How do you solve #2y = 4 – 6x # and #3y + 9x = 6#?

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Answer 1 · 2017-03-29T12:45:55

y = 2

#3y + 9x = 6#

Lets solve for y first

Add 3x to both sides

3y + 9x + 3y = 6 + 3y

9x = 6 + 3y

x = 6 + 3y/9

using the value of x solve the other equation

#2y = 4-6* ((6+3y)/9)#

Solve for normal division that is #(6+3)/9# but differently

#(3y)/9 + 6/9# = #1/3y + 2/3#

Plug it into the equation

#2y = 4-6*1/3y + 2/3#

#2y = 4-6/18y + 12/18#

Reduce the fractions to lower terms

#2y = 4-1/3y + 2/3#

#2y =1/3y + 2/3#

Simplify

#4 -1/3y + 2/3 = 4+(-1/3y) +2/3#

#2y = 4+(-1/3y)+2/3#

#2y = -1/3y +(4+2/3)#

#2y = -1/3y + 14/3#

Add #1/3y# to both sides

#2y +1/3y = -1/3y + 14/3 + 1/3y#

#-1/3y + 1/3y = 0#

#2y +1/3y = 14/3 #

#7/3y = 14/3#

#y = (14/3)/(7/3) = 2#