Solve for x (e(x+c))/b=2

$\frac{e (x + c)}{b} = 2$

Step 1

Multiply both sides by $b$ .

$\frac{e (x + c)}{b} b = 2 b$

Step 2

Simplify the left side.

Simplify $\frac{e (x + c)}{b} b$ .

Cancel the common factor of $b$ .

Cancel the common factor.

$\frac{e (x + c)}{b} b = 2 b$

Rewrite the expression.

$e (x + c) = 2 b$

$e (x + c) = 2 b$

Apply the distributive property.

$e x + e c = 2 b$

Simplify with commuting.

Reorder $e$ and $x$ .

$x e + e c = 2 b$

Reorder $e$ and $c$ .

$x e + c e = 2 b$

Reorder $x e$ and $c e$ .

$c e + x e = 2 b$

$c e + x e = 2 b$

$c e + x e = 2 b$

$c e + x e = 2 b$

Step 3

Solve for $x$ .

Subtract $c e$ from both sides of the equation.

$x e = 2 b - c e$

Divide each term in $x e = 2 b - c e$ by $e$ and simplify.

Divide each term in $x e = 2 b - c e$ by $e$ .

$\frac{x e}{e} = \frac{2 b}{e} + \frac{- c e}{e}$

Simplify the left side.

Cancel the common factor of $e$ .

Cancel the common factor.

$\frac{x e}{e} = \frac{2 b}{e} + \frac{- c e}{e}$

Divide $x$ by $1$ .

$x = \frac{2 b}{e} + \frac{- c e}{e}$

$x = \frac{2 b}{e} + \frac{- c e}{e}$

$x = \frac{2 b}{e} + \frac{- c e}{e}$

Simplify the right side.

Cancel the common factor of $e$ .

Cancel the common factor.

$x = \frac{2 b}{e} + \frac{- c e}{e}$

Divide $- c$ by $1$ .

$x = \frac{2 b}{e} - c$

$x = \frac{2 b}{e} - c$

$x = \frac{2 b}{e} - c$

$x = \frac{2 b}{e} - c$

$x = \frac{2 b}{e} - c$

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