How do you solve the system #y=3x# and #2x+y=15# using substitution?

1 Answer
May 20, 2017

Answer:

See a solution process below:

Explanation:

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

#2x + y = 15# becomes:

#2x + 3x = 15#

#(2 + 3)x = 15#

#5x = 15#

#(5x)/color(red)(5) = 15/color(red)(5)#

#(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 3#

#x = 3#

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

#y = 3x# becomes:

#y = 3 xx 3#

#y = 9#

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