Algebra Examples

$\frac{4}{5 x} + \frac{1}{10} < \frac{3}{2 x}$

Step 1

Multiply both sides by $2 x$ .

$(\frac{4}{5 x} + \frac{1}{10}) (2 x) = \frac{3}{2 x} (2 x)$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(\frac{4}{5 x} + \frac{1}{10}) (2 x)$ .

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Step 2.1.1.1

Simplify terms.

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Step 2.1.1.1.1

Apply the distributive property.

$\frac{4}{5 x} (2 x) + \frac{1}{10} (2 x) = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.2

Rewrite using the commutative property of multiplication.

$2 \frac{4}{5 x} x + \frac{1}{10} (2 x) = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.3

Cancel the common factor of $2$ .

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Step 2.1.1.1.3.1

Factor $2$ out of $10$ .

$2 \frac{4}{5 x} x + \frac{1}{2 (5)} (2 x) = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.3.2

Factor $2$ out of $2 x$ .

$2 \frac{4}{5 x} x + \frac{1}{2 (5)} (2 (x)) = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.3.3

Cancel the common factor.

$2 \frac{4}{5 x} x + \frac{1}{2 \cdot 5} (2 x) = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.3.4

Rewrite the expression.

$2 \frac{4}{5 x} x + \frac{1}{5} x = \frac{3}{2 x} (2 x)$

Step 2.1.1.1.4

Combine $\frac{1}{5}$ and $x$ .

$2 \frac{4}{5 x} x + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.2

Simplify each term.

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Step 2.1.1.2.1

Multiply $2 \frac{4}{5 x}$ .

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Step 2.1.1.2.1.1

Combine $2$ and $\frac{4}{5 x}$ .

$\frac{2 \cdot 4}{5 x} x + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.2.1.2

Multiply $2$ by $4$ .

$\frac{8}{5 x} x + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.2.2

Cancel the common factor of $x$ .

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Step 2.1.1.2.2.1

Factor $x$ out of $5 x$ .

$\frac{8}{x \cdot 5} x + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.2.2.2

Cancel the common factor.

$\frac{8}{x \cdot 5} x + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.2.2.3

Rewrite the expression.

$\frac{8}{5} + \frac{x}{5} = \frac{3}{2 x} (2 x)$

Step 2.1.1.3

Reorder $\frac{8}{5}$ and $\frac{x}{5}$ .

$\frac{x}{5} + \frac{8}{5} = \frac{3}{2 x} (2 x)$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $\frac{3}{2 x} (2 x)$ .

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Step 2.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{x}{5} + \frac{8}{5} = 2 \frac{3}{2 x} x$

Step 2.2.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.2.1

Factor $2$ out of $2 x$ .

$\frac{x}{5} + \frac{8}{5} = 2 \frac{3}{2 (x)} x$

Step 2.2.1.2.2

Cancel the common factor.

$\frac{x}{5} + \frac{8}{5} = 2 \frac{3}{2 x} x$

Step 2.2.1.2.3

Rewrite the expression.

$\frac{x}{5} + \frac{8}{5} = \frac{3}{x} x$

Step 2.2.1.3

Cancel the common factor of $x$ .

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Step 2.2.1.3.1

Cancel the common factor.

$\frac{x}{5} + \frac{8}{5} = \frac{3}{x} x$

Step 2.2.1.3.2

Rewrite the expression.

$\frac{x}{5} + \frac{8}{5} = 3$

Step 3

Solve for $x$ .

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Step 3.1

Move all terms not containing $x$ to the right side of the equation.

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Step 3.1.1

Subtract $\frac{8}{5}$ from both sides of the equation.

$\frac{x}{5} = 3 - \frac{8}{5}$

Step 3.1.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{x}{5} = 3 \cdot \frac{5}{5} - \frac{8}{5}$

Step 3.1.3

Combine $3$ and $\frac{5}{5}$ .

$\frac{x}{5} = \frac{3 \cdot 5}{5} - \frac{8}{5}$

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{x}{5} = \frac{3 \cdot 5 - 8}{5}$

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Multiply $3$ by $5$ .

$\frac{x}{5} = \frac{15 - 8}{5}$

Step 3.1.5.2

Subtract $8$ from $15$ .

$\frac{x}{5} = \frac{7}{5}$

Step 3.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = 7$

Step 4

Find the domain of $- \frac{7}{10 x} + \frac{1}{10}$ .

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Step 4.1

Set the denominator in $\frac{7}{10 x}$ equal to $0$ to find where the expression is undefined.

$10 x = 0$

Step 4.2

Divide each term in $10 x = 0$ by $10$ and simplify.

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Step 4.2.1

Divide each term in $10 x = 0$ by $10$ .

$\frac{10 x}{10} = \frac{0}{10}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $10$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{10 x}{10} = \frac{0}{10}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{10}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $0$ by $10$ .

$x = 0$

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.

$x < 0$

$0 < x < 7$

$x > 7$

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.

$\frac{4}{5 (- 2)} + \frac{1}{10} < \frac{3}{2 (- 2)}$

Step 6.1.3

The left side $- 0.3$ is not less than the right side $- 0.75$ , which means that the given statement is false.

False

Step 6.2

Test a value on the interval $0 < x < 7$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $0 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 6.2.2

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

$\frac{4}{5 (4)} + \frac{1}{10} < \frac{3}{2 (4)}$

Step 6.2.3

The left side $0.3$ is less than the right side $0.375$ , which means that the given statement is always true.

True

Step 6.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 6.3.2

Replace $x$ with $10$ in the original inequality.

$\frac{4}{5 (10)} + \frac{1}{10} < \frac{3}{2 (10)}$

Step 6.3.3

The left side $0.18$ is not less than the right side $0.15$ , which means that the given statement is false.

False

Step 6.4

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

$x < 0$ False

$0 < x < 7$ True

$x > 7$ False

$x < 0$ False

$0 < x < 7$ True

$x > 7$ False

Step 7

The solution consists of all of the true intervals.

$0 < x < 7$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$0 < x < 7$

Interval Notation:

$(0, 7)$

Step 9