Solve for x 1+x+(x^2)/(1-x)=1/(1-x)

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

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.

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

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

$1$

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

$(1 - x) = 1 - x$

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

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

$1 - x$

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

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

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

Simplify the left side.

Simplify each term.

Multiply $1 - x$ by $1$ .

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- x$ and $x$ .

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

Add $1$ and $0$ .

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

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

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

Add $1$ and $0$ .

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

Simplify the right side.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

$1 = 1$

Since $1 = 1$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x 1+x+(x^2)/(1-x)=1/(1-x)

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