How do you solve #\frac { r - - 1} { 3} = 2#?

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Answer 1 · 2017-05-21T23:07:20

See a solution process below:

First, rewrite the numerator of the fraction. Remember, minus a minus is a plus:

#(r - - 1)/3 = 2#

#(r + 1)/3 = 2#

Next, multiply each side of the equation by #color(red)(3)# to eliminate the fraction while keeping the equation balanced:

#color(red)(3) xx (r + 1)/3 = color(red)(3) xx 2#

#cancel(color(red)(3)) xx (r + 1)/color(red)(cancel(color(black)(3))) = 6#

#r + 1 = 6#

Now, subtract #color(red)(1)# from each side of the equation to solve for #r# while keeping the equation balanced:

#r + 1 - color(red)(1) = 6 - color(red)(1)#

#r + 0 = 5#

#r = 5#