Solve for x f(x-1)=2x+3

$f (x - 1) = 2 x + 3$

Step 1

Simplify $f (x - 1)$ .

Rewrite.

$0 + 0 + f (x - 1) = 2 x + 3$

Simplify by multiplying through.

Apply the distributive property.

$f x + f \cdot - 1 = 2 x + 3$

Move $- 1$ to the left of $f$ .

$f x - 1 \cdot f = 2 x + 3$

Rewrite $- 1 f$ as $- f$ .

$f x - f = 2 x + 3$

Step 2

Subtract $2 x$ from both sides of the equation.

$f x - f - 2 x = 3$

Step 3

Add $f$ to both sides of the equation.

$f x - 2 x = 3 + f$

Step 4

Factor $x$ out of $f x - 2 x$ .

Factor $x$ out of $f x$ .

$x f - 2 x = 3 + f$

Factor $x$ out of $- 2 x$ .

$x f + x \cdot - 2 = 3 + f$

Factor $x$ out of $x f + x \cdot - 2$ .

$x (f - 2) = 3 + f$

Step 5

Divide each term in $x (f - 2) = 3 + f$ by $f - 2$ and simplify.

Divide each term in $x (f - 2) = 3 + f$ by $f - 2$ .

$\frac{x (f - 2)}{f - 2} = \frac{3}{f - 2} + \frac{f}{f - 2}$

Simplify the left side.

Cancel the common factor of $f - 2$ .

Cancel the common factor.

$\frac{x (f - 2)}{f - 2} = \frac{3}{f - 2} + \frac{f}{f - 2}$

Divide $x$ by $1$ .

$x = \frac{3}{f - 2} + \frac{f}{f - 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{3 + f}{f - 2}$