Algebra Examples

$\frac{x^{2} + x - 6}{x - 7} \geq 0$

Step 1

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

$x^{2} + x - 6 = 0$

$x - 7 = 0$

Step 2

Factor $x^{2} + x - 6$ using the AC method.

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Step 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.2

Write the factored form using these integers.

$(x - 2) (x + 3) = 0$

Step 3

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

$x - 2 = 0$

$x + 3 = 0$

Step 4

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

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Step 4.1

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

$x - 2 = 0$

Step 4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 7

Add $7$ to both sides of the equation.

$x = 7$

Step 8

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

$x = 2, - 3$

$x = 7$

Step 9

Consolidate the solutions.

$x = 2, - 3, 7$

Step 10

Find the domain of $\frac{x^{2} + x - 6}{x - 7}$ .

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Step 10.1

Set the denominator in $\frac{x^{2} + x - 6}{x - 7}$ equal to $0$ to find where the expression is undefined.

$x - 7 = 0$

Step 10.2

Add $7$ to both sides of the equation.

$x = 7$

Step 10.3

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

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

Step 11

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$2 < x < 7$

$x > 7$

Step 12

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 12.1

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

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Step 12.1.1

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

$x = - 6$

Step 12.1.2

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

$\frac{{(- 6)}^{2} - 6 - 6}{(- 6) - 7} \geq 0$

Step 12.1.3

The left side $- 1.\overline{846153}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.2

Test a value on the interval $- 3 < x < 2$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $- 3 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 12.2.2

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

$\frac{{(0)}^{2} + 0 - 6}{(0) - 7} \geq 0$

Step 12.2.3

The left side $0.\overline{857142}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $2 < x < 7$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $2 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 12.3.2

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

$\frac{{(4)}^{2} + 4 - 6}{(4) - 7} \geq 0$

Step 12.3.3

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.4

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

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Step 12.4.1

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

$x = 10$

Step 12.4.2

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

$\frac{{(10)}^{2} + 10 - 6}{(10) - 7} \geq 0$

Step 12.4.3

The left side $34.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.5

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

$x < - 3$ False

$- 3 < x < 2$ True

$2 < x < 7$ False

$x > 7$ True

$x < - 3$ False

$- 3 < x < 2$ True

$2 < x < 7$ False

$x > 7$ True

Step 13

The solution consists of all of the true intervals.

$- 3 \leq x \leq 2$ or $x > 7$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x \leq 2 or x > 7$

Interval Notation:

$[- 3, 2] \cup (7, \infty)$

Step 15