How do you solve #2x-y=4# and #7x+3y=27# using substitution?

1 Answer
Sep 1, 2017

Answer:

See a solution process below:

Explanation:

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

#2x - y = 4#

#-color(red)(2x) + 2x - y = -color(red)(2x) + 4#

#0 - y = -2x + 4#

#-y = -2x + 4#

#color(red)(-1) xxx -y = color(red)(-1)(-2x + 4)#

#y = (color(red)(-1) xx -2x) + (color(red)(-1) xx 4)#

#y = 2x - 4#

Step 2) Substitute #(2x - 4)# for #y# in the second equation and solve for #x#:

#7x + 3y = 27# becomes:

#7x + 3(2x - 4) = 27#

#7x + (3 xx 2x) - (3 xx 4) = 27#

#7x + 6x - 12 = 27#

#(7 + 6)x - 12 = 27#

#13x - 12 = 27#

#13x - 12 + color(red)(12) = 27 + color(red)(12)#

#13x - 0 = 39#

#13x = 39#

#(13x)/color(red)(13) = 39/color(red)(13)#

#(color(red)(cancel(color(black)(13)))x)/cancel(color(red)(13)) = 3#

#x = 3#

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

#y = 2x - 4# becomes:

#y = (2 xx 3) - 4#

#y = 6 - 4#

#y = 2#

The Solution Is: #x = 3# and #y = 2# or #(3, 2)#