Algebra Examples

$\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42} \geq 0$

Step 1

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

$x^{2} - 3 x - 18 = 0$

$13 x - x^{2} - 42 = 0$

Step 2

Factor $x^{2} - 3 x - 18$ 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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Step 2.2

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

Step 3

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

$x - 6 = 0$

$x + 3 = 0$

Step 4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.2

Add $6$ to both sides of the equation.

$x = 6$

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 $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Step 7

Factor the left side of the equation.

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Step 7.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$13 u - u^{2} - 42 = 0$

Step 7.2

Factor by grouping.

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Step 7.2.1

Reorder terms.

$- u^{2} + 13 u - 42 = 0$

Step 7.2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 42 = 42$ and whose sum is $b = 13$ .

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Step 7.2.2.1

Factor $13$ out of $13 u$ .

$- u^{2} + 13 (u) - 42 = 0$

Step 7.2.2.2

Rewrite $13$ as $6$ plus $7$

$- u^{2} + (6 + 7) u - 42 = 0$

Step 7.2.2.3

Apply the distributive property.

$- u^{2} + 6 u + 7 u - 42 = 0$

Step 7.2.3

Factor out the greatest common factor from each group.

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Step 7.2.3.1

Group the first two terms and the last two terms.

$(- u^{2} + 6 u) + 7 u - 42 = 0$

Step 7.2.3.2

Factor out the greatest common factor (GCF) from each group.

$u (- u + 6) - 7 (- u + 6) = 0$

Step 7.2.4

Factor the polynomial by factoring out the greatest common factor, $- u + 6$ .

$(- u + 6) (u - 7) = 0$

Step 7.3

Replace all occurrences of $u$ with $x$ .

$(- x + 6) (x - 7) = 0$

Step 8

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

$- x + 6 = 0$

$x - 7 = 0$

Step 9

Set $- x + 6$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Step 9.2

Solve $- x + 6 = 0$ for $x$ .

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Step 9.2.1

Subtract $6$ from both sides of the equation.

$- x = - 6$

Step 9.2.2

Divide each term in $- x = - 6$ by $- 1$ and simplify.

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Step 9.2.2.1

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Step 9.2.2.2

Simplify the left side.

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Step 9.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Step 9.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Step 9.2.2.3

Simplify the right side.

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Step 9.2.2.3.1

Divide $- 6$ by $- 1$ .

$x = 6$

Step 10

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

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Step 10.1

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

$x - 7 = 0$

Step 10.2

Add $7$ to both sides of the equation.

$x = 7$

Step 11

The final solution is all the values that make $(- x + 6) (x - 7) = 0$ true.

$x = 6, 7$

Step 12

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

$x = 6, - 3$

$x = 6, 7$

Step 13

Consolidate the solutions.

$x = 6, - 3, 6, 7$

Step 14

Find the domain of $\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42}$ .

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Step 14.1

Set the denominator in $\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42}$ equal to $0$ to find where the expression is undefined.

$13 x - x^{2} - 42 = 0$

Step 14.2

Solve for $x$ .

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Step 14.2.1

Factor the left side of the equation.

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Step 14.2.1.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$13 u - u^{2} - 42 = 0$

Step 14.2.1.2

Factor by grouping.

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Step 14.2.1.2.1

Reorder terms.

$- u^{2} + 13 u - 42 = 0$

Step 14.2.1.2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 42 = 42$ and whose sum is $b = 13$ .

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Step 14.2.1.2.2.1

Factor $13$ out of $13 u$ .

$- u^{2} + 13 (u) - 42 = 0$

Step 14.2.1.2.2.2

Rewrite $13$ as $6$ plus $7$

$- u^{2} + (6 + 7) u - 42 = 0$

Step 14.2.1.2.2.3

Apply the distributive property.

$- u^{2} + 6 u + 7 u - 42 = 0$

Step 14.2.1.2.3

Factor out the greatest common factor from each group.

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Step 14.2.1.2.3.1

Group the first two terms and the last two terms.

$(- u^{2} + 6 u) + 7 u - 42 = 0$

Step 14.2.1.2.3.2

Factor out the greatest common factor (GCF) from each group.

$u (- u + 6) - 7 (- u + 6) = 0$

Step 14.2.1.2.4

Factor the polynomial by factoring out the greatest common factor, $- u + 6$ .

$(- u + 6) (u - 7) = 0$

Step 14.2.1.3

Replace all occurrences of $u$ with $x$ .

$(- x + 6) (x - 7) = 0$

Step 14.2.2

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

$- x + 6 = 0$

$x - 7 = 0$

Step 14.2.3

Set $- x + 6$ equal to $0$ and solve for $x$ .

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Step 14.2.3.1

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Step 14.2.3.2

Solve $- x + 6 = 0$ for $x$ .

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Step 14.2.3.2.1

Subtract $6$ from both sides of the equation.

$- x = - 6$

Step 14.2.3.2.2

Divide each term in $- x = - 6$ by $- 1$ and simplify.

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Step 14.2.3.2.2.1

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Step 14.2.3.2.2.2

Simplify the left side.

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Step 14.2.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Step 14.2.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Step 14.2.3.2.2.3

Simplify the right side.

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Step 14.2.3.2.2.3.1

Divide $- 6$ by $- 1$ .

$x = 6$

Step 14.2.4

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

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Step 14.2.4.1

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

$x - 7 = 0$

Step 14.2.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 14.2.5

The final solution is all the values that make $(- x + 6) (x - 7) = 0$ true.

$x = 6, 7$

Step 14.3

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

$(- \infty, 6) \cup (6, 7) \cup (7, \infty)$

Step 15

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 6$

$6 < x < 7$

$x > 7$

Step 16

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 16.1

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

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Step 16.1.1

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

$x = - 6$

Step 16.1.2

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

$\frac{{(- 6)}^{2} - 3 \cdot - 6 - 18}{13 (- 6) - {(- 6)}^{2} - 42} \geq 0$

Step 16.1.3

The left side $- 0.\overline{230769}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 16.2

Test a value on the interval $- 3 < x < 6$ to see if it makes the inequality true.

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Step 16.2.1

Choose a value on the interval $- 3 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

Step 16.2.2

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

$\frac{{(0)}^{2} - 3 \cdot 0 - 18}{13 (0) - {(0)}^{2} - 42} \geq 0$

Step 16.2.3

The left side $0.\overline{428571}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.3

Test a value on the interval $6 < x < 7$ to see if it makes the inequality true.

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Step 16.3.1

Choose a value on the interval $6 < x < 7$ and see if this value makes the original inequality true.

$x = 6.5$

Step 16.3.2

Replace $x$ with $6.5$ in the original inequality.

$\frac{{(6.5)}^{2} - 3 \cdot 6.5 - 18}{13 (6.5) - {(6.5)}^{2} - 42} \geq 0$

Step 16.3.3

The left side $19$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.4

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

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Step 16.4.1

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

$x = 10$

Step 16.4.2

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

$\frac{{(10)}^{2} - 3 \cdot 10 - 18}{13 (10) - {(10)}^{2} - 42} \geq 0$

Step 16.4.3

The left side $- 4.\overline{3}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 16.5

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

$x < - 3$ False

$- 3 < x < 6$ True

$6 < x < 7$ True

$x > 7$ False

$x < - 3$ False

$- 3 < x < 6$ True

$6 < x < 7$ True

$x > 7$ False

Step 17

The solution consists of all of the true intervals.

$- 3 \leq x < 6$ or $6 < x < 7$

Step 18

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x < 6 or 6 < x < 7$

Interval Notation:

$[- 3, 6) \cup (6, 7)$

Step 19