How do you solve the system of equations #y+2x=3# and #-y+4=-x#?

1 Answer
Dec 2, 2016

#x=-1/3# and #y=11/3# or #3 2/3#

Explanation:

#y+2x=3#
#-y+4=-x#

In the first equation, we can determine a value for #y#.

#y+2x=3#

Subtract #2x# from both sides.

#y=3-2x#

In the second equation, substitute #y# with #color(red)((3-2x))#.

#-color(red)((3-2x))+4=-x#

Open the brackets and simplify. The product of two negatives is a positive, and that of a negative and positive is a negative.

#-3+2x+4=-x#

#2x+1=-x#

Add #x# to both sides.

#3x+1=0#

Subtract #1# from both sides.

#3x=-1#

Divide both sides by #3#.

#x=-1/3#

In the first equation, substitute #x# with #-1/3#.

#y+2x=3#

#y+2(-1/3)=3#

Open the brackets and simplify.

#y-2/3=3#

Add #2/3# to both sides.

#y=3+2/3#

#y=3 2/3# or #y=11/3#