Finite Math Examples

$\frac{x + 1}{2 - x} < \frac{x}{33 + x}$

Step 1

Subtract $\frac{x}{33 + x}$ from both sides of the inequality.

$\frac{x + 1}{2 - x} - \frac{x}{33 + x} < 0$

Step 2

Simplify $\frac{x + 1}{2 - x} - \frac{x}{33 + x}$ .

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Step 2.1

To write $\frac{x + 1}{2 - x}$ as a fraction with a common denominator, multiply by $\frac{33 + x}{33 + x}$ .

$\frac{x + 1}{2 - x} \cdot \frac{33 + x}{33 + x} - \frac{x}{33 + x} < 0$

Step 2.2

To write $- \frac{x}{33 + x}$ as a fraction with a common denominator, multiply by $\frac{2 - x}{2 - x}$ .

$\frac{x + 1}{2 - x} \cdot \frac{33 + x}{33 + x} - \frac{x}{33 + x} \cdot \frac{2 - x}{2 - x} < 0$

Step 2.3

Write each expression with a common denominator of $(2 - x) (33 + x)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{x + 1}{2 - x}$ by $\frac{33 + x}{33 + x}$ .

$\frac{(x + 1) (33 + x)}{(2 - x) (33 + x)} - \frac{x}{33 + x} \cdot \frac{2 - x}{2 - x} < 0$

Step 2.3.2

Multiply $\frac{x}{33 + x}$ by $\frac{2 - x}{2 - x}$ .

$\frac{(x + 1) (33 + x)}{(2 - x) (33 + x)} - \frac{x (2 - x)}{(33 + x) (2 - x)} < 0$

Step 2.3.3

Reorder the factors of $(33 + x) (2 - x)$ .

$\frac{(x + 1) (33 + x)}{(2 - x) (33 + x)} - \frac{x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{(x + 1) (33 + x) - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5

Simplify the numerator.

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Step 2.5.1

Expand $(x + 1) (33 + x)$ using the FOIL Method.

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Step 2.5.1.1

Apply the distributive property.

$\frac{x (33 + x) + 1 (33 + x) - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.1.2

Apply the distributive property.

$\frac{x \cdot 33 + x \cdot x + 1 (33 + x) - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.1.3

Apply the distributive property.

$\frac{x \cdot 33 + x \cdot x + 1 \cdot 33 + 1 x - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.2

Simplify and combine like terms.

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Step 2.5.2.1

Simplify each term.

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Step 2.5.2.1.1

Move $33$ to the left of $x$ .

$\frac{33 \cdot x + x \cdot x + 1 \cdot 33 + 1 x - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.2.1.2

Multiply $x$ by $x$ .

$\frac{33 x + x^{2} + 1 \cdot 33 + 1 x - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.2.1.3

Multiply $33$ by $1$ .

$\frac{33 x + x^{2} + 33 + 1 x - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.2.1.4

Multiply $x$ by $1$ .

$\frac{33 x + x^{2} + 33 + x - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.2.2

Add $33 x$ and $x$ .

$\frac{34 x + x^{2} + 33 - x (2 - x)}{(2 - x) (33 + x)} < 0$

Step 2.5.3

Apply the distributive property.

$\frac{34 x + x^{2} + 33 - x \cdot 2 - x (- x)}{(2 - x) (33 + x)} < 0$

Step 2.5.4

Multiply $2$ by $- 1$ .

$\frac{34 x + x^{2} + 33 - 2 x - x (- x)}{(2 - x) (33 + x)} < 0$

Step 2.5.5

Rewrite using the commutative property of multiplication.

$\frac{34 x + x^{2} + 33 - 2 x - 1 \cdot - 1 x \cdot x}{(2 - x) (33 + x)} < 0$

Step 2.5.6

Simplify each term.

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Step 2.5.6.1

Multiply $x$ by $x$ by adding the exponents.

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Step 2.5.6.1.1

Move $x$ .

$\frac{34 x + x^{2} + 33 - 2 x - 1 \cdot - 1 (x \cdot x)}{(2 - x) (33 + x)} < 0$

Step 2.5.6.1.2

Multiply $x$ by $x$ .

