Solve for x x-(x+7)/(x-1)=2

$x - \frac{x + 7}{x - 1} = 2$

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.

$1, x - 1, 1$

Remove parentheses.

$1, x - 1, 1$

The LCM of one and any expression is the expression.

$x - 1$

Multiply each term in $x - \frac{x + 7}{x - 1} = 2$ by $x - 1$ to eliminate the fractions.

Multiply each term in $x - \frac{x + 7}{x - 1} = 2$ by $x - 1$ .

$x (x - 1) - \frac{x + 7}{x - 1} (x - 1) = 2 (x - 1)$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x \cdot x + x \cdot - 1 - \frac{x + 7}{x - 1} (x - 1) = 2 (x - 1)$

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - \frac{x + 7}{x - 1} (x - 1) = 2 (x - 1)$

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

$x^{2} - 1 \cdot x - \frac{x + 7}{x - 1} (x - 1) = 2 (x - 1)$

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

$x^{2} - x - \frac{x + 7}{x - 1} (x - 1) = 2 (x - 1)$

Cancel the common factor of $x - 1$ .

Move the leading negative in $- \frac{x + 7}{x - 1}$ into the numerator.

$x^{2} - x + \frac{- (x + 7)}{x - 1} (x - 1) = 2 (x - 1)$

Cancel the common factor.

$x^{2} - x + \frac{- (x + 7)}{x - 1} (x - 1) = 2 (x - 1)$

Rewrite the expression.

$x^{2} - x - (x + 7) = 2 (x - 1)$

Apply the distributive property.

$x^{2} - x - x - 1 \cdot 7 = 2 (x - 1)$

Multiply $- 1$ by $7$ .

$x^{2} - x - x - 7 = 2 (x - 1)$

Subtract $x$ from $- x$ .

$x^{2} - 2 x - 7 = 2 (x - 1)$

Simplify the right side.

Apply the distributive property.

$x^{2} - 2 x - 7 = 2 x + 2 \cdot - 1$

Multiply $2$ by $- 1$ .

$x^{2} - 2 x - 7 = 2 x - 2$

Solve the equation.

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

Subtract $2 x$ from both sides of the equation.

$x^{2} - 2 x - 7 - 2 x = - 2$

Subtract $2 x$ from $- 2 x$ .

$x^{2} - 4 x - 7 = - 2$

Add $2$ to both sides of the equation.

$x^{2} - 4 x - 7 + 2 = 0$

Add $- 7$ and $2$ .

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

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.

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

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-(x+7)/(x-1)=2

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