Algebra Examples

$x + 4 < \frac{5}{x}$

Step 1

Multiply both sides by $x$ .

$(x + 4) x = \frac{5}{x} x$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(x + 4) x$ .

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Step 2.1.1.1

Apply the distributive property.

$x \cdot x + 4 x = \frac{5}{x} x$

Step 2.1.1.2

Multiply $x$ by $x$ .

$x^{2} + 4 x = \frac{5}{x} x$

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.

$x^{2} + 4 x = \frac{5}{x} x$

Step 2.2.1.2

Rewrite the expression.

$x^{2} + 4 x = 5$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $5$ from both sides of the equation.

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

Step 3.2

Factor $x^{2} + 4 x - 5$ using the AC method.

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Step 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 3.2.2

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

Step 3.3

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

$x - 1 = 0$

$x + 5 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 1 = 0$

Step 3.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.5

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 3.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3.6

The final solution is all the values that make $(x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Step 4

Find the domain of $x + 4 - \frac{5}{x}$ .

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Step 4.1

Set the denominator in $\frac{5}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 4.2

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.

$x < - 5$

$- 5 < x < 0$

$0 < x < 1$

$x > 1$

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 $x < - 5$ to see if it makes the inequality true.

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Step 6.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 6.1.2

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

$(- 8) + 4 < \frac{5}{- 8}$

Step 6.1.3

The left side $- 4$ is less than the right side $- 0.625$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $- 5 < x < 0$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $- 5 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 6.2.2

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

$(- 2) + 4 < \frac{5}{- 2}$

Step 6.2.3

The left side $2$ is not less than the right side $- 2.5$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 6.3.2

Replace $x$ with $0.5$ in the original inequality.

$(0.5) + 4 < \frac{5}{0.5}$

Step 6.3.3

The left side $4.5$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 6.4

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 6.4.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 6.4.2

Replace $x$ with $4$ in the original inequality.

$(4) + 4 < \frac{5}{4}$

Step 6.4.3

The left side $8$ is not less than the right side $1.25$ , which means that the given statement is false.

False

Step 6.5

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

$x < - 5$ True

$- 5 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

$x < - 5$ True

$- 5 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

Step 7

The solution consists of all of the true intervals.

$x < - 5$ or $0 < x < 1$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or 0 < x < 1$

Interval Notation:

$(- \infty, - 5) \cup (0, 1)$

Step 9