Solve for x x/(x-1)+5x=1/(x-1)

$\frac{x}{x - 1} + 5 x = \frac{1}{x - 1}$

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.

$x - 1, 1, x - 1$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

The LCM of $x - 1, x - 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x - 1$

Multiply each term in $\frac{x}{x - 1} + 5 x = \frac{1}{x - 1}$ by $x - 1$ to eliminate the fractions.

Multiply each term in $\frac{x}{x - 1} + 5 x = \frac{1}{x - 1}$ by $x - 1$ .

$\frac{x}{x - 1} (x - 1) + 5 x (x - 1) = \frac{1}{x - 1} (x - 1)$

Simplify the left side.

Simplify each term.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{x}{x - 1} (x - 1) + 5 x (x - 1) = \frac{1}{x - 1} (x - 1)$

Rewrite the expression.

$x + 5 x (x - 1) = \frac{1}{x - 1} (x - 1)$

Apply the distributive property.

$x + 5 x \cdot x + 5 x \cdot - 1 = \frac{1}{x - 1} (x - 1)$

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

Move $x$ .

$x + 5 (x \cdot x) + 5 x \cdot - 1 = \frac{1}{x - 1} (x - 1)$

Multiply $x$ by $x$ .

$x + 5 x^{2} + 5 x \cdot - 1 = \frac{1}{x - 1} (x - 1)$

Multiply $- 1$ by $5$ .

$x + 5 x^{2} - 5 x = \frac{1}{x - 1} (x - 1)$

Subtract $5 x$ from $x$ .

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

Simplify the right side.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$5 x^{2} - 4 x = 1$

Solve the equation.

Subtract $1$ from both sides of the equation.

$5 x^{2} - 4 x - 1 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = - 4$ .

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

$5 x^{2} - 4 x - 1 = 0$

Rewrite $- 4$ as $1$ plus $- 5$

$5 x^{2} + (1 - 5) x - 1 = 0$

Apply the distributive property.

$5 x^{2} + 1 x - 5 x - 1 = 0$

Multiply $x$ by $1$ .

$5 x^{2} + x - 5 x - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{2} + x) - 5 x - 1 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

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

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

$5 x + 1 = 0$

$x - 1 = 0$

Set $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(5 x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{5}, 1$

Exclude the solutions that do not make $\frac{x}{x - 1} + 5 x = \frac{1}{x - 1}$ true.

$x = - \frac{1}{5}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{5}$

Decimal Form:

$x = - 0.2$

Solve for x x/(x-1)+5x=1/(x-1)

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