How do you solve #96= 12( 2y + 2)#?

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Answer 1 · 2016-12-27T12:59:21

#y = 3#

First, expand the terms within the parenthesis by multiplying each individual term in the parenthesis by #color(red)(12)#.

#96 = (color(red)(12) xx 2y) + (color(red)(12) xx 2)#

#96 = 24y + 24#

Next, we can subtract #color(blue)(24)# from each side to isolate the #y# term and to keep the equation balanced:

#96 - color(blue)(24) = 24y + 24 - color(blue)(24)#

#72 = 24y + 0#

#72 = 24y#

Now, we can divide each side by #color(green)(24)# to solve for #y# while keeping the equation balanced:

#72/color(green)(24) = (24y)/color(green)(24)#

#3 = (color(green)(cancel(color(black)(24)))y)/cancel(color(green)(24))#

#3 = y# or #y = 3#