Solve for x (2x+8)/6+8=x-4

$\frac{2 x + 8}{6} + 8 = x - 4$

Simplify $\frac{2 x + 8}{6} + 8$ .

Cancel the common factor of $2 x + 8$ and $6$ .

Factor $2$ out of $2 x$ .

$\frac{2 (x) + 8}{6} + 8 = x - 4$

Factor $2$ out of $8$ .

$\frac{2 x + 2 \cdot 4}{6} + 8 = x - 4$

Factor $2$ out of $2 x + 2 \cdot 4$ .

$\frac{2 (x + 4)}{6} + 8 = x - 4$

Cancel the common factors.

Factor $2$ out of $6$ .

$\frac{2 (x + 4)}{2 (3)} + 8 = x - 4$

Cancel the common factor.

$\frac{2 (x + 4)}{2 \cdot 3} + 8 = x - 4$

Rewrite the expression.

$\frac{x + 4}{3} + 8 = x - 4$

To write $8$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x + 4}{3} + 8 \cdot \frac{3}{3} = x - 4$

Simplify terms.

Combine $8$ and $\frac{3}{3}$ .

$\frac{x + 4}{3} + \frac{8 \cdot 3}{3} = x - 4$

Combine the numerators over the common denominator.

$\frac{x + 4 + 8 \cdot 3}{3} = x - 4$

Simplify the numerator.

Multiply $8$ by $3$ .

$\frac{x + 4 + 24}{3} = x - 4$

Add $4$ and $24$ .

$\frac{x + 28}{3} = x - 4$

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

Subtract $x$ from both sides of the equation.

$\frac{x + 28}{3} - x = - 4$

To write $- x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x + 28}{3} - x \cdot \frac{3}{3} = - 4$

Combine $- x$ and $\frac{3}{3}$ .

$\frac{x + 28}{3} + \frac{- x \cdot 3}{3} = - 4$

Combine the numerators over the common denominator.

$\frac{x + 28 - x \cdot 3}{3} = - 4$

Simplify the numerator.

Multiply $3$ by $- 1$ .

$\frac{x + 28 - 3 x}{3} = - 4$

Subtract $3 x$ from $x$ .

$\frac{- 2 x + 28}{3} = - 4$

Factor $2$ out of $- 2 x + 28$ .

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

$\frac{2 (- x) + 28}{3} = - 4$

Factor $2$ out of $28$ .

$\frac{2 (- x) + 2 (14)}{3} = - 4$

Factor $2$ out of $2 (- x) + 2 (14)$ .

$\frac{2 (- x + 14)}{3} = - 4$

Factor $- 1$ out of $- x$ .

$\frac{2 (- (x) + 14)}{3} = - 4$

Rewrite $14$ as $- 1 (- 14)$ .

$\frac{2 (- (x) - 1 (- 14))}{3} = - 4$

Factor $- 1$ out of $- (x) - 1 (- 14)$ .

$\frac{2 (- (x - 14))}{3} = - 4$

Rewrite $- (x - 14)$ as $- 1 (x - 14)$ .

$\frac{2 (- 1 (x - 14))}{3} = - 4$

Move the negative in front of the fraction.

$- \frac{2 (x - 14)}{3} = - 4$

Multiply both sides of the equation by $- \frac{3}{2}$ .

$- \frac{3}{2} (- \frac{2 (x - 14)}{3}) = - \frac{3}{2} \cdot - 4$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{3}{2} (- \frac{2 (x - 14)}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$\frac{- 3}{2} (- \frac{2 (x - 14)}{3}) = - \frac{3}{2} \cdot - 4$

Move the leading negative in $- \frac{2 (x - 14)}{3}$ into the numerator.

$\frac{- 3}{2} \cdot \frac{- 2 (x - 14)}{3} = - \frac{3}{2} \cdot - 4$

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{2} \cdot \frac{- 2 (x - 14)}{3} = - \frac{3}{2} \cdot - 4$

Cancel the common factor.

$\frac{3 \cdot - 1}{2} \cdot \frac{- 2 (x - 14)}{3} = - \frac{3}{2} \cdot - 4$

Rewrite the expression.

$\frac{- 1}{2} (- 2 (x - 14)) = - \frac{3}{2} \cdot - 4$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2 (x - 14)$ .

$\frac{- 1}{2} (2 (- (x - 14))) = - \frac{3}{2} \cdot - 4$

Cancel the common factor.

$\frac{- 1}{2} (2 (- (x - 14))) = - \frac{3}{2} \cdot - 4$

Rewrite the expression.

$- - (x - 14) = - \frac{3}{2} \cdot - 4$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (x - 14) = - \frac{3}{2} \cdot - 4$

Multiply $x - 14$ by $1$ .

$x - 14 = - \frac{3}{2} \cdot - 4$

Simplify the right side.

Simplify $- \frac{3}{2} \cdot - 4$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$x - 14 = \frac{- 3}{2} \cdot - 4$

Factor $2$ out of $- 4$ .

$x - 14 = \frac{- 3}{2} \cdot (2 (- 2))$

Cancel the common factor.

$x - 14 = \frac{- 3}{2} \cdot (2 \cdot - 2)$

Rewrite the expression.

$x - 14 = - 3 \cdot - 2$

Multiply $- 3$ by $- 2$ .

$x - 14 = 6$

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

Add $14$ to both sides of the equation.

$x = 6 + 14$

Add $6$ and $14$ .

$x = 20$

Solve for x (2x+8)/6+8=x-4

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