How do you solve #7y - 2x = 10# and #-3y + x = -3# using substitution?

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Answer 1 · 2017-10-17T12:16:55

See a solution process below:

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

#-3y + x = -3#

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

#0 + x = 3y - 3#

#x = 3y - 3#

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

#7y - 2x = 10# becomes:

#7y - 2(3y - 3) = 10#

#7y - (2 * 3y) - (2 * -3) = 10#

#7y - 6y - (-6) = 10#

#7y - 6y + 6 = 10#

#(7 - 6)y + 6 = 10#

#1y + 6 = 10#

#y + 6 - color(red)(6) = 10 - color(red)(6)#

#y + 0 = 4#

#y = 4#

Step 3: Substitute #4# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 3y - 3# becomes:

#x = (3 * 4) - 3#

#x = 12 - 3#

#x = 9#

The Solution Is: #x = 9# and #y = 4# or #(9, 4)#