How do you solve #12(6m+2)=-6(3m+67) #?

1 Answer
Mar 16, 2018

Answer:

#m=-71/15=-4 11/15#

Explanation:

Solve:

#12(6m+2)=-6(3m+67)#

Divide both sides by #-6#.

#(color(red)cancel(color(black)(12))^2(6m+2))/color(red)cancel(color(black)(-6))^1=(-color(red)cancel(color(black)(6))^1(3m+67))/color(red)cancel(color(black)(-6))^1#

Simplify.

#-2(6m+2)=3m+67#

Expand.

#-12m-4=3m+67#

Add #4# to both sides.

#-12m-4+4=3m+67+4#

Simplify.

#-12m-0=3m+71#

#-12m=3m+71#

Subtract #3m# from both sides.

#-12m-3m=3m-3m+71#

Simplify.

#-15m=0+71#

#-15m=71#

Divide both sides by #-15#.

#(color(red)cancel(color(black)(-15))^1m)/(color(red)cancel(color(black)(-15))^1)=71/(-15)#

Simplify.

#m=-71/15#

We can convert this improper fraction to a mixed number #color(red)(a)color(blue) b/color(green)(c)#. Divide the numerator by the denominator by long division. The whole number quotient is the whole number #(color(red)a)# of the mixed number, the remainder is the numerator #(color(blue)b)#, and the denominator (divisor) stays the same #(color(green)c)#.

#-71-:15=-"4 remainder 11"#

#m=-71/15=-color(red)(4) color(blue)(11)/color(green)(15#