How do you solve the system #x=3y# and #3x-5y=12# using substitution?

Question

5823 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-28T14:35:07

See the entire solution process below:

Step 1) Because the first equation is already solved for #x# we can substitute #3y# for #x# in the second equation and solve for #y#:

#3x - 5y = 12# becomes:

#(3 xx 3y) - 5y = 12#

#9y - 5y = 12#

#(9 - 5)y = 12#

#4y = 12#

#(4y)/color(red)(4) = 12/color(red)(4)#

#(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = 3#

#y = 3#

Step 2) Substitute #3# for #y# in the first equation and calculate #x#:

#x = 3y# becomes:

#x = 3 xx 3#

#x = 9#

The solution is: #x = 9# and #y = 3# or #(9, 3)#