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

Question

Genesis Allen · Accepted Answer

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

Resolve:

#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#

Make it simple.

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

Extend.

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

Add #4# to both sides.

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

Make it simple.

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

#-12m=3m+71#

Subtract #3m# from both sides.

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

Make it simple.

#-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)#

Make it simple.

#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#

Eva Abramson · Answer

undefined To solve the equation 12(6m+2)=-6(3m+67), you would first distribute the terms inside the parentheses, then combine like terms, and finally isolate the variable "m" by moving constants to one side of the equation and variables to the other side. The solution to this equation is $ m = -\frac{335}{27} $.