Solve for x (7x)/6-3=(5x)/4

$\frac{7 x}{6} - 3 = \frac{5 x}{4}$

Multiply both sides by $4$ .

$(\frac{7 x}{6} - 3) \cdot 4 = \frac{5 x}{4} \cdot 4$

Simplify.

Simplify the left side.

Simplify $(\frac{7 x}{6} - 3) \cdot 4$ .

Apply the distributive property.

$\frac{7 x}{6} \cdot 4 - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$\frac{7 x}{2 (3)} \cdot 4 - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Factor $2$ out of $4$ .

$\frac{7 x}{2 \cdot 3} \cdot (2 \cdot 2) - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Cancel the common factor.

$\frac{7 x}{2 \cdot 3} \cdot (2 \cdot 2) - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Rewrite the expression.

$\frac{7 x}{3} \cdot 2 - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Combine $\frac{7 x}{3}$ and $2$ .

$\frac{7 x \cdot 2}{3} - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Multiply.

Multiply $2$ by $7$ .

$\frac{14 x}{3} - 3 \cdot 4 = \frac{5 x}{4} \cdot 4$

Multiply $- 3$ by $4$ .

$\frac{14 x}{3} - 12 = \frac{5 x}{4} \cdot 4$

Simplify the right side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{14 x}{3} - 12 = \frac{5 x}{4} \cdot 4$

Rewrite the expression.

$\frac{14 x}{3} - 12 = 5 x$

Solve for $x$ .

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

Subtract $5 x$ from both sides of the equation.

$\frac{14 x}{3} - 12 - 5 x = 0$

To write $- 5 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{14 x}{3} - 5 x \cdot \frac{3}{3} - 12 = 0$

Combine $- 5 x$ and $\frac{3}{3}$ .

$\frac{14 x}{3} + \frac{- 5 x \cdot 3}{3} - 12 = 0$

Combine the numerators over the common denominator.

$\frac{14 x - 5 x \cdot 3}{3} - 12 = 0$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $14 x - 5 x \cdot 3$ .

Factor $x$ out of $14 x$ .

$\frac{x \cdot 14 - 5 x \cdot 3}{3} - 12 = 0$

Factor $x$ out of $- 5 x \cdot 3$ .

$\frac{x \cdot 14 + x (- 5 \cdot 3)}{3} - 12 = 0$

Factor $x$ out of $x \cdot 14 + x (- 5 \cdot 3)$ .

$\frac{x (14 - 5 \cdot 3)}{3} - 12 = 0$

Multiply $- 5$ by $3$ .

$\frac{x (14 - 15)}{3} - 12 = 0$

Subtract $15$ from $14$ .

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

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

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

Move the negative in front of the fraction.

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

Add $12$ to both sides of the equation.

$- \frac{x}{3} = 12$

Multiply both sides of the equation by $- 3$ .

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

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

$- - x = - 3 \cdot 12$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x = - 3 \cdot 12$

Multiply $x$ by $1$ .

$x = - 3 \cdot 12$

Simplify the right side.

Multiply $- 3$ by $12$ .

$x = - 36$

Solve for x (7x)/6-3=(5x)/4

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