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

$x + 1 = \frac{4}{x + 1}$

Subtract $1$ from both sides of the equation.

$x = \frac{4}{x + 1} - 1$

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.

$1, x + 1, 1$

Remove parentheses.

$1, x + 1, 1$

The LCM of one and any expression is the expression.

$x + 1$

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

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

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

Simplify the left side.

Apply the distributive property.

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

Simplify the expression.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $4$ .

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

Solve the equation.

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

Add $x$ to both sides of the equation.

$x^{2} + x + x = 3$

Add $x$ and $x$ .

$x^{2} + 2 x = 3$

Subtract $3$ from both sides of the equation.

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

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

$- 1, 3$

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

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.

$x = 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

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

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