How do you solve #-9+ 8r = r + 5#?

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Answer 1 · 2017-04-06T21:40:39

See the entire solution process below:

First, add #color(red)(9)# and subtract #color(red)(r)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#-9 + 8r + color(red)(9) - color(red)(r) = r + 5 + color(red)(9) - color(red)(r)#

#-9 + color(red)(9) + 8r - color(red)(r) = r - color(red)(r) + 5 + color(red)(9)#

#0 + 8r - color(red)(1r) = 0 + 14#

#(8 - color(red)(1))r = 14#

#7r = 14#

Now, divide each side of the equation by #color(red)(7)# to solve for #r# while keeping the equation balanced:

#(7r)/color(red)(7) = 14/color(red)(7)#

#(color(red)(cancel(color(black)(7)))r)/cancel(color(red)(7)) = 2#

#r = 2#