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
