Pre-Algebra Examples

$\frac{x + 1}{x - 4} + \frac{x - 2}{x + 4} < \frac{- 2 x^{2} + x + 32}{x^{2} - 16}$

Step 1

Subtract $\frac{- 2 x^{2} + x + 32}{x^{2} - 16}$ from both sides of the inequality.

$\frac{x + 1}{x - 4} + \frac{x - 2}{x + 4} - \frac{- 2 x^{2} + x + 32}{x^{2} - 16} < 0$

Step 2

Simplify $\frac{x + 1}{x - 4} + \frac{x - 2}{x + 4} - \frac{- 2 x^{2} + x + 32}{x^{2} - 16}$ .

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Step 2.1

Simplify the denominator.

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Step 2.1.1

Rewrite $16$ as $4^{2}$ .

$\frac{x + 1}{x - 4} + \frac{x - 2}{x + 4} - \frac{- 2 x^{2} + x + 32}{x^{2} - 4^{2}} < 0$

Step 2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$\frac{x + 1}{x - 4} + \frac{x - 2}{x + 4} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.2

To write $\frac{x + 1}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

$\frac{x + 1}{x - 4} \cdot \frac{x + 4}{x + 4} + \frac{x - 2}{x + 4} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.3

To write $\frac{x - 2}{x + 4}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{x + 1}{x - 4} \cdot \frac{x + 4}{x + 4} + \frac{x - 2}{x + 4} \cdot \frac{x - 4}{x - 4} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.4

Write each expression with a common denominator of $(x - 4) (x + 4)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{x + 1}{x - 4}$ by $\frac{x + 4}{x + 4}$ .

$\frac{(x + 1) (x + 4)}{(x - 4) (x + 4)} + \frac{x - 2}{x + 4} \cdot \frac{x - 4}{x - 4} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.4.2

Multiply $\frac{x - 2}{x + 4}$ by $\frac{x - 4}{x - 4}$ .

$\frac{(x + 1) (x + 4)}{(x - 4) (x + 4)} + \frac{(x - 2) (x - 4)}{(x + 4) (x - 4)} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.4.3

Reorder the factors of $(x - 4) (x + 4)$ .

$\frac{(x + 1) (x + 4)}{(x + 4) (x - 4)} + \frac{(x - 2) (x - 4)}{(x + 4) (x - 4)} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{(x + 1) (x + 4) + (x - 2) (x - 4)}{(x + 4) (x - 4)} - \frac{- 2 x^{2} + x + 32}{(x + 4) (x - 4)} < 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{(x + 1) (x + 4) + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7

Simplify each term.

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Step 2.7.1

Expand $(x + 1) (x + 4)$ using the FOIL Method.

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Step 2.7.1.1

Apply the distributive property.

$\frac{x (x + 4) + 1 (x + 4) + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.1.2

Apply the distributive property.

$\frac{x \cdot x + x \cdot 4 + 1 (x + 4) + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.1.3

Apply the distributive property.

$\frac{x \cdot x + x \cdot 4 + 1 x + 1 \cdot 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.2

Simplify and combine like terms.

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Step 2.7.2.1

Simplify each term.

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Step 2.7.2.1.1

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 4 + 1 x + 1 \cdot 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.2.1.2

Move $4$ to the left of $x$ .

$\frac{x^{2} + 4 \cdot x + 1 x + 1 \cdot 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.2.1.3

Multiply $x$ by $1$ .

$\frac{x^{2} + 4 x + x + 1 \cdot 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.2.1.4

Multiply $4$ by $1$ .

$\frac{x^{2} + 4 x + x + 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.2.2

Add $4 x$ and $x$ .

$\frac{x^{2} + 5 x + 4 + (x - 2) (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.3

Expand $(x - 2) (x - 4)$ using the FOIL Method.

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Step 2.7.3.1

Apply the distributive property.

$\frac{x^{2} + 5 x + 4 + x (x - 4) - 2 (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.3.2

Apply the distributive property.

$\frac{x^{2} + 5 x + 4 + x \cdot x + x \cdot - 4 - 2 (x - 4) - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.3.3

Apply the distributive property.

$\frac{x^{2} + 5 x + 4 + x \cdot x + x \cdot - 4 - 2 x - 2 \cdot - 4 - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.4

Simplify and combine like terms.

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Step 2.7.4.1

Simplify each term.

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Step 2.7.4.1.1

Multiply $x$ by $x$ .

$\frac{x^{2} + 5 x + 4 + x^{2} + x \cdot - 4 - 2 x - 2 \cdot - 4 - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.4.1.2

Move $- 4$ to the left of $x$ .

