How do you solve #3x+ 4y=27# and #-8x+ y=33# using substitution?

1 Answer
Apr 6, 2018

Answer:

The solution is #(-3,9)#.

Explanation:

Solve the system:

#"Equation 1":# #3x+4y=27#

#"Equation 2":# #-8x+y=33#

Both equations are linear equations in standard form. The solution to the system is the point that the two lines have in common. The #x# and #y# coordinates will be determined by substitution.

Solve Equation 1 for #x#.

#3x+4y=27#

#3x=27-4y#

Divide both sides by #3#.

#x=27/3-(4y)/3#

Simplify.

#x=9-(4y)/3#

Substitute #9-4y/3# for #x# in Equation 2. Solve for #y#.

#-8x+y=33#

#-8(9-(4y)/3)+y=33#

#-72+(32y)/3+y=33#

Multiply #y# by #3/3# to create an equivalent fraction with #3# in the denominator.

#-72+(32y)/3+yxx3/3=33#

#-72+(32y)/3+(3y)/3=33#

#-72+(35y)/3=33#

Add #72# to both sides.

#(35y)/3=33+72#

Simplify.

#(35y)/3=105#

Multiply both sides by #3#.

#35y=105xx3#

#35y=315#

Divide both sides by #35#

#y=315/35#

#y=9#

Substitute #9# for #y# in Equation 1. Solve for #x#.

#3x+4(9)=27#

#3x+36=27#

#3x=27-36#

#3x=-9#

#x=(-9)/3#

#x=-3#

The solution is the point of intersection for the two lines, which is #(-3,9)#.

graph{(3x+4y-27)(y-8x-33)=0 [-14.24, 11.07, 1.37, 14.03]}