Precalculus Examples

$\frac{3 x + 5}{x^{2} + 2 x - 35} \geq 0$

Step 1

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

$3 x + 5 = 0$

$x^{2} + 2 x - 35 = 0$

Step 2

Subtract $5$ from both sides of the equation.

$3 x = - 5$

Step 3

Divide each term in $3 x = - 5$ by $3$ and simplify.

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Step 3.1

Divide each term in $3 x = - 5$ by $3$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $3$ .

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Step 3.2.1.1

Cancel the common factor.

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

Step 3.2.1.2

Divide $x$ by $1$ .

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

Step 3.3

Simplify the right side.

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Step 3.3.1

Move the negative in front of the fraction.

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

Step 4

Factor $x^{2} + 2 x - 35$ using the AC method.

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Step 4.1

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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Step 4.2

Write the factored form using these integers.

$(x - 5) (x + 7) = 0$

Step 5

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

$x - 5 = 0$

$x + 7 = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

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

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Step 7.1

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

$x + 7 = 0$

Step 7.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

Step 9

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

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

$x = 5, - 7$

Step 10

Consolidate the solutions.

$x = - \frac{5}{3}, 5, - 7$

Step 11

Find the domain of $\frac{3 x + 5}{x^{2} + 2 x - 35}$ .

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Step 11.1

Set the denominator in $\frac{3 x + 5}{x^{2} + 2 x - 35}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 35 = 0$

Step 11.2

Solve for $x$ .

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Step 11.2.1

Factor $x^{2} + 2 x - 35$ using the AC method.

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Step 11.2.1.1

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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Step 11.2.1.2

Write the factored form using these integers.

$(x - 5) (x + 7) = 0$

Step 11.2.2

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

$x - 5 = 0$

$x + 7 = 0$

Step 11.2.3

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 11.2.3.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 11.2.3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 11.2.4

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

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Step 11.2.4.1

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

$x + 7 = 0$

Step 11.2.4.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 11.2.5

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

Step 11.3

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

$(- \infty, - 7) \cup (- 7, 5) \cup (5, \infty)$

Step 12

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - \frac{5}{3}$

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

$x > 5$

Step 13

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 13.1

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

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Step 13.1.1

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

$x = - 10$

Step 13.1.2

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

$\frac{3 (- 10) + 5}{{(- 10)}^{2} + 2 (- 10) - 35} \geq 0$

Step 13.1.3

The left side $- 0.\overline{5}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 13.2

Test a value on the interval $- 7 < x < - \frac{5}{3}$ to see if it makes the inequality true.

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Step 13.2.1

Choose a value on the interval $- 7 < x < - \frac{5}{3}$ and see if this value makes the original inequality true.

$x = - 4$

Step 13.2.2

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

$\frac{3 (- 4) + 5}{{(- 4)}^{2} + 2 (- 4) - 35} \geq 0$

Step 13.2.3

The left side $0.\overline{259}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.3

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

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Step 13.3.1

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

$x = 0$

Step 13.3.2

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

$\frac{3 (0) + 5}{{(0)}^{2} + 2 (0) - 35} \geq 0$

Step 13.3.3

The left side $- 0.\overline{142857}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 13.4

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

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Step 13.4.1

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

$x = 8$

Step 13.4.2

Replace $x$ with $8$ in the original inequality.

$\frac{3 (8) + 5}{{(8)}^{2} + 2 (8) - 35} \geq 0$

Step 13.4.3

The left side $0.6\overline{4}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.5

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

$x < - 7$ False

$- 7 < x < - \frac{5}{3}$ True

$- \frac{5}{3} < x < 5$ False

$x > 5$ True

$x < - 7$ False

$- 7 < x < - \frac{5}{3}$ True

$- \frac{5}{3} < x < 5$ False

$x > 5$ True

Step 14

The solution consists of all of the true intervals.

$- 7 < x \leq - \frac{5}{3}$ or $x > 5$

Step 15

Convert the inequality to interval notation.

$(- 7, - \frac{5}{3}] \cup (5, \infty)$

Step 16