How do you solve #\frac { 5r + 1} { 8} = - 3#?

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Answer 1 · 2017-04-02T14:42:16

See the entire solution process below:

First, multiply each side of the equation by #color(red)(8)# to eliminate the fraction while keeping the equation balanced:

#color(red)(8) xx (5r + 1)/8 = color(red)(8) xx -3#

#cancel(color(red)(8)) xx (5r + 1)/color(red)(cancel(color(black)(8))) = -24#

#5r + 1 = -24#

Next, subtract #color(red)(1)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#5r + 1 - color(red)(1) = -24 - color(red)(1)#

#5r + 0 = -25#

#5r = -25#

Now, divide each side of the equation by #color(red)(5)# to solve for #r# while keeping the equation balanced:

#(5r)/color(red)(5) = -25/color(red)(5)#

#(color(red)(cancel(color(black)(5)))r)/cancel(color(red)(5)) = -5#

#r = -5#