How do you solve the system of equations #x+ 4y = 4# and #- 5x + 3y = 3#?

1 Answer
May 23, 2018

Answer:

#x = 0#
#y = 1#

Explanation:

#x + 4y = 4#
#-5x + 3y = 3#

Solving by Substitution

First, let's use one of these equations and simplify for the equation of a value. I think #x# will be a good one to solve first because it looks easy to simplify. Let's begin:

#x + 4y = 4#

Subtract #4y# from both sides to get an equation for #x#. You should now have:

#x = -4y + 4#

This will be our #x# value that we'll be substituting in the second equation. Let's plug this term in:

#-5x + 3y = 3#
#-5(-4y + 4) + 3y = 3

Distribute. #-5y * -4y# becomes #20y# because two positives make a negative, and #-5 * 4# will become #-20# because only one negative is present.

#20y - 20 + 3y = 3#

Combine like terms.

#23y - 20 = 3#

Now, it's a two-step equation. Add #20# to both sides to cancel out #-20# in order to get the division step. You should now have:

#23y = 23#

Divide by #23# to isolate for #y#.

#y = 1#

Now knowing what #y# is, return to your simplified equation for the value of #x# and substitute the value of #y# for #y#:

#x = -4y + 4#
#x = -4(1) + 4#
#x = -4 + 4#
#x = 0#