How do you solve #18-2(3m-1)=(6+2(m+3))/2#?

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Answer 1 · 2017-06-30T01:23:55

#m=2#

Start by simplifying both left and right sides:

#18-2(3m-1)=(6+2(m+3))/2#

#18-6m+2=(6+2m+6)/2#

#20-6m=(12+2m)/2#

#20-6m=6+m#

Bring the variable to one side by subtracting #m# to both sides while also subtracting #20# to both sides to isolate the variable.

#cancel(20-20)-6m-m=6+cancel(m-m)-20#

#-7m=-14#

Solve for #m# by dividing #-7# from both sides:

#cancel(-7)/cancel(-7)m=(-14)/-7#

#m=2#