How do you solve the system #2x-3y=12 and x=4y+1#?

1 Answer
Jun 2, 2018

Answer:

#x = 9#
#y = 2#

Explanation:

#2x - 3y = 12#
#x = 4y + 1#

Solving by Substitution

First, we're going to use an equation for the value of a variable in order to plug it into the opposite equation of the system. Because #x = 4y + 1# is already an equation for the value of a variable, we'll be using it. In the other equation of the system, plug in #x#'s value where #x# is. So:

#2(4y + 1) - 3y = 12#

Next, you'll be distributing. What this means is that you'll be multiplying the outside number, #2#, by the terms in the parentheses, #4y# and #1#. So:

#2 * 4y = 8y#
#2 * 1 = 2#

Re-write your equation.

#8y + 2 - 3y = 12#

Combine like terms. #8y - 3y = 5y#, so:

#5y + 2 = 12#

This is a two-step equation. To solve it, subtract 2 from both sides to isolate for #y#. You should now have:

#5y = 10#

Divide by #5# to isolate for #y#:

#y = 2#

Plug the value of #y# back into the equation for the value of #x#:

#x = 4y + 1#
#x = 4(2) + 1#
#x = 8 + 1#
#x = 9#

To truly prove that #x# is 9 and #y# is 2:

#2x - 3y = 12#
#2(9) - 3(2) = 12#
#18 - 6 = 12#
#12 = 12#