Solve for x c(1+bx)=a

$c (1 + b x) = a$

Divide each term in $c (1 + b x) = a$ by $c$ and simplify.

Divide each term in $c (1 + b x) = a$ by $c$ .

$\frac{c (1 + b x)}{c} = \frac{a}{c}$

Simplify the left side.

Cancel the common factor of $c$ .

Cancel the common factor.

$\frac{c (1 + b x)}{c} = \frac{a}{c}$

Divide $1 + b x$ by $1$ .

$1 + b x = \frac{a}{c}$

Subtract $1$ from both sides of the equation.

$b x = \frac{a}{c} - 1$

Divide each term in $b x = \frac{a}{c} - 1$ by $b$ and simplify.

Divide each term in $b x = \frac{a}{c} - 1$ by $b$ .

$\frac{b x}{b} = \frac{\frac{a}{c}}{b} + \frac{- 1}{b}$

Simplify the left side.

Cancel the common factor of $b$ .

Cancel the common factor.

$\frac{b x}{b} = \frac{\frac{a}{c}}{b} + \frac{- 1}{b}$

Divide $x$ by $1$ .

$x = \frac{\frac{a}{c}}{b} + \frac{- 1}{b}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{a}{c} \cdot \frac{1}{b} + \frac{- 1}{b}$

Multiply $\frac{a}{c}$ by $\frac{1}{b}$ .

$x = \frac{a}{c b} + \frac{- 1}{b}$

Move the negative in front of the fraction.

$x = \frac{a}{c b} - \frac{1}{b}$