$\frac{34 x + x^{2} + 33 - 2 x - 1 \cdot - 1 x^{2}}{(2 - x) (33 + x)} < 0$

Step 2.5.6.2

Multiply $- 1$ by $- 1$ .

$\frac{34 x + x^{2} + 33 - 2 x + 1 x^{2}}{(2 - x) (33 + x)} < 0$

Step 2.5.6.3

Multiply $x^{2}$ by $1$ .

$\frac{34 x + x^{2} + 33 - 2 x + x^{2}}{(2 - x) (33 + x)} < 0$

Step 2.5.7

Subtract $2 x$ from $34 x$ .

$\frac{32 x + x^{2} + 33 + x^{2}}{(2 - x) (33 + x)} < 0$

Step 2.5.8

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

$\frac{32 x + 2 x^{2} + 33}{(2 - x) (33 + x)} < 0$

Step 2.5.9

Reorder terms.

$\frac{2 x^{2} + 32 x + 33}{(2 - x) (33 + x)} < 0$

Step 3

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

$2 x^{2} + 32 x + 33 = 0$

$2 - x = 0$

$33 + x = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 2$ , $b = 32$ , and $c = 33$ into the quadratic formula and solve for $x$ .

$\frac{- 32 \pm \sqrt{32^{2} - 4 \cdot (2 \cdot 33)}}{2 \cdot 2}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $32$ to the power of $2$ .

$x = \frac{- 32 \pm \sqrt{1024 - 4 \cdot 2 \cdot 33}}{2 \cdot 2}$

Step 6.1.2

Multiply $- 4 \cdot 2 \cdot 33$ .

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Step 6.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 32 \pm \sqrt{1024 - 8 \cdot 33}}{2 \cdot 2}$

Step 6.1.2.2

Multiply $- 8$ by $33$ .

$x = \frac{- 32 \pm \sqrt{1024 - 264}}{2 \cdot 2}$

Step 6.1.3

Subtract $264$ from $1024$ .

$x = \frac{- 32 \pm \sqrt{760}}{2 \cdot 2}$

Step 6.1.4

Rewrite $760$ as $2^{2} \cdot 190$ .

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Step 6.1.4.1

Factor $4$ out of $760$ .

$x = \frac{- 32 \pm \sqrt{4 (190)}}{2 \cdot 2}$

Step 6.1.4.2

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

$x = \frac{- 32 \pm \sqrt{2^{2} \cdot 190}}{2 \cdot 2}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{- 32 \pm 2 \sqrt{190}}{2 \cdot 2}$

Step 6.2

Multiply $2$ by $2$ .

$x = \frac{- 32 \pm 2 \sqrt{190}}{4}$

Step 6.3

Simplify $\frac{- 32 \pm 2 \sqrt{190}}{4}$ .

$x = \frac{- 16 \pm \sqrt{190}}{2}$

Step 7

The final answer is the combination of both solutions.

$x = - \frac{16 - \sqrt{190}}{2}, - \frac{16 + \sqrt{190}}{2}$

Step 8

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 9

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 9.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 9.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 10

Subtract $33$ from both sides of the equation.

$x = - 33$

Step 11

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

$x = - \frac{16 - \sqrt{190}}{2}, - \frac{16 + \sqrt{190}}{2}$

$x = 2$

$x = - 33$

Step 12

Consolidate the solutions.

$x = - \frac{16 - \sqrt{190}}{2}, - \frac{16 + \sqrt{190}}{2}, 2, - 33$

Step 13

Find the domain of $\frac{2 x^{2} + 32 x + 33}{(2 - x) (33 + x)}$ .

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Step 13.1

Set the denominator in $\frac{2 x^{2} + 32 x + 33}{(2 - x) (33 + x)}$ equal to $0$ to find where the expression is undefined.

$(2 - x) (33 + x) = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

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

$2 - x = 0$

$33 + x = 0$

Step 13.2.2

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

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Step 13.2.2.1

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

$2 - x = 0$

Step 13.2.2.2

Solve $2 - x = 0$ for $x$ .

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Step 13.2.2.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 13.2.2.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 13.2.2.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 13.2.2.2.2.2

Simplify the left side.

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Step 13.2.2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 13.2.2.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 13.2.2.2.2.3

Simplify the right side.

