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

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

Step 1

Simplify $f (x + 3)$ .

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Move $3$ to the left of $f$ .

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

Step 2

Simplify $2 (x - 3) - 1$ .

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $- 3$ .

$f x + 3 f = 2 x - 6 - 1$

Subtract $1$ from $- 6$ .

$f x + 3 f = 2 x - 7$

Step 3

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

$f x + 3 f - 2 x = - 7$

Step 4

Subtract $3 f$ from both sides of the equation.

$f x - 2 x = - 7 - 3 f$

Step 5

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

Factor $x$ out of $f x$ .

$x f - 2 x = - 7 - 3 f$

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

$x f + x \cdot - 2 = - 7 - 3 f$

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

$x (f - 2) = - 7 - 3 f$

Step 6

Divide each term in $x (f - 2) = - 7 - 3 f$ by $f - 2$ and simplify.

Divide each term in $x (f - 2) = - 7 - 3 f$ by $f - 2$ .

$\frac{x (f - 2)}{f - 2} = \frac{- 7}{f - 2} + \frac{- 3 f}{f - 2}$

Simplify the left side.

Cancel the common factor of $f - 2$ .

Cancel the common factor.

$\frac{x (f - 2)}{f - 2} = \frac{- 7}{f - 2} + \frac{- 3 f}{f - 2}$

Divide $x$ by $1$ .

$x = \frac{- 7}{f - 2} + \frac{- 3 f}{f - 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{- 7 - 3 f}{f - 2}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 (7) - 3 f}{f - 2}$

Factor $- 1$ out of $- 3 f$ .

$x = \frac{- 1 (7) - (3 f)}{f - 2}$

Factor $- 1$ out of $- 1 (7) - (3 f)$ .

$x = \frac{- 1 (7 + 3 f)}{f - 2}$

Move the negative in front of the fraction.

$x = - \frac{7 + 3 f}{f - 2}$

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

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