How do you solve the system of equations #2x + y = 8# and #3x - y = 7#?

1 Answer
Apr 11, 2018

See a solution process below:

Explanation:

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

#2x + y = 8#

#2x - color(red)(2x) + y = 8 - color(red)(2x)#

#0 + y = 8 - 2x#

#y = 8 - 2x#

Step 2) Substitute #(8 - 2x)# for #y# in the second equation and solve for #x#:

#3x - y = 7# becomes:

#3x - (8 - 2x) = 7#

#3x - 8 + 2x = 7#

#3x - 8 + color(red)(8) + 2x = 7 + color(red)(8)#

#3x - 0 + 2x = 15#

#3x + 2x = 15#

#(3 + 2)x = 15#

#5x = 15#

#(5x)/color(red)(5) = 15/color(red)(5)#

#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 = 8 - 2x# becomes:

#y = 8 - (2 xx 3)#

#y = 8 - 6#

#y = 2#

**The Solution Is:

#x = 3# and #y = 2#

Or

#(3, 2)#