How do you solve for #m# in #3m - 5r = 2n + 7#?

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Answer 1 · 2017-03-11T22:40:48

See the entire solution process below:

First, add #color(red)(5r)# to each side of the equation to isolate the #m# term while keeping the equation balanced:

#3m - 5r + color(red)(5r) = 2n + 7 + color(red)(5r)#

#3m - 0 = 2n + color(red)(5r) + 7#

#3m = 2n + 5r + 7#

Now, divide each side of the equation by #color(red)(3)# to solve for #m# while keeping the equation balanced:

#(3m)/color(red)(3) = (2n + 5r + 7)/color(red)(3)#

#(color(red)(cancel(color(black)(3)))m)/cancel(color(red)(3)) = (2n + 5r + 7)/3#

#m = (2n + 5r + 7)/3#