$\frac{x^{2} + 5 x + 4 + x^{2} - 4 \cdot x - 2 x - 2 \cdot - 4 - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.4.1.3

Multiply $- 2$ by $- 4$ .

$\frac{x^{2} + 5 x + 4 + x^{2} - 4 x - 2 x + 8 - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.4.2

Subtract $2 x$ from $- 4 x$ .

$\frac{x^{2} + 5 x + 4 + x^{2} - 6 x + 8 - (- 2 x^{2} + x + 32)}{(x + 4) (x - 4)} < 0$

Step 2.7.5

Apply the distributive property.

$\frac{x^{2} + 5 x + 4 + x^{2} - 6 x + 8 - (- 2 x^{2}) - x - 1 \cdot 32}{(x + 4) (x - 4)} < 0$

Step 2.7.6

Simplify.

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Step 2.7.6.1

Multiply $- 2$ by $- 1$ .

$\frac{x^{2} + 5 x + 4 + x^{2} - 6 x + 8 + 2 x^{2} - x - 1 \cdot 32}{(x + 4) (x - 4)} < 0$

Step 2.7.6.2

Multiply $- 1$ by $32$ .

$\frac{x^{2} + 5 x + 4 + x^{2} - 6 x + 8 + 2 x^{2} - x - 32}{(x + 4) (x - 4)} < 0$

Step 2.8

Add $x^{2}$ and $x^{2}$ .

$\frac{2 x^{2} + 5 x + 4 - 6 x + 8 + 2 x^{2} - x - 32}{(x + 4) (x - 4)} < 0$

Step 2.9

Add $2 x^{2}$ and $2 x^{2}$ .

$\frac{4 x^{2} + 5 x + 4 - 6 x + 8 - x - 32}{(x + 4) (x - 4)} < 0$

Step 2.10

Subtract $6 x$ from $5 x$ .

$\frac{4 x^{2} - x + 4 + 8 - x - 32}{(x + 4) (x - 4)} < 0$

Step 2.11

Subtract $x$ from $- x$ .

$\frac{4 x^{2} - 2 x + 4 + 8 - 32}{(x + 4) (x - 4)} < 0$

Step 2.12

Add $4$ and $8$ .

$\frac{4 x^{2} - 2 x + 12 - 32}{(x + 4) (x - 4)} < 0$

Step 2.13

Subtract $32$ from $12$ .

$\frac{4 x^{2} - 2 x - 20}{(x + 4) (x - 4)} < 0$

Step 2.14

Simplify the numerator.

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Step 2.14.1

Factor $2$ out of $4 x^{2} - 2 x - 20$ .

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Step 2.14.1.1

Factor $2$ out of $4 x^{2}$ .

$\frac{2 (2 x^{2}) - 2 x - 20}{(x + 4) (x - 4)} < 0$

Step 2.14.1.2

Factor $2$ out of $- 2 x$ .

$\frac{2 (2 x^{2}) + 2 (- x) - 20}{(x + 4) (x - 4)} < 0$

Step 2.14.1.3

Factor $2$ out of $- 20$ .

$\frac{2 (2 x^{2}) + 2 (- x) + 2 (- 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.1.4

Factor $2$ out of $2 (2 x^{2}) + 2 (- x)$ .

$\frac{2 (2 x^{2} - x) + 2 (- 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.1.5

Factor $2$ out of $2 (2 x^{2} - x) + 2 (- 10)$ .

$\frac{2 (2 x^{2} - x - 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.2

Factor by grouping.

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Step 2.14.2.1

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 $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = - 1$ .

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Step 2.14.2.1.1

Factor $- 1$ out of $- x$ .

$\frac{2 (2 x^{2} - (x) - 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.2.1.2

Rewrite $- 1$ as $4$ plus $- 5$

$\frac{2 (2 x^{2} + (4 - 5) x - 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.2.1.3

Apply the distributive property.

$\frac{2 (2 x^{2} + 4 x - 5 x - 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.2.2

Factor out the greatest common factor from each group.

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Step 2.14.2.2.1

Group the first two terms and the last two terms.

$\frac{2 ((2 x^{2} + 4 x) - 5 x - 10)}{(x + 4) (x - 4)} < 0$

Step 2.14.2.2.2

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

$\frac{2 (2 x (x + 2) - 5 (x + 2))}{(x + 4) (x - 4)} < 0$

Step 2.14.2.3

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$\frac{2 ((x + 2) (2 x - 5))}{(x + 4) (x - 4)} < 0$

$\frac{2 (x + 2) (2 x - 5)}{(x + 4) (x - 4)} < 0$

Step 3

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

$x + 2 = 0$

$2 x - 5 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 4

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Add $5$ to both sides of the equation.

