Solve for x (6x)/(x+4)=(x-1)

$\frac{6 x}{x + 4} = (x - 1)$

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.

$x + 4, 1, 1$

Remove parentheses.

$x + 4, 1, 1$

The LCM of one and any expression is the expression.

$x + 4$

Multiply each term in $\frac{6 x}{x + 4} = x - 1$ by $x + 4$ to eliminate the fractions.

Multiply each term in $\frac{6 x}{x + 4} = x - 1$ by $x + 4$ .

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

Simplify the left side.

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

$6 x = x (x + 4) - (x + 4)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$6 x = x \cdot x + x \cdot 4 - (x + 4)$

Multiply $x$ by $x$ .

$6 x = x^{2} + x \cdot 4 - (x + 4)$

Move $4$ to the left of $x$ .

$6 x = x^{2} + 4 \cdot x - (x + 4)$

Apply the distributive property.

$6 x = x^{2} + 4 x - x - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$6 x = x^{2} + 4 x - x - 4$

Subtract $x$ from $4 x$ .

$6 x = x^{2} + 3 x - 4$

Solve the equation.

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 3 x - 4 = 6 x$

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

Subtract $6 x$ from both sides of the equation.

$x^{2} + 3 x - 4 - 6 x = 0$

Subtract $6 x$ from $3 x$ .

$x^{2} - 3 x - 4 = 0$

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve for x (6x)/(x+4)=(x-1)

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