How do you solve the following system: #-2x + 5y = 20, x +4y = 16#?

1 Answer
Mar 21, 2018

Let's use substitution:

#-2x + 5y = 20#

#x + 4y = 16#

We need to solve for #x# in the second equation

#x = 16 - 4y#

Now we substitute #(16 - 4y)# for #x# in the first equation

#-2(16 - 4y) + 5y = 20#

distribute the #-2#

#-32 + 8y + 5y = 20#

Solve for #y#. Add #32# to both sides

#13y = 52#

Divide by #13# on both sides

#y = 4#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

We have #y#, let's find #x#:

#x = 16 - 4y#

#x = 16 - 4(4)#

#x = 16 - 16#

#x = 0#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

To check our work, let's plug our values for #x# and #y# into the first equation and see if it equals #20#:

#-2(0) + 5(4)#

#0 + 20#

#20# EQUALS #20#! We were right