Algebra Examples

$\frac{1}{2} + \frac{12}{x^{2}} > \frac{5}{x}$

Step 1

Multiply both sides by $x$ .

$(\frac{1}{2} + \frac{12}{x^{2}}) x = \frac{5}{x} x$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(\frac{1}{2} + \frac{12}{x^{2}}) x$ .

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Step 2.1.1.1

Apply the distributive property.

$\frac{1}{2} x + \frac{12}{x^{2}} x = \frac{5}{x} x$

Step 2.1.1.2

Combine $\frac{1}{2}$ and $x$ .

$\frac{x}{2} + \frac{12}{x^{2}} x = \frac{5}{x} x$

Step 2.1.1.3

Cancel the common factor of $x$ .

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Step 2.1.1.3.1

Factor $x$ out of $x^{2}$ .

$\frac{x}{2} + \frac{12}{x \cdot x} x = \frac{5}{x} x$

Step 2.1.1.3.2

Cancel the common factor.

$\frac{x}{2} + \frac{12}{x \cdot x} x = \frac{5}{x} x$

Step 2.1.1.3.3

Rewrite the expression.

$\frac{x}{2} + \frac{12}{x} = \frac{5}{x} x$

Step 2.2

Simplify the right 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.

$\frac{x}{2} + \frac{12}{x} = \frac{5}{x} x$

Step 2.2.1.2

Rewrite the expression.

$\frac{x}{2} + \frac{12}{x} = 5$

Step 3

Solve for $x$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

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

$2, x, 1$

Step 3.1.2

Since $2, x, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $2, 1, 1$ then find LCM for the variable part $x^{1}$ .

Step 3.1.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.1.4

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 3.1.5

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 3.1.6

The LCM of $2, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 3.1.7

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 3.1.8

The LCM of $x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x$

Step 3.1.9

The LCM for $2, x, 1$ is the numeric part $2$ multiplied by the variable part.

$2 x$

Step 3.2

Multiply each term in $\frac{x}{2} + \frac{12}{x} = 5$ by $2 x$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{x}{2} + \frac{12}{x} = 5$ by $2 x$ .

$\frac{x}{2} (2 x) + \frac{12}{x} (2 x) = 5 (2 x)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{x}{2} x + \frac{12}{x} (2 x) = 5 (2 x)$

Step 3.2.2.1.2

Cancel the common factor of $2$ .

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Step 3.2.2.1.2.1

Cancel the common factor.

$2 \frac{x}{2} x + \frac{12}{x} (2 x) = 5 (2 x)$

Step 3.2.2.1.2.2

Rewrite the expression.

$x \cdot x + \frac{12}{x} (2 x) = 5 (2 x)$

Step 3.2.2.1.3

Multiply $x$ by $x$ .

$x^{2} + \frac{12}{x} (2 x) = 5 (2 x)$

Step 3.2.2.1.4

Rewrite using the commutative property of multiplication.

$x^{2} + 2 \frac{12}{x} x = 5 (2 x)$

Step 3.2.2.1.5

Multiply $2 \frac{12}{x}$ .

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Step 3.2.2.1.5.1

Combine $2$ and $\frac{12}{x}$ .

$x^{2} + \frac{2 \cdot 12}{x} x = 5 (2 x)$

Step 3.2.2.1.5.2

Multiply $2$ by $12$ .

$x^{2} + \frac{24}{x} x = 5 (2 x)$

Step 3.2.2.1.6

Cancel the common factor of $x$ .

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Step 3.2.2.1.6.1

Cancel the common factor.

$x^{2} + \frac{24}{x} x = 5 (2 x)$

Step 3.2.2.1.6.2

Rewrite the expression.

$x^{2} + 24 = 5 (2 x)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Multiply $2$ by $5$ .

$x^{2} + 24 = 10 x$

Step 3.3

Solve the equation.

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Step 3.3.1

Subtract $10 x$ from both sides of the equation.

$x^{2} + 24 - 10 x = 0$

Step 3.3.2

Factor $x^{2} + 24 - 10 x$ using the AC method.

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Step 3.3.2.1

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 $24$ and whose sum is $- 10$ .

$- 6, - 4$

Step 3.3.2.2

Write the factored form using these integers.

$(x - 6) (x - 4) = 0$

Step 3.3.3

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

$x - 6 = 0$

$x - 4 = 0$

Step 3.3.4

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

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Step 3.3.4.1

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

$x - 6 = 0$

Step 3.3.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.3.5

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

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Step 3.3.5.1

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

$x - 4 = 0$

Step 3.3.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.3.6

The final solution is all the values that make $(x - 6) (x - 4) = 0$ true.

$x = 6, 4$

Step 4

Find the domain of $\frac{1}{2} + \frac{12}{x^{2}} - \frac{5}{x}$ .

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Step 4.1

Set the denominator in $\frac{12}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 4.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 0) \cup (0, \infty)$

Step 5

Use each root to create test intervals.

$x < 0$

$0 < x < 4$

$4 < x < 6$

$x > 6$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 6.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 6.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 6.1.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{1}{2} + \frac{12}{{(- 2)}^{2}} > \frac{5}{- 2}$

Step 6.1.3

The left side $3.5$ is greater than the right side $- 2.5$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 6.2.2

Replace $x$ with $2$ in the original inequality.

$\frac{1}{2} + \frac{12}{{(2)}^{2}} > \frac{5}{2}$

Step 6.2.3

The left side $3.5$ is greater than the right side $2.5$ , which means that the given statement is always true.

True

Step 6.3

Test a value on the interval $4 < x < 6$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $4 < x < 6$ and see if this value makes the original inequality true.

$x = 5$

Step 6.3.2

Replace $x$ with $5$ in the original inequality.

$\frac{1}{2} + \frac{12}{{(5)}^{2}} > \frac{5}{5}$

Step 6.3.3

The left side $0.98$ is not greater than the right side $1$ , which means that the given statement is false.

False

Step 6.4

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 6.4.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 6.4.2

Replace $x$ with $8$ in the original inequality.

$\frac{1}{2} + \frac{12}{{(8)}^{2}} > \frac{5}{8}$

Step 6.4.3

The left side $0.6875$ is greater than the right side $0.625$ , which means that the given statement is always true.

True

Step 6.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ True

$0 < x < 4$ True

$4 < x < 6$ False

$x > 6$ True

$x < 0$ True

$0 < x < 4$ True

$4 < x < 6$ False

$x > 6$ True

Step 7

The solution consists of all of the true intervals.

$x < 0$ or $0 < x < 4$ or $x > 6$

Step 8

Convert the inequality to interval notation.

$(- \infty, 0) \cup (0, 4) \cup (6, \infty)$

Step 9