How do you solve #2(a+3)=-12# using the distributive property?

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Answer 1 · 2017-04-08T18:32:08

See the entire solution process below:

First, multiply each term within the parenthesis by the term outside the parenthesis on the left side of the equation:

#color(red)(2)(a + 3) = -12#

#(color(red)(2) xx a) + (color(red)(2) xx 3) = -12#

#2a + 6 = -12#

Next, subtract #color(red)(6)# from each side of the equation to isolate the #a# term while keeping the equation balanced:

#2a + 6 - color(red)(6) = -12 - color(red)(6)#

#2a + 0 = -18#

#2a = -18#

Now, divide each side of the equation by #color(red)(2)# to solve for #a# while keeping the equation balanced:

#(2a)/color(red)(2) = -18/color(red)(2)#

#(color(red)(cancel(color(black)(2)))a)/cancel(color(red)(2)) = -9#

#a = -9#