Precalculus Examples

$\frac{x + 2}{x + 3} < \frac{x - 1}{x - 2}$

Step 1

Subtract $\frac{x - 1}{x - 2}$ from both sides of the inequality.

$\frac{x + 2}{x + 3} - \frac{x - 1}{x - 2} < 0$

Step 2

Simplify $\frac{x + 2}{x + 3} - \frac{x - 1}{x - 2}$ .

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Step 2.1

To write $\frac{x + 2}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{x + 2}{x + 3} \cdot \frac{x - 2}{x - 2} - \frac{x - 1}{x - 2} < 0$

Step 2.2

To write $- \frac{x - 1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{x + 2}{x + 3} \cdot \frac{x - 2}{x - 2} - \frac{x - 1}{x - 2} \cdot \frac{x + 3}{x + 3} < 0$

Step 2.3

Write each expression with a common denominator of $(x + 3) (x - 2)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{x + 2}{x + 3}$ by $\frac{x - 2}{x - 2}$ .

$\frac{(x + 2) (x - 2)}{(x + 3) (x - 2)} - \frac{x - 1}{x - 2} \cdot \frac{x + 3}{x + 3} < 0$

Step 2.3.2

Multiply $\frac{x - 1}{x - 2}$ by $\frac{x + 3}{x + 3}$ .

$\frac{(x + 2) (x - 2)}{(x + 3) (x - 2)} - \frac{(x - 1) (x + 3)}{(x - 2) (x + 3)} < 0$

Step 2.3.3

Reorder the factors of $(x - 2) (x + 3)$ .

$\frac{(x + 2) (x - 2)}{(x + 3) (x - 2)} - \frac{(x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{(x + 2) (x - 2) - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5

Simplify the numerator.

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Step 2.5.1

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

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Step 2.5.1.1

Apply the distributive property.

$\frac{x (x - 2) + 2 (x - 2) - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.1.2

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 2 + 2 (x - 2) - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.1.3

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.2

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 2.5.2.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$\frac{x \cdot x - 2 x + 2 x + 2 \cdot - 2 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.2.2

Add $- 2 x$ and $2 x$ .

$\frac{x \cdot x + 0 + 2 \cdot - 2 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.2.3

Add $x \cdot x$ and $0$ .

$\frac{x \cdot x + 2 \cdot - 2 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.3

Simplify each term.

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Step 2.5.3.1

Multiply $x$ by $x$ .

$\frac{x^{2} + 2 \cdot - 2 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.3.2

Multiply $2$ by $- 2$ .

$\frac{x^{2} - 4 - (x - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.4

Apply the distributive property.

$\frac{x^{2} - 4 + (- x - - 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.5

Multiply $- 1$ by $- 1$ .

$\frac{x^{2} - 4 + (- x + 1) (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.6

Expand $(- x + 1) (x + 3)$ using the FOIL Method.

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Step 2.5.6.1

Apply the distributive property.

$\frac{x^{2} - 4 - x (x + 3) + 1 (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.6.2

Apply the distributive property.

$\frac{x^{2} - 4 - x \cdot x - x \cdot 3 + 1 (x + 3)}{(x + 3) (x - 2)} < 0$

Step 2.5.6.3

Apply the distributive property.

$\frac{x^{2} - 4 - x \cdot x - x \cdot 3 + 1 x + 1 \cdot 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7

Simplify and combine like terms.

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Step 2.5.7.1

Simplify each term.

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Step 2.5.7.1.1

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

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Step 2.5.7.1.1.1

Move $x$ .

$\frac{x^{2} - 4 - (x \cdot x) - x \cdot 3 + 1 x + 1 \cdot 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7.1.1.2

Multiply $x$ by $x$ .

$\frac{x^{2} - 4 - x^{2} - x \cdot 3 + 1 x + 1 \cdot 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7.1.2

Multiply $3$ by $- 1$ .

$\frac{x^{2} - 4 - x^{2} - 3 x + 1 x + 1 \cdot 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7.1.3

Multiply $x$ by $1$ .

$\frac{x^{2} - 4 - x^{2} - 3 x + x + 1 \cdot 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7.1.4

Multiply $3$ by $1$ .

$\frac{x^{2} - 4 - x^{2} - 3 x + x + 3}{(x + 3) (x - 2)} < 0$

Step 2.5.7.2

Add $- 3 x$ and $x$ .

