Question

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

Answer 1

`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`