Precalculus Examples

$x - \frac{10}{x - 1} = 4$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x - 1, 1$

Step 1.2

Remove parentheses.

$1, x - 1, 1$

Step 1.3

The LCM of one and any expression is the expression.

$x - 1$

Step 2

Multiply each term in $x - \frac{10}{x - 1} = 4$ by $x - 1$ to eliminate the fractions.

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Step 2.1

Multiply each term in $x - \frac{10}{x - 1} = 4$ by $x - 1$ .

$x (x - 1) - \frac{10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Apply the distributive property.

$x \cdot x + x \cdot - 1 - \frac{10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.2

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - \frac{10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.3

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

$x^{2} - 1 \cdot x - \frac{10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.4

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

$x^{2} - x - \frac{10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.5

Cancel the common factor of $x - 1$ .

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Step 2.2.1.5.1

Move the leading negative in $- \frac{10}{x - 1}$ into the numerator.

$x^{2} - x + \frac{- 10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.5.2

Cancel the common factor.

$x^{2} - x + \frac{- 10}{x - 1} (x - 1) = 4 (x - 1)$

Step 2.2.1.5.3

Rewrite the expression.

$x^{2} - x - 10 = 4 (x - 1)$

Step 2.3

Simplify the right side.

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Step 2.3.1

Apply the distributive property.

$x^{2} - x - 10 = 4 x + 4 \cdot - 1$

Step 2.3.2

Multiply $4$ by $- 1$ .

$x^{2} - x - 10 = 4 x - 4$

Step 3

Solve the equation.

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Step 3.1

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

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Step 3.1.1

Subtract $4 x$ from both sides of the equation.

$x^{2} - x - 10 - 4 x = - 4$

Step 3.1.2

Subtract $4 x$ from $- x$ .

$x^{2} - 5 x - 10 = - 4$

Step 3.2

Add $4$ to both sides of the equation.

$x^{2} - 5 x - 10 + 4 = 0$

Step 3.3

Add $- 10$ and $4$ .

$x^{2} - 5 x - 6 = 0$

Step 3.4

Factor $x^{2} - 5 x - 6$ using the AC method.

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Step 3.4.1

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 $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Step 3.4.2

Write the factored form using these integers.

$(x - 6) (x + 1) = 0$

Step 3.5

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

$x - 6 = 0$

$x + 1 = 0$

Step 3.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 3.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.7

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

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Step 3.7.1

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

$x + 1 = 0$

Step 3.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.8

The final solution is all the values that make $(x - 6) (x + 1) = 0$ true.

$x = 6, - 1$