How do you solve #3/8m + 7/8 = 2m#?

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Answer 1 · 2016-03-02T12:58:36

You can do this in several ways. I usually try to get rid of fractions first.

Multiply everything by #8#:
#->cancel8*3/cancel8m+cancel8*7/cancel8=8*2m#

#->3m+7=16m#

Now get the #m#'s to one side and the numbers to the other:

#-3m# on both sides:

#->cancel(3m)-cancel(3m)+7=16m-3m#

#->13m=7->m=7/13#

Answer 2 · 2016-03-02T13:20:32

#m=7/13#

#3/8 m+7/8=2m#

Multiply each term by #8# to get rid of the denominator

#rarrcancel8*3/cancel8 m+cancel8*7/cancel8=8*2m#

#rarr3m+7=16m#

Rewrite

#rarr16m=3m+7#

#rarr16m-3m=7#

#rarr13m=7#

#rArrm=7/13#