Solve for x x-(3x)/(x+6)=2

$x - \frac{3 x}{x + 6} = 2$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x + 6, 1$

Remove parentheses.

$1, x + 6, 1$

The LCM of one and any expression is the expression.

$x + 6$

Multiply each term in $x - \frac{3 x}{x + 6} = 2$ by $x + 6$ to eliminate the fractions.

Multiply each term in $x - \frac{3 x}{x + 6} = 2$ by $x + 6$ .

$x (x + 6) - \frac{3 x}{x + 6} (x + 6) = 2 (x + 6)$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x \cdot x + x \cdot 6 - \frac{3 x}{x + 6} (x + 6) = 2 (x + 6)$

Multiply $x$ by $x$ .

$x^{2} + x \cdot 6 - \frac{3 x}{x + 6} (x + 6) = 2 (x + 6)$

Move $6$ to the left of $x$ .

$x^{2} + 6 \cdot x - \frac{3 x}{x + 6} (x + 6) = 2 (x + 6)$

Cancel the common factor of $x + 6$ .

Move the leading negative in $- \frac{3 x}{x + 6}$ into the numerator.

$x^{2} + 6 x + \frac{- 3 x}{x + 6} (x + 6) = 2 (x + 6)$

Cancel the common factor.

$x^{2} + 6 x + \frac{- 3 x}{x + 6} (x + 6) = 2 (x + 6)$

Rewrite the expression.

$x^{2} + 6 x - 3 x = 2 (x + 6)$

Subtract $3 x$ from $6 x$ .

$x^{2} + 3 x = 2 (x + 6)$

Simplify the right side.

Apply the distributive property.

$x^{2} + 3 x = 2 x + 2 \cdot 6$

Multiply $2$ by $6$ .

$x^{2} + 3 x = 2 x + 12$

Solve the equation.

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

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

$x^{2} + 3 x - 2 x = 12$

Subtract $2 x$ from $3 x$ .

$x^{2} + x = 12$

Subtract $12$ from both sides of the equation.

$x^{2} + x - 12 = 0$

Factor $x^{2} + x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(x - 3) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 4 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Solve for x x-(3x)/(x+6)=2

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