$2 x = 5$

Step 6

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 6.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 7

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

Add $4$ to both sides of the equation.

$x = 4$

Step 9

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

$x = - 2$

$x = \frac{5}{2}$

$x = - 4$

$x = 4$

Step 10

Consolidate the solutions.

$x = - 2, \frac{5}{2}, - 4, 4$

Step 11

Find the domain of $\frac{2 (x + 2) (2 x - 5)}{(x + 4) (x - 4)}$ .

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Step 11.1

Set the denominator in $\frac{2 (x + 2) (2 x - 5)}{(x + 4) (x - 4)}$ equal to $0$ to find where the expression is undefined.

$(x + 4) (x - 4) = 0$

Step 11.2

Solve for $x$ .

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Step 11.2.1

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

$x + 4 = 0$

$x - 4 = 0$

Step 11.2.2

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 11.2.2.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 11.2.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 11.2.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 11.2.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 11.2.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 11.2.4

The final solution is all the values that make $(x + 4) (x - 4) = 0$ true.

$x = - 4, 4$

Step 11.3

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

$(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)$

Step 12

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$- 2 < x < \frac{5}{2}$

$\frac{5}{2} < x < 4$

$x > 4$

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 $x < - 4$ to see if it makes the inequality true.

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Step 13.1.1

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

$x = - 6$

Step 13.1.2

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

$\frac{(- 6) + 1}{(- 6) - 4} + \frac{(- 6) - 2}{(- 6) + 4} < \frac{- 2 {(- 6)}^{2} - 6 + 32}{{(- 6)}^{2} - 16}$

Step 13.1.3

The left side $4.5$ is not less than the right side $- 2.3$ , which means that the given statement is false.

False

Step 13.2

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

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Step 13.2.1

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

Step 13.2.2

Replace $x$ with $- 3$ in the original inequality.

$\frac{(- 3) + 1}{(- 3) - 4} + \frac{(- 3) - 2}{(- 3) + 4} < \frac{- 2 {(- 3)}^{2} - 3 + 32}{{(- 3)}^{2} - 16}$

Step 13.2.3

The left side $- 4.\overline{714285}$ is less than the right side $- 1.\overline{571428}$ , which means that the given statement is always true.

True

Step 13.3

Test a value on the interval $- 2 < x < \frac{5}{2}$ to see if it makes the inequality true.

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Step 13.3.1

Choose a value on the interval $- 2 < x < \frac{5}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 13.3.2

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

$\frac{(0) + 1}{(0) - 4} + \frac{(0) - 2}{(0) + 4} < \frac{- 2 {(0)}^{2} + 0 + 32}{{(0)}^{2} - 16}$

Step 13.3.3

The left side $- 0.75$ is not less than the right side $- 2$ , which means that the given statement is false.

False

Step 13.4

Test a value on the interval $\frac{5}{2} < x < 4$ to see if it makes the inequality true.

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Step 13.4.1

Choose a value on the interval $\frac{5}{2} < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Step 13.4.2

Replace $x$ with $3$ in the original inequality.

$\frac{(3) + 1}{(3) - 4} + \frac{(3) - 2}{(3) + 4} < \frac{- 2 {(3)}^{2} + 3 + 32}{{(3)}^{2} - 16}$

Step 13.4.3

The left side $- 3.\overline{857142}$ is less than the right side $- 2.\overline{428571}$ , which means that the given statement is always true.

True

Step 13.5

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

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Step 13.5.1

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

$x = 6$

Step 13.5.2

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

$\frac{(6) + 1}{(6) - 4} + \frac{(6) - 2}{(6) + 4} < \frac{- 2 {(6)}^{2} + 6 + 32}{{(6)}^{2} - 16}$

Step 13.5.3

The left side $3.9$ is not less than the right side $- 1.7$ , which means that the given statement is false.

False

Step 13.6

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

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < \frac{5}{2}$ False

$\frac{5}{2} < x < 4$ True

$x > 4$ False

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < \frac{5}{2}$ False

$\frac{5}{2} < x < 4$ True

$x > 4$ False

Step 14

The solution consists of all of the true intervals.

$- 4 < x < - 2$ or $\frac{5}{2} < x < 4$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$- 4 < x < - 2 or \frac{5}{2} < x < 4$

Interval Notation:

$(- 4, - 2) \cup (\frac{5}{2}, 4)$

Step 16