Write as a Set of Linear Factors x/12-7=x/4

$\frac{x}{12} - 7 = \frac{x}{4}$

Subtract $\frac{x}{4}$ from both sides of the equation.

$\frac{x}{12} - 7 - \frac{x}{4} = 0$

Simplify $\frac{x}{12} - 7 - \frac{x}{4}$ .

To write $- \frac{x}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x}{12} - \frac{x}{4} \cdot \frac{3}{3} - 7 = 0$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{4}$ by $\frac{3}{3}$ .

$\frac{x}{12} - \frac{x \cdot 3}{4 \cdot 3} - 7 = 0$

Multiply $4$ by $3$ .

$\frac{x}{12} - \frac{x \cdot 3}{12} - 7 = 0$

Combine the numerators over the common denominator.

$\frac{x - x \cdot 3}{12} - 7 = 0$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $x - x \cdot 3$ .

Raise $x$ to the power of $1$ .

$\frac{x^{1} - x \cdot 3}{12} - 7 = 0$

Factor $x$ out of $x^{1}$ .

$\frac{x \cdot 1 - x \cdot 3}{12} - 7 = 0$

Factor $x$ out of $- x \cdot 3$ .

$\frac{x \cdot 1 + x (- 1 \cdot 3)}{12} - 7 = 0$

Factor $x$ out of $x \cdot 1 + x (- 1 \cdot 3)$ .

$\frac{x (1 - 1 \cdot 3)}{12} - 7 = 0$

Multiply $- 1$ by $3$ .

$\frac{x (1 - 3)}{12} - 7 = 0$

Subtract $3$ from $1$ .

$\frac{x \cdot - 2}{12} - 7 = 0$

Cancel the common factor of $- 2$ and $12$ .

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

$\frac{2 (x \cdot - 1)}{12} - 7 = 0$

Cancel the common factors.

Factor $2$ out of $12$ .

$\frac{2 (x \cdot - 1)}{2 (6)} - 7 = 0$

Cancel the common factor.

$\frac{2 (x \cdot - 1)}{2 \cdot 6} - 7 = 0$

Rewrite the expression.

$\frac{x \cdot - 1}{6} - 7 = 0$

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

$\frac{- 1 \cdot x}{6} - 7 = 0$

Move the negative in front of the fraction.

$- \frac{x}{6} - 7 = 0$

Factor $- 1$ out of $- \frac{x}{6} - 7$ .

Factor $- 1$ out of $- \frac{x}{6}$ .

$- (\frac{x}{6}) - 7 = 0$

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

$- (\frac{x}{6}) - 1 \cdot 7 = 0$

Factor $- 1$ out of $- (\frac{x}{6}) - 1 (7)$ .

$- (\frac{x}{6} + 7) = 0$

Write as a Set of Linear Factors x/12-7=x/4

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