How do you solve #1/2(4+x)+2=16#?

1 Answer
smendyka
Jun 2, 2017

See a solution process below:

Explanation:

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

#1/2(4 + x) + 2 - color(red)(2) = 16 - color(red)(2)#

#1/2(4 + x) + 0 = 14#

#1/2(4 + x) = 14#

Next, multiply each side of the equation by #color(red)(2)# to eliminate the fraction and the need for parenthesis while keeping the equation balanced:

#color(red)(2) xx 1/2(4 + x) = color(red)(2) xx 14#

#cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))(4 + x) = 28#

#4 + x = 28#

Now, subtract #color(red)(4)# from each side of the equation to solve for #x# while keeping the equation balanced:

#-color(red)(4) + 4 + x = -color(red)(4) + 28#

#0 + x = 24#

#x = 24#

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