Solve Using the Quadratic Formula (2x-1)/(x+1)=(x+1)/(x-1)

$\frac{2 x - 1}{x + 1} = \frac{x + 1}{x - 1}$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $\frac{2 x - 1}{x + 1}$ .

Split the fraction $\frac{2 x - 1}{x + 1}$ into two fractions.

$\frac{2 x}{x + 1} + \frac{- 1}{x + 1} = \frac{x + 1}{x - 1}$

Move the negative in front of the fraction.

$\frac{2 x}{x + 1} - \frac{1}{x + 1} = \frac{x + 1}{x - 1}$

Simplify the right side.

Split the fraction $\frac{x + 1}{x - 1}$ into two fractions.

$\frac{2 x}{x + 1} - \frac{1}{x + 1} = \frac{x}{x - 1} + \frac{1}{x - 1}$

Move all the expressions to the left side of the equation.

Subtract $\frac{x}{x - 1}$ from both sides of the equation.

$\frac{2 x}{x + 1} - \frac{1}{x + 1} - \frac{x}{x - 1} = \frac{1}{x - 1}$

Subtract $\frac{1}{x - 1}$ from both sides of the equation.

$\frac{2 x}{x + 1} - \frac{1}{x + 1} - \frac{x}{x - 1} - \frac{1}{x - 1} = 0$

Step 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.

$x + 1, x + 1, x - 1, x - 1, 1$

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

$(x - 1) = x - 1$

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

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

$(x + 1) (x - 1)$

Step 3

Multiply each term in $\frac{2 x}{x + 1} - \frac{1}{x + 1} - \frac{x}{x - 1} - \frac{1}{x - 1} = 0$ by $(x + 1) (x - 1)$ to eliminate the fractions.

Multiply each term in $\frac{2 x}{x + 1} - \frac{1}{x + 1} - \frac{x}{x - 1} - \frac{1}{x - 1} = 0$ by $(x + 1) (x - 1)$ .

$\frac{2 x}{x + 1} ((x + 1) (x - 1)) - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{2 x}{x + 1} ((x + 1) (x - 1)) - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Rewrite the expression.

$2 x (x - 1) - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 1 - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

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

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 1 - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 1 - \frac{1}{x + 1} ((x + 1) (x - 1)) - \frac{x}{x - 1} ((x + 1) (x - 1)) - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Multiply $- 1$ by $2$ .

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

Cancel the common factor of $x + 1$ .

Move the leading negative in $- \frac{1}{x + 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $- 1$ by $- 1$ .

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

Cancel the common factor of $x - 1$ .

Move the leading negative in $- \frac{x}{x - 1}$ into the numerator.

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

Factor $x - 1$ out of $(x + 1) (x - 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

$2 x^{2} - 2 x - x + 1 - x \cdot x - x \cdot 1 - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

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

Move $x$ .

$2 x^{2} - 2 x - x + 1 - (x \cdot x) - x \cdot 1 - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Multiply $x$ by $x$ .

$2 x^{2} - 2 x - x + 1 - x^{2} - x \cdot 1 - \frac{1}{x - 1} ((x + 1) (x - 1)) = 0 ((x + 1) (x - 1))$

Multiply $- 1$ by $1$ .

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

Cancel the common factor of $x - 1$ .

Move the leading negative in $- \frac{1}{x - 1}$ into the numerator.

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

Factor $x - 1$ out of $(x + 1) (x - 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Simplify by adding terms.

Combine the opposite terms in $2 x^{2} - 2 x - x + 1 - x^{2} - x - x - 1$ .

Subtract $1$ from $1$ .

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

Add $2 x^{2} - 2 x - x - x^{2} - x - x$ and $0$ .

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

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

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

Subtract $x$ from $- 2 x$ .

$x^{2} - 3 x - x - x = 0 ((x + 1) (x - 1))$

Subtract $x$ from $- 3 x$ .

$x^{2} - 4 x - x = 0 ((x + 1) (x - 1))$

Subtract $x$ from $- 4 x$ .

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

Simplify the right side.

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Add $- x$ and $x$ .

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

Add $x^{2}$ and $0$ .

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

Multiply $0$ by $x^{2} - 1$ .

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

Step 4

Solve the equation.

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

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

$x \cdot x - 5 x = 0$

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

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

$x (x - 5) = 0$

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

$x = 0$

$x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 5$

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

$x = 0, 5$

Solve Using the Quadratic Formula (2x-1)/(x+1)=(x+1)/(x-1)

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