Algebra Examples

$\frac{(x + 5) (x - 4)}{(x - 3) (x + 7)} \leq 0$

Step 1

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

$x + 5 = 0$

$x - 4 = 0$

$x - 3 = 0$

$x + 7 = 0$

Step 2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3

Add $4$ to both sides of the equation.

$x = 4$

Step 4

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6

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

$x = - 5$

$x = 4$

$x = 3$

$x = - 7$

Step 7

Consolidate the solutions.

$x = - 5, 4, 3, - 7$

Step 8

Find the domain of $\frac{(x + 5) (x - 4)}{(x - 3) (x + 7)}$ .

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Step 8.1

Set the denominator in $\frac{(x + 5) (x - 4)}{(x - 3) (x + 7)}$ equal to $0$ to find where the expression is undefined.

$(x - 3) (x + 7) = 0$

Step 8.2

Solve for $x$ .

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Step 8.2.1

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

$x - 3 = 0$

$x + 7 = 0$

Step 8.2.2

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

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Step 8.2.2.1

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

$x - 3 = 0$

Step 8.2.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.2.3

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 8.2.3.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 8.2.3.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8.2.4

The final solution is all the values that make $(x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Step 8.3

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

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

Step 9

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - 5$

$- 5 < x < 3$

$3 < x < 4$

$x > 4$

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 $x < - 7$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = - 10$

Step 10.1.2

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

$\frac{((- 10) + 5) ((- 10) - 4)}{((- 10) - 3) ((- 10) + 7)} \leq 0$

Step 10.1.3

The left side $1.\overline{794871}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 7 < x < - 5$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 7 < x < - 5$ and see if this value makes the original inequality true.

$x = - 6$

Step 10.2.2

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

$\frac{((- 6) + 5) ((- 6) - 4)}{((- 6) - 3) ((- 6) + 7)} \leq 0$

Step 10.2.3

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

True

Step 10.3

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 10.3.2

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

$\frac{((0) + 5) ((0) - 4)}{((0) - 3) ((0) + 7)} \leq 0$

Step 10.3.3

The left side $0.\overline{952380}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.4

Test a value on the interval $3 < x < 4$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Step 10.4.2

Replace $x$ with $3.5$ in the original inequality.

$\frac{((3.5) + 5) ((3.5) - 4)}{((3.5) - 3) ((3.5) + 7)} \leq 0$

Step 10.4.3

The left side $- 0.\overline{809523}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.5

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 10.5.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 10.5.2

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

$\frac{((6) + 5) ((6) - 4)}{((6) - 3) ((6) + 7)} \leq 0$

Step 10.5.3

The left side $0.\overline{564102}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.6

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

$x < - 7$ False

$- 7 < x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

$x < - 7$ False

$- 7 < x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

Step 11

The solution consists of all of the true intervals.

$- 7 < x \leq - 5$ or $3 < x \leq 4$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 7 < x \leq - 5 or 3 < x \leq 4$

Interval Notation:

$(- 7, - 5] \cup (3, 4]$

Step 13