Finite Math Examples

$| \frac{x - 5}{x} | (x^{2} - x - 12) = 0$

Step 1

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

$| \frac{x - 5}{x} | = 0$

$x^{2} - x - 12 = 0$

Step 2

Set $| \frac{x - 5}{x} |$ equal to $0$ and solve for $x$ .

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Step 2.1

Set $| \frac{x - 5}{x} |$ equal to $0$ .

$| \frac{x - 5}{x} | = 0$

Step 2.2

Solve $| \frac{x - 5}{x} | = 0$ for $x$ .

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Step 2.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$\frac{x - 5}{x} = \pm 0$

Step 2.2.2

Plus or minus $0$ is $0$ .

$\frac{x - 5}{x} = 0$

Step 2.2.3

Set the numerator equal to zero.

$x - 5 = 0$

Step 2.2.4

Add $5$ to both sides of the equation.

$x = 5$

Step 3

Set $x^{2} - x - 12$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{2} - x - 12$ equal to $0$ .

$x^{2} - x - 12 = 0$

Step 3.2

Solve $x^{2} - x - 12 = 0$ for $x$ .

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Step 3.2.1

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

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Step 3.2.1.1

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 3.2.1.2

Write the factored form using these integers.

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

Step 3.2.2

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

$x - 4 = 0$

$x + 3 = 0$

Step 3.2.3

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

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Step 3.2.3.1

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

$x - 4 = 0$

Step 3.2.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.2.4

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

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Step 3.2.4.1

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

$x + 3 = 0$

Step 3.2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.2.5

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

$x = 4, - 3$

Step 4

The final solution is all the values that make $| \frac{x - 5}{x} | (x^{2} - x - 12) = 0$ true.

$x = 5, 4, - 3$