How do you solve #-4r + 5- 6r = - 32#?

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Answer 1 · 2017-08-09T17:23:53

See a solution process below:

First, group and combine like terms on the left side of the equation:

#-4r - 6r + 5 = -32#

#(-4 - 6)r + 5 = -32#

#-10r + 5 = -32#

Next, subtract #color(red)(5)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#-10r + 5 - color(red)(5) = -32 - color(red)(5)#

#-10r + 0 = -37#

#-10r = -37#

Now, divide each side of the equation by #color(red)(-10)# to solve for #r# while keeping the equation balanced:

#(-10r)/color(red)(-10) = (-37)/color(red)(-10)#

#(color(red)(cancel(color(black)(-10)))r)/cancel(color(red)(-10)) = 37/10#

#r = 37/10# or #r = 3.7#