How do you solve the following linear system: #x+y=4, -3x+y=-8 #?

1 Answer
May 31, 2018

Answer:

See a solution process below:

Explanation:

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

#x + y = 4#

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

#x + 0 = 4 - y#

#x = 4 - y#

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

#-3x + y = -8# becomes:

#-3(4 - y) + y = -8#

#(-3 xx 4) + (3 xx y) + y = -8#

#-12 + 3y + y = -8#

#-12 + color(red)(12) + 3y + y = -8 + color(red)(12)#

#0 + 3y + y = 4#

#3y + y = 4#

#3y + 1y = 4#

#(3 + 1)y = 4#

#4y = 4#

#(4y)/color(red)(4) = 4/color(red)(4)#

#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 = 4 - y# becomes:

#x = 4 - 1#

#x = 3#

The Solution Is:

#x = 3# and #y = 1#

Or

#(3, 1)#