How do you solve #-2y - 22= 2( y + 3)#?

1 Answer
smendyka
Jun 2, 2017

See a solution process below:

Explanation:

First, expand the terms in the parenthesis on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#-2y - 22 = color(red)(2)(y + 3)#

#-2y - 22 = (color(red)(2) xx y) + (color(red)(2) xx 3)#

#-2y - 22 = 2y + 6#

Next, add #color(red)(2y)# and subtract #color(blue)(6)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#color(red)(2y) - 2y - 22 - color(blue)(6) = color(red)(2y) + 2y + 6 - color(blue)(6)#

#0 - 28 = (color(red)(2) + 2)y + 0#

#-28 = 4y#

Now, divide each side of the equation by #color(red)(4)# to solve for #y# while keeping the equation balanced:

#-28/color(red)(4) = (4y)/color(red)(4)#

#-7 = (color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4))#

#-7 = y#

#y = -7#

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