Solve for B T=C(2+AB)

$T = C (2 + A B)$

Rewrite the equation as $C (2 + A B) = T$ .

$C (2 + A B) = T$

Divide each term in $C (2 + A B) = T$ by $C$ and simplify.

Divide each term in $C (2 + A B) = T$ by $C$ .

$\frac{C (2 + A B)}{C} = \frac{T}{C}$

Simplify the left side.

Cancel the common factor of $C$ .

Cancel the common factor.

$\frac{C (2 + A B)}{C} = \frac{T}{C}$

Divide $2 + A B$ by $1$ .

$2 + A B = \frac{T}{C}$

Subtract $2$ from both sides of the equation.

$A B = \frac{T}{C} - 2$

Divide each term in $A B = \frac{T}{C} - 2$ by $A$ and simplify.

Divide each term in $A B = \frac{T}{C} - 2$ by $A$ .

$\frac{A B}{A} = \frac{\frac{T}{C}}{A} + \frac{- 2}{A}$

Simplify the left side.

Cancel the common factor of $A$ .

Cancel the common factor.

$\frac{A B}{A} = \frac{\frac{T}{C}}{A} + \frac{- 2}{A}$

Divide $B$ by $1$ .

$B = \frac{\frac{T}{C}}{A} + \frac{- 2}{A}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$B = \frac{T}{C} \cdot \frac{1}{A} + \frac{- 2}{A}$

Multiply $\frac{T}{C}$ by $\frac{1}{A}$ .

$B = \frac{T}{C A} + \frac{- 2}{A}$

Move the negative in front of the fraction.

$B = \frac{T}{C A} - \frac{2}{A}$