Solve for b fb+3b=d

$f b + 3 b = d$

Step 1

Factor $b$ out of $f b + 3 b$ .

Factor $b$ out of $f b$ .

$b f + 3 b = d$

Factor $b$ out of $3 b$ .

$b f + b \cdot 3 = d$

Factor $b$ out of $b f + b \cdot 3$ .

$b (f + 3) = d$

Step 2

Divide each term in $b (f + 3) = d$ by $f + 3$ and simplify.

Divide each term in $b (f + 3) = d$ by $f + 3$ .

$\frac{b (f + 3)}{f + 3} = \frac{d}{f + 3}$

Simplify the left side.

Cancel the common factor of $f + 3$ .

Cancel the common factor.

$\frac{b (f + 3)}{f + 3} = \frac{d}{f + 3}$

Divide $b$ by $1$ .

$b = \frac{d}{f + 3}$