How do you solve # 2(r-3)+6=14#?

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Answer 1 · 2018-03-20T16:22:37

See a solution process below:

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

#2(r - 3) + 6 - color(red)(6) = 14 - color(red)(6)#

#2(r - 3) + 0 = 8#

#2(r - 3) = 8#

Next, divide each side of the equation by #color(red)(2)# to eliminate the parenthesis while keeping the equation balanced:

#(2(r - 3))/color(red)(2) = 8/color(red)(2)#

#(color(red)(cancel(color(black)(2)))(r - 3))/cancel(color(red)(2)) = 4#

#r - 3 = 4#

Now, add #color(red)(3)# from each side of the equation to solve for #r# while keeping the equation balanced:

#r - 3 + color(red)(3) = 4 + color(red)(3)#

#r - 0 = 7#

#r = 7#