 9/9 = 1#

Second: #2x-y=-2#

#2(-1/3)-4/3=-2 -> -2/3-4/3 = -2 -> -18/9 = -2#