How do you solve #2x + 3y = 6# and #x + y = 3# using substitution?

1 Answer
Mar 23, 2018

Answer:

#x = 3, y = 0#

Explanation:

#2x + 3y = 6#

#x + y = 3#

Let's solve for #x# in the second equation

#x = 3 - y#

Now let's plug #(3-y)# in for #x# in the first equation

#2(3-y) + 3y = 6#

#6 - 2y + 3y = 6#

#y = 0#

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

Now we have #y#, let's find #x#

#x = 3 - y#

#x = 3 - 0#

#x = 3#

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

Let's double check our work and use our #x# and #y# values in the first equation and solve it. If our numbers are correct, the equation should equal #6#

#2(3) + 3(0)

#6 + 0#

#6 = 6#! We were right!