How do you solve by substitution #1/2x+2y=12# and #x-2y=6#?

3 Answers
Jul 10, 2018

Answer:

#x=12,y=3#

Explanation:

Plugging the second equation

#x=6+2y# in the first one we get

#1/2*(6+2y)+2y=12#

#1/2*6+1/2*2y+2y=12#
simplifying we get

#3+y+2y=12#

#3+3y=12#
so

#3y=9#

#y=3#

so #x=6+2*3=6+6=12#

Answer:

#x=12, \ y=3#

Explanation:

The given linear equations

#1/2x+2y=12\ ..........(1)#

#x-2y=6\ ..........(2)#

substituting the value of #x=6+2y# from (2) in (1) as follows

#1/2(6+2y)+2y=12#

#3+y+2y=12#

#3y=9#

#y=3#

substituting #y=3# in (1) we get

#1/2x+2(3)=12#

#1/2x+6=12#

#1/2x=6#

#x=12#

Jul 10, 2018

Answer:

The solution is the point #(12,3)#.

Explanation:

Solve system of equations:

Equation 1: #1/2x+2y=12#

Equation 2: #x-2y=6#

Solve Equation 1 for #x#.

#1/2x+2y=12#

Multiply both sides by #2#.

#x+4y=24#

Subtract #4y# from both sides.

#x=24-4y#

Substitute #24-4y# for #x# in Equation 2 and solve for #y#.

#x-2y=6#

#24-4y-2y=6#

Simplify.

#24-6y=6#

Subtract #24# from both sides.

#-6y=6-24#

Simplify.

#-6y=-18#

Divide both sides by #-6#.

#y=(-18)/(-6)#

#y=3#

Substitute #3# for #y# in Equation 1.

#1/2x+2y=12#

#1/2x+2(3)=12#

#1/2x+6=12#

Subtract #6# from both sides.

#1/2x=12-6#

Simplify.

#1/2x=6#

Multiply both sides by #2#.

#x=12#

The solution is the point #(12,3)#.

graph{(1/2x+2y-12)(x-2y-6)=0 [3.5, 23.5, -3.2, 6.8]}