How do you solve #3x - 7y = 27# and #- 5x + 4y = - 45# using substitution?

1 Answer
Meave60
May 7, 2018

The solution is #(9,0)#.

Explanation:

Solve the system of equations:

#"Equation 1":# #3x-7y=27#

#"Equation 2":# #-5x+4y=-45#

These are two linear equations in standard form. The solution to the system is the point they have in common, which is the point where they intersect.

Solve Equation 1 for #x#.

#3x-7y=27#

Add #7y# to both sides of the equation.

#3x=7y+27#

Divide both sides by #3#.

#x=7/3y+27/3#

#x=7/3y+9#

Substitute #7/3y+9# for #x# in Equation 2 and solve for #y#.

#-5x+4y=-45#

#-5(7/3y+9)+4y=-45#

Expand.

#-35/3y-45+4y=-45#

Simplify #-35/3y# to #(-35y)/3#.

Add #45# to both sides.

#(-35y)/3+4y=-45+45#

#(-35y)/3+4y=0#

Multiply #4y# by #3/3# to get an equivalent fraction with #3# as the denominator.

#(-35y)/3+4yxx3/3=0#

#(-35y)/3+(12y)/3=0#

Simplify.

#(-23y)/3=0#

Multiply both sides by #3#.

#-23y=0xx3#

#-23y=0#

Divide both sides by #-23#.

#y=0/(-23)#

#y=0#

Substitute #0# for #y# in Equation 1 and solve for #x#.

#3x-7y=27#

#3x-7(0)=27#

#3x=27#

Divide both sides by #3#.

#x=27/3#

#x=9#

The solution is #(9,0)#.

graph{(3x-7y-27)(-5x+4y+45)=0 [-10, 10, -5, 5]}

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