Solve for x S=bx-bc+cx

$S = b x - b c + c x$

Rewrite the equation as $b x - b c + c x = S$ .

$b x - b c + c x = S$

Add $b c$ to both sides of the equation.

$b x + c x = S + b c$

Factor $x$ out of $b x + c x$ .

Factor $x$ out of $b x$ .

$x b + c x = S + b c$

Factor $x$ out of $c x$ .

$x b + x c = S + b c$

Factor $x$ out of $x b + x c$ .

$x (b + c) = S + b c$

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

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

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

Simplify the left side.

Cancel the common factor of $b + c$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{S}{b + c} + \frac{b c}{b + c}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{S + b c}{b + c}$