Solve for f 2f+3-5f=6+2(4-f)

$2 f + 3 - 5 f = 6 + 2 (4 - f)$

Step 1

Subtract $5 f$ from $2 f$ .

$- 3 f + 3 = 6 + 2 (4 - f)$

Step 2

Simplify $6 + 2 (4 - f)$ .

Simplify each term.

Apply the distributive property.

$- 3 f + 3 = 6 + 2 \cdot 4 + 2 (- f)$

Multiply $2$ by $4$ .

$- 3 f + 3 = 6 + 8 + 2 (- f)$

Multiply $- 1$ by $2$ .

$- 3 f + 3 = 6 + 8 - 2 f$

Add $6$ and $8$ .

$- 3 f + 3 = 14 - 2 f$

Step 3

Move all terms containing $f$ to the left side of the equation.

Add $2 f$ to both sides of the equation.

$- 3 f + 3 + 2 f = 14$

Add $- 3 f$ and $2 f$ .

$- f + 3 = 14$

Step 4

Move all terms not containing $f$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$- f = 14 - 3$

Subtract $3$ from $14$ .

$- f = 11$

Step 5

Divide each term in $- f = 11$ by $- 1$ and simplify.

Divide each term in $- f = 11$ by $- 1$ .

$\frac{- f}{- 1} = \frac{11}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{f}{1} = \frac{11}{- 1}$

Divide $f$ by $1$ .

$f = \frac{11}{- 1}$

Simplify the right side.

Divide $11$ by $- 1$ .

$f = - 11$