How do you solve the following system using substitution?: #x=4-4y, -x+2y=2#

1 Answer
Nov 5, 2015

Answer:

y = 1
x=0

Explanation:

First replace all values of x in the second equation with #4-4y# (this can be done for y as well but you would need to isolate the variable first)

Which gives you:
#-(4-4y)+2y=2#

Then solve using algebra:
Distribute the #-#
#-4+4y+2y=2#

Combine all all the y values
#-4+6y=2#

Add 4 to both sides
#(-4+6y)+4=(2)+4rarr6y=6#

Finally divide both sides by 6
#(6y)/6=6/6rarry=1#

Thus your y value is equal to 1. To find the x value, substitute this 1 into one of the equations for all y's
#x=4-4(1)#

Then simplify
#x=4-4rarrx=0#

thus x is equal to 0

You could also check your answer by substituting both values into the other equation
#-(0)+2(1)=2#

#0+2=2#

#2=2#

Your answers are correct!