Algebra Examples

$x + 1 < \frac{5 x - 3}{x - 3}$

Step 1

Subtract $\frac{5 x - 3}{x - 3}$ from both sides of the inequality.

$x + 1 - \frac{5 x - 3}{x - 3} < 0$

Step 2

Simplify $x + 1 - \frac{5 x - 3}{x - 3}$ .

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Step 2.1

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

$\frac{x (x - 3)}{x - 3} - \frac{5 x - 3}{x - 3} + 1 < 0$

Step 2.2

Combine the numerators over the common denominator.

$\frac{x (x - 3) - (5 x - 3)}{x - 3} + 1 < 0$

Step 2.3

Simplify the numerator.

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Step 2.3.1

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 3 - (5 x - 3)}{x - 3} + 1 < 0$

Step 2.3.2

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 3 - (5 x - 3)}{x - 3} + 1 < 0$

Step 2.3.3

Move $- 3$ to the left of $x$ .

$\frac{x^{2} - 3 \cdot x - (5 x - 3)}{x - 3} + 1 < 0$

Step 2.3.4

Apply the distributive property.

$\frac{x^{2} - 3 x - (5 x) - - 3}{x - 3} + 1 < 0$

Step 2.3.5

Multiply $5$ by $- 1$ .

$\frac{x^{2} - 3 x - 5 x - - 3}{x - 3} + 1 < 0$

Step 2.3.6

Multiply $- 1$ by $- 3$ .

$\frac{x^{2} - 3 x - 5 x + 3}{x - 3} + 1 < 0$

Step 2.3.7

Subtract $5 x$ from $- 3 x$ .

$\frac{x^{2} - 8 x + 3}{x - 3} + 1 < 0$

Step 2.4

Write $1$ as a fraction with a common denominator.

$\frac{x^{2} - 8 x + 3}{x - 3} + \frac{x - 3}{x - 3} < 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{x^{2} - 8 x + 3 + x - 3}{x - 3} < 0$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Add $- 8 x$ and $x$ .

$\frac{x^{2} - 7 x + 3 - 3}{x - 3} < 0$

Step 2.6.2

Subtract $3$ from $3$ .

$\frac{x^{2} - 7 x + 0}{x - 3} < 0$

Step 2.6.3

Add $x^{2} - 7 x$ and $0$ .

$\frac{x^{2} - 7 x}{x - 3} < 0$

Step 2.6.4

Factor $x$ out of $x^{2} - 7 x$ .

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Step 2.6.4.1

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

$\frac{x \cdot x - 7 x}{x - 3} < 0$

Step 2.6.4.2

Factor $x$ out of $- 7 x$ .

$\frac{x \cdot x + x \cdot - 7}{x - 3} < 0$

Step 2.6.4.3

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$\frac{x (x - 7)}{x - 3} < 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

$x - 7 = 0$

$x - 3 = 0$

Step 4

Add $7$ to both sides of the equation.

$x = 7$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 7$

$x = 3$

Step 7

Consolidate the solutions.

$x = 0, 7, 3$

Step 8

Find the domain of $\frac{x (x - 7)}{x - 3}$ .

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Step 8.1

Set the denominator in $\frac{x (x - 7)}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Step 8.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.3

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

$(- \infty, 3) \cup (3, \infty)$

Step 9

Use each root to create test intervals.

$x < 0$

$0 < x < 3$

$3 < x < 7$

$x > 7$

Step 10

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 10.1

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

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Step 10.1.1

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

$x = - 2$

Step 10.1.2

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

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

Step 10.1.3

The left side $- 1$ is less than the right side $2.6$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 10.2.2

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

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

Step 10.2.3

The left side $3$ is not less than the right side $- 7$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $3 < x < 7$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $3 < x < 7$ and see if this value makes the original inequality true.

$x = 5$

Step 10.3.2

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

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

Step 10.3.3

The left side $6$ is less than the right side $11$ , which means that the given statement is always true.

True

Step 10.4

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

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Step 10.4.1

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

$x = 10$

Step 10.4.2

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

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

Step 10.4.3

The left side $11$ is not less than the right side $6.\overline{714285}$ , which means that the given statement is false.

False

Step 10.5

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

$x < 0$ True

$0 < x < 3$ False

$3 < x < 7$ True

$x > 7$ False

$x < 0$ True

$0 < x < 3$ False

$3 < x < 7$ True

$x > 7$ False

Step 11

The solution consists of all of the true intervals.

$x < 0$ or $3 < x < 7$

Step 12

Convert the inequality to interval notation.

$(- \infty, 0) \cup (3, 7)$

Step 13