$\frac{x^{2} - 4 - x^{2} - 2 x + 3}{(x + 3) (x - 2)} < 0$

Step 2.5.8

Subtract $x^{2}$ from $x^{2}$ .

$\frac{0 - 4 - 2 x + 3}{(x + 3) (x - 2)} < 0$

Step 2.5.9

Subtract $4$ from $0$ .

$\frac{- 4 - 2 x + 3}{(x + 3) (x - 2)} < 0$

Step 2.5.10

Add $- 4$ and $3$ .

$\frac{- 2 x - 1}{(x + 3) (x - 2)} < 0$

Step 2.6

Factor $- 1$ out of $- 2 x$ .

$\frac{- (2 x) - 1}{(x + 3) (x - 2)} < 0$

Step 2.7

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{- (2 x) - 1 (1)}{(x + 3) (x - 2)} < 0$

Step 2.8

Factor $- 1$ out of $- (2 x) - 1 (1)$ .

$\frac{- (2 x + 1)}{(x + 3) (x - 2)} < 0$

Step 2.9

Rewrite $- (2 x + 1)$ as $- 1 (2 x + 1)$ .

$\frac{- 1 (2 x + 1)}{(x + 3) (x - 2)} < 0$

Step 2.10

Move the negative in front of the fraction.

$- \frac{2 x + 1}{(x + 3) (x - 2)} < 0$

Step 3

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

$2 x + 1 = 0$

$x + 3 = 0$

$x - 2 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $2$ .

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Step 5.2.1.1

Cancel the common factor.

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

Step 5.2.1.2

Divide $x$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Move the negative in front of the fraction.

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

Step 6

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

Add $2$ to both sides of the equation.

$x = 2$

Step 8

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

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

$x = - 3$

$x = 2$

Step 9

Consolidate the solutions.

$x = - \frac{1}{2}, - 3, 2$

Step 10

Find the domain of $- \frac{2 x + 1}{(x + 3) (x - 2)}$ .

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Step 10.1

Set the denominator in $\frac{2 x + 1}{(x + 3) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x + 3) (x - 2) = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

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

$x + 3 = 0$

$x - 2 = 0$

Step 10.2.2

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

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Step 10.2.2.1

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

$x + 3 = 0$

Step 10.2.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 10.2.3

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

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Step 10.2.3.1

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

$x - 2 = 0$

Step 10.2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 10.2.4

The final solution is all the values that make $(x + 3) (x - 2) = 0$ true.

$x = - 3, 2$

Step 10.3

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

$(- \infty, - 3) \cup (- 3, 2) \cup (2, \infty)$

Step 11

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - \frac{1}{2}$

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

$x > 2$

Step 12

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 12.1

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

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Step 12.1.1

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

$x = - 6$

Step 12.1.2

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

$\frac{(- 6) + 2}{(- 6) + 3} < \frac{(- 6) - 1}{(- 6) - 2}$

Step 12.1.3

The left side $1.\overline{3}$ is not less than the right side $0.875$ , which means that the given statement is false.

False

Step 12.2

Test a value on the interval $- 3 < x < - \frac{1}{2}$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $- 3 < x < - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = - 2$

Step 12.2.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{(- 2) + 2}{(- 2) + 3} < \frac{(- 2) - 1}{(- 2) - 2}$

Step 12.2.3

The left side $0$ is less than the right side $0.75$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $- \frac{1}{2} < x < 2$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $- \frac{1}{2} < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 12.3.2

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

$\frac{(0) + 2}{(0) + 3} < \frac{(0) - 1}{(0) - 2}$

Step 12.3.3

The left side $0.\overline{6}$ is not less than the right side $0.5$ , which means that the given statement is false.

False

Step 12.4

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

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Step 12.4.1

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

$x = 4$

Step 12.4.2

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

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

Step 12.4.3

The left side $0.\overline{857142}$ is less than the right side $1.5$ , which means that the given statement is always true.

True

Step 12.5

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

$x < - 3$ False

$- 3 < x < - \frac{1}{2}$ True

$- \frac{1}{2} < x < 2$ False

$x > 2$ True

$x < - 3$ False

$- 3 < x < - \frac{1}{2}$ True

$- \frac{1}{2} < x < 2$ False

$x > 2$ True

Step 13

The solution consists of all of the true intervals.

$- 3 < x < - \frac{1}{2}$ or $x > 2$

Step 14

Convert the inequality to interval notation.

$(- 3, - \frac{1}{2}) \cup (2, \infty)$

Step 15