How do you solve #2- 4a = 16#?

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Answer 1 · 2017-05-12T13:30:12

See a solution process below:

First, subtract #color(red)(2)# from each side of the equation to isolate the #a# term while keeping the equation balanced:

#-color(red)(2) + 2 - 4a = -color(red)(2) + 16#

#0 - 4a = 14#

#-4a = 14#

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

#(-4a)/color(red)(-4) = 14/color(red)(-4)#

#(color(red)(cancel(color(black)(-4)))a)/cancel(color(red)(-4)) = (2 xx 7)/color(red)(2 xx -2)#

#a = (color(red)(cancel(color(black)(2))) xx 7)/color(red)(color(black)(cancel(color(red)(2))) xx -2)#

#a = -7/2#

Or

#a = -3.5#