How do you solve #(2(2b+1))/3=3(b-2)#?

1 Answer
Jan 10, 2018

#b=4#

Explanation:

#\frac{2(2b+1)}{3}=3(b-2)#

Let's remove the denominator from the left multiplying both side with the same quantity, in this case #3# so we can simplify
#\color(blue)\cancel{3}*\frac{2(2b+1)}{\cancel{3}}=3(b-2)\color(blue){*3}#

#2(2b+1)=9(b-2)#

Now let's do the multiplication
#4b+2=9b-18#

And move all the terms with #b# to the left and all the constants (terms without letter) to the right. Every time we move something to the other side let's change the sign.

#4b-9b=-2-18#
#-5b = -20#

Multiply with #-1# to get a positive #b#
#\color(blue){(-1)}-5b = -20 \color(blue){(-1)}#
#5b=20#

#\frac{\cancel{5}}{\cancel{5}}b=\frac{20}{5}#
#b=4#