Solve for x x+5/(x-4)=2x

$x + \frac{5}{x - 4} = 2 x$

Step 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.

$1, x - 4, 1$

Remove parentheses.

$1, x - 4, 1$

The LCM of one and any expression is the expression.

$x - 4$

Step 2

Multiply each term in $x + \frac{5}{x - 4} = 2 x$ by $x - 4$ to eliminate the fractions.

Multiply each term in $x + \frac{5}{x - 4} = 2 x$ by $x - 4$ .

$x (x - 4) + \frac{5}{x - 4} (x - 4) = 2 x (x - 4)$

Simplify the left side.

Simplify each term.

Apply the distributive property.

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

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 4 + \frac{5}{x - 4} (x - 4) = 2 x (x - 4)$

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

$x^{2} - 4 \cdot x + \frac{5}{x - 4} (x - 4) = 2 x (x - 4)$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$x^{2} - 4 x + \frac{5}{x - 4} (x - 4) = 2 x (x - 4)$

Rewrite the expression.

$x^{2} - 4 x + 5 = 2 x (x - 4)$

Simplify the right side.

Apply the distributive property.

$x^{2} - 4 x + 5 = 2 x \cdot x + 2 x \cdot - 4$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x^{2} - 4 x + 5 = 2 (x \cdot x) + 2 x \cdot - 4$

Multiply $x$ by $x$ .

$x^{2} - 4 x + 5 = 2 x^{2} + 2 x \cdot - 4$

Multiply $- 4$ by $2$ .

$x^{2} - 4 x + 5 = 2 x^{2} - 8 x$

Step 3

Solve the equation.

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

Subtract $2 x^{2}$ from both sides of the equation.

$x^{2} - 4 x + 5 - 2 x^{2} = - 8 x$

Add $8 x$ to both sides of the equation.

$x^{2} - 4 x + 5 - 2 x^{2} + 8 x = 0$

Subtract $2 x^{2}$ from $x^{2}$ .

$- x^{2} - 4 x + 5 + 8 x = 0$

Add $- 4 x$ and $8 x$ .

$- x^{2} + 4 x + 5 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 4 x + 5$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 4 x + 5 = 0$

Factor $- 1$ out of $4 x$ .

$- (x^{2}) - (- 4 x) + 5 = 0$

Rewrite $5$ as $- 1 (- 5)$ .

$- (x^{2}) - (- 4 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 4 x)$ .

$- (x^{2} - 4 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (x^{2} - 4 x) - 1 (- 5)$ .

$- (x^{2} - 4 x - 5) = 0$

Factor.

Factor $x^{2} - 4 x - 5$ 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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Write the factored form using these integers.

$- ((x - 5) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 5) (x + 1) = 0$

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

$x - 5 = 0$

$x + 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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 $- (x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Solve for x x+5/(x-4)=2x

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