How do you solve #-2x+3y=15# and #-6x+6y=18# using substitution?

Question

2755 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T11:33:26

See a solution process below:

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

#-6x + 6y = 18#

#(-6x + 6y)/color(red)(6) = 18/color(red)(6)#

#(-6x)/color(red)(6) + (6y)/color(red)(6) = 3#

#-x + y = 3#

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

#0 + y = 3 + x#

#y = 3 + x#

Step 2) Substitute #(3 + x)# for #y# in the first equation and solve for #x#:

#-2x + 3y = 15# becomes:

#-2x + 3(3 + x) = 15#

#-2x + (3 xx 3) + (3 xx x) = 15#

#-2x + 9 + 3x = 15#

#-2x + 9 - color(red)(9) + 3x = 15 - color(red)(9)#

#-2x + 0 + 3x = 6#

#-2x + 3x = 6#

#(-2 + 3)x = 6#

#1x = 6#

#x = 6#

Step 3) Substitute #6# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = 3 + x# becomes:

#y = 3 + 6#

#y = 9#

The Solution Is:

#x = 6# and #y = 9#

Or

#(6, 9)#