How do you solve #(-2m-2)/3 + (2m-3)/6 = m/8#?

Question

1151 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-07T12:55:09

#m = -28/11#

Step 1) Multiple each term by the necessary form of #1# to get to a common denominator. In this case a common denominator is #4#:

#8/8((-2m - 2)/3) + 4/4((2m - 3)/6) = 3/3(m/8)#

#(-16m - 16)/24 + (8m - 12)/24 = (3m)/24#

Step 2) Keep the equation balanced and allow the right side of the equation to equal #0#:

#(-16m - 16)/24 + (8m - 12)/24 - (3m)/ 24 = (3m)/24 - (3m)/24#

#(-16m - 16)/24 + (8m - 12)/24 - (3m)/ 24 = 0#

Step 3) Combine like terms in the numerator because they are all over a common denominator:

#(-16m + 8m - 3m - 16 - 12)/24 = 0#

#(-11m - 28)/24 = 0#

Step 3) solve for #m# while keeping the equation balanced:

#24((-11m - 28)/24) = 24*0#

#-11m - 28 = 0#

#-11m - 28 + 28 = 0 + 28#

#-11m = 28#

#(-11m)/-11 = 28/-11#

#m = -28/11#