Pre-Algebra Examples

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

Step 1

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

$x - 3 = 0$

$x - 1 = 0$

$x + 2 = 0$

Step 2

Add $3$ to both sides of the equation.

$x = 3$

Step 3

Add $1$ to both sides of the equation.

$x = 1$

Step 4

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

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

$x = 3$

$x = 1$

$x = - 2$

Step 6

Consolidate the solutions.

$x = 3, 1, - 2$

Step 7

Find the domain of $\frac{x - 3}{(x - 1) (x + 2)}$ .

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Step 7.1

Set the denominator in $\frac{x - 3}{(x - 1) (x + 2)}$ equal to $0$ to find where the expression is undefined.

$(x - 1) (x + 2) = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

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

$x - 1 = 0$

$x + 2 = 0$

Step 7.2.2

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

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Step 7.2.2.1

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

$x - 1 = 0$

Step 7.2.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 7.2.3

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

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Step 7.2.3.1

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

$x + 2 = 0$

Step 7.2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7.2.4

The final solution is all the values that make $(x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 7.3

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

$(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$

Step 8

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < 3$

$x > 3$

Step 9

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 9.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 9.1.2

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

$\frac{(- 4) - 3}{((- 4) - 1) ((- 4) + 2)} < 0$

Step 9.1.3

The left side $- 0.7$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- 2 < x < 1$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 2 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

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

Step 9.2.3

The left side $1.5$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 9.3.2

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

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

Step 9.3.3

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.4

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

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Step 9.4.1

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

$x = 6$

Step 9.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{(6) - 3}{((6) - 1) ((6) + 2)} < 0$

Step 9.4.3

The left side $0.075$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.5

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

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

Step 10

The solution consists of all of the true intervals.

$x < - 2$ or $1 < x < 3$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or 1 < x < 3$

Interval Notation:

$(- \infty, - 2) \cup (1, 3)$

Step 12