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

Jan 10, 2018

$b = 4$

Explanation:

$\setminus \frac{2 \left(2 b + 1\right)}{3} = 3 \left(b - 2\right)$

Let's remove the denominator from the left multiplying both side with the same quantity, in this case $3$ so we can simplify
$\setminus \textcolor{b l u e}{\setminus} \cancel{3} \cdot \setminus \frac{2 \left(2 b + 1\right)}{\setminus \cancel{3}} = 3 \left(b - 2\right) \setminus \textcolor{b l u e}{\cdot 3}$

$2 \left(2 b + 1\right) = 9 \left(b - 2\right)$

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

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.

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

Multiply with $- 1$ to get a positive $b$
$\setminus \textcolor{b l u e}{\left(- 1\right)} - 5 b = - 20 \setminus \textcolor{b l u e}{\left(- 1\right)}$
$5 b = 20$

$\setminus \frac{\setminus \cancel{5}}{\setminus \cancel{5}} b = \setminus \frac{20}{5}$
$b = 4$