Algebra Examples

$\frac{1}{x - 4} + \frac{1}{x + 3} = \frac{1}{x^{2} - x - 12}$

Step 1

Subtract $\frac{1}{x^{2} - x - 12}$ from both sides of the equation.

$\frac{1}{x - 4} + \frac{1}{x + 3} - \frac{1}{x^{2} - x - 12} = 0$

Step 2

Simplify $\frac{1}{x - 4} + \frac{1}{x + 3} - \frac{1}{x^{2} - x - 12}$ .

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Step 2.1

Factor $x^{2} - x - 12$ using the AC method.

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Step 2.1.1

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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 2.1.2

Write the factored form using these integers.

$\frac{1}{x - 4} + \frac{1}{x + 3} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.2

To write $\frac{1}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{1}{x - 4} \cdot \frac{x + 3}{x + 3} + \frac{1}{x + 3} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.3

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{1}{x - 4} \cdot \frac{x + 3}{x + 3} + \frac{1}{x + 3} \cdot \frac{x - 4}{x - 4} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.4

Write each expression with a common denominator of $(x - 4) (x + 3)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{1}{x - 4}$ by $\frac{x + 3}{x + 3}$ .

$\frac{x + 3}{(x - 4) (x + 3)} + \frac{1}{x + 3} \cdot \frac{x - 4}{x - 4} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.4.2

Multiply $\frac{1}{x + 3}$ by $\frac{x - 4}{x - 4}$ .

$\frac{x + 3}{(x - 4) (x + 3)} + \frac{x - 4}{(x + 3) (x - 4)} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.4.3

Reorder the factors of $(x - 4) (x + 3)$ .

$\frac{x + 3}{(x + 3) (x - 4)} + \frac{x - 4}{(x + 3) (x - 4)} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{x + 3 + x - 4}{(x + 3) (x - 4)} - \frac{1}{(x - 4) (x + 3)} = 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{x + 3 + x - 4 - 1}{(x + 3) (x - 4)} = 0$

Step 2.7

Add $x$ and $x$ .

$\frac{2 x + 3 - 4 - 1}{(x + 3) (x - 4)} = 0$

Step 2.8

Subtract $4$ from $3$ .

$\frac{2 x - 1 - 1}{(x + 3) (x - 4)} = 0$

Step 2.9

Subtract $1$ from $- 1$ .

$\frac{2 x - 2}{(x + 3) (x - 4)} = 0$

Step 2.10

Factor $2$ out of $2 x - 2$ .

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Step 2.10.1

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 2}{(x + 3) (x - 4)} = 0$

Step 2.10.2

Factor $2$ out of $- 2$ .

$\frac{2 (x) + 2 (- 1)}{(x + 3) (x - 4)} = 0$

Step 2.10.3

Factor $2$ out of $2 (x) + 2 (- 1)$ .

$\frac{2 (x - 1)}{(x + 3) (x - 4)} = 0$

Step 3

Set the denominator in $\frac{2 (x - 1)}{(x + 3) (x - 4)}$ equal to $0$ to find where the expression is undefined.

$(x + 3) (x - 4) = 0$

Step 4

Solve for $x$ .

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Step 4.1

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

$x + 3 = 0$

$x - 4 = 0$

Step 4.2

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

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Step 4.2.1

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

$x + 3 = 0$

Step 4.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 4.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 4.4

The final solution is all the values that make $(x + 3) (x - 4) = 0$ true.

$x = - 3, 4$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 3, x = 4$

Step 6