How do you solve the system #1/2x+2y=12# and #x-2y=6# using substitution?

1 Answer
May 3, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Solve the second equation for #x#:

#x - 2y = 6#

#x - 2y + color(red)(2y) = 6 + color(red)(2y)#

#x - 0 = 6 + 2y#

#x = 6 + 2y#

Step 2) Substitute #6 + 2y# for #x# in the first equation and solve for #y#:

#1/2x + 2y = 12# becomes:

#1/2(6 + 2y) + 2y = 12#

#(1/2 * 6) + (1/2 * 2y) + 2y = 12#

#6/2+ (1/color(red)(cancel(color(black)(2))) * color(red)(cancel(color(black)(2)))y) + 2y = 12#

#3+ 1y + 2y = 12#

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

#3+ 3y = 12#

#-color(red)(3) + 3+ 3y = -color(red)(3) + 12#

#0+ 3y = 9#

#3y = 9#

#(3y)/color(red)(3) = 9/color(red)(3)#

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

#y = 3#

Step 3) Substitute #3# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 6 + 2y# becomes:

#x = 6 + (2 * 3)#

#x = 6 + 6#

#x = 12#

The solution is: #x = 12# and #y = 3# or #(12, 3)#