Finite Math Examples

$\frac{14}{x} = x - 5$

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.

$x, 1, 1$

Step 1.2

The LCM of one and any expression is the expression.

$x$

Step 2

Multiply each term in $\frac{14}{x} = x - 5$ by $x$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{14}{x} = x - 5$ by $x$ .

$\frac{14}{x} x = x \cdot x - 5 x$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $x$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{14}{x} x = x \cdot x - 5 x$

Step 2.2.1.2

Rewrite the expression.

$14 = x \cdot x - 5 x$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply $x$ by $x$ .

$14 = x^{2} - 5 x$

Step 3

Solve the equation.

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Step 3.1

Rewrite the equation as $x^{2} - 5 x = 14$ .

$x^{2} - 5 x = 14$

Step 3.2

Subtract $14$ from both sides of the equation.

$x^{2} - 5 x - 14 = 0$

Step 3.3

Factor $x^{2} - 5 x - 14$ using the AC method.

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Step 3.3.1

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 $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Step 3.3.2

Write the factored form using these integers.

$(x - 7) (x + 2) = 0$

Step 3.4

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

$x - 7 = 0$

$x + 2 = 0$

Step 3.5

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3.5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 3.6

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 3.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.7

The final solution is all the values that make $(x - 7) (x + 2) = 0$ true.

$x = 7, - 2$