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Step 13.2.2.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 13.2.3

Set $33 + x$ equal to $0$ and solve for $x$ .

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Step 13.2.3.1

Set $33 + x$ equal to $0$ .

$33 + x = 0$

Step 13.2.3.2

Subtract $33$ from both sides of the equation.

$x = - 33$

Step 13.2.4

The final solution is all the values that make $(2 - x) (33 + x) = 0$ true.

$x = 2, - 33$

Step 13.3

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

$(- \infty, - 33) \cup (- 33, 2) \cup (2, \infty)$

Step 14

Use each root to create test intervals.

$x < - 33$

$- 33 < x < - \frac{16 + \sqrt{190}}{2}$

$- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$

$- \frac{16 - \sqrt{190}}{2} < x < 2$

$x > 2$

Step 15

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 15.1

Test a value on the interval $x < - 33$ to see if it makes the inequality true.

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Step 15.1.1

Choose a value on the interval $x < - 33$ and see if this value makes the original inequality true.

$x = - 36$

Step 15.1.2

Replace $x$ with $- 36$ in the original inequality.

$\frac{(- 36) + 1}{2 - (- 36)} < \frac{- 36}{33 - 36}$

Step 15.1.3

The left side $- 0.92105263$ is less than the right side $12$ , which means that the given statement is always true.

True

Step 15.2

Test a value on the interval $- 33 < x < - \frac{16 + \sqrt{190}}{2}$ to see if it makes the inequality true.

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Step 15.2.1

Choose a value on the interval $- 33 < x < - \frac{16 + \sqrt{190}}{2}$ and see if this value makes the original inequality true.

$x = - 17$

Step 15.2.2

Replace $x$ with $- 17$ in the original inequality.

$\frac{(- 17) + 1}{2 - (- 17)} < \frac{- 17}{33 - 17}$

Step 15.2.3

The left side $- 0.84210526$ is not less than the right side $- 1.0625$ , which means that the given statement is false.

False

Step 15.3

Test a value on the interval $- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$ to see if it makes the inequality true.

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Step 15.3.1

Choose a value on the interval $- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$ and see if this value makes the original inequality true.

$x = - 4$

Step 15.3.2

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

$\frac{(- 4) + 1}{2 - (- 4)} < \frac{- 4}{33 - 4}$

Step 15.3.3

The left side $- 0.5$ is less than the right side $- 0.13793103$ , which means that the given statement is always true.

True

Step 15.4

Test a value on the interval $- \frac{16 - \sqrt{190}}{2} < x < 2$ to see if it makes the inequality true.

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Step 15.4.1

Choose a value on the interval $- \frac{16 - \sqrt{190}}{2} < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 15.4.2

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

$\frac{(0) + 1}{2 - (0)} < \frac{0}{33 + 0}$

Step 15.4.3

The left side $0.5$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 15.5

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

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Step 15.5.1

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

$x = 4$

Step 15.5.2

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

$\frac{(4) + 1}{2 - (4)} < \frac{4}{33 + 4}$

Step 15.5.3

The left side $- 2.5$ is less than the right side $0.\overline{108}$ , which means that the given statement is always true.

True

Step 15.6

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

$x < - 33$ True

$- 33 < x < - \frac{16 + \sqrt{190}}{2}$ False

$- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$ True

$- \frac{16 - \sqrt{190}}{2} < x < 2$ False

$x > 2$ True

$x < - 33$ True

$- 33 < x < - \frac{16 + \sqrt{190}}{2}$ False

$- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$ True

$- \frac{16 - \sqrt{190}}{2} < x < 2$ False

$x > 2$ True

Step 16

The solution consists of all of the true intervals.

$x < - 33$ or $- \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2}$ or $x > 2$

Step 17

The result can be shown in multiple forms.

Inequality Form:

$x < - 33 or - \frac{16 + \sqrt{190}}{2} < x < - \frac{16 - \sqrt{190}}{2} or x > 2$

Interval Notation:

$(- \infty, - 33) \cup (- \frac{16 + \sqrt{190}}{2}, - \frac{16 - \sqrt{190}}{2}) \cup (2, \infty)$

Step 18