How do you solve #2x+6y=32# and #5x-2y=12# using substitution?

2 Answers
Mar 11, 2018

Answer:

#therefore# #color(red)(x=4 and y=4#

Explanation:

#2x+6y=32#

#=> x + 3y = 16#

Multiplying both sides by 2

#x=16-3y#

Given that,

#5x−2y=12#

Replacing #x=16-3y#

#5(16-3y)-2y=12#

#80-15y-2y=12#

#80-17y=12#

#17y=80-12#

#17y=68#

#y=68/17#

#color(magenta)(y=4#

#therefore x=16-3y#

#x=16-3xx4#

#x=16-12#

#color(magenta)(x=4#

#therefore# #color(red)(x=4 and y=4#

~Hope this helps! :)

Mar 11, 2018

Answer:

#x=4, y=4#

Explanation:

#2x+6y=32#
#5x-2y=12#

In order to use substitution, rearrange one of the equations

#2x+6y=32#
#2x= -6y+32# (Subtracted 6y from both sides)
#x=-3y+16# (Divided both sides by 2)

Now that we rearrange the equation, we plug it into #5x-2y=12#

#5(-3y+16)-2y=12#
#-15y+80-2y=12# (Distributed 5 to the terms in bracket)
#-17y+80=12# (Combined like terms)
#-17y = -68# (Subtract 80 from both sides)
#y=4# (Divided both sides by -17)

Now that we have the y term, we can plug it into either equation to find x

#2x+6(4) = 32#
#2x+24=32# (Multiplied 6*4)
#2x= 8# (Subtracted 24 from both sides)
#x=4# (Divided both sides by 2)

Check your answer y plugging in y=4 and x=4 into either equation.