How do you solve the system of linear equations #x + 3y = 8# and #2x - y = 9#?

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Answer 1 · 2017-05-18T11:05:50

See a solution process below:

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

#x + 3y = 8#

#x + 3y - color(red)(3y) = 8 - color(red)(3y)#

#x + 0 = 8 - 3y#

#x = 8 - 3y#

Step 2) Substitute #8 - 3y# for #x# in the second equation and solve for #y#:

#2x - y = 9# becomes:

#2(8 - 3y) - y = 9#

#(2 * 8) - (2 * 3y) - y = 9#

#16 - 6y - y = 9#

#16 - 6y - 1y = 9#

#16 - 7y = 9#

#-color(red)(16) + 16 - 7y = -color(red)(16) + 9#

#0 - 7y = -7#

#-7y = -7#

#(-7y)/color(red)(-7) = (-7)/color(red)(-7)#

#(color(red)(cancel(color(black)(-7)))y)/cancel(color(red)(-7)) = 1#

#y = 1#

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

#x = 8 - 3y# becomes:

#x = 8 - (3 * 1)#

#x = 8 - 3#

#x = 5#

The solution is: #x = 5# and #y = 1# or #(5, 1)#