Algebra Examples

$\frac{x - 1}{x + 2} > \frac{x - 5}{x - 3}$

Step 1

Subtract $\frac{x - 5}{x - 3}$ from both sides of the inequality.

$\frac{x - 1}{x + 2} - \frac{x - 5}{x - 3} > 0$

Step 2

Simplify $\frac{x - 1}{x + 2} - \frac{x - 5}{x - 3}$ .

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Step 2.1

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

$\frac{x - 1}{x + 2} \cdot \frac{x - 3}{x - 3} - \frac{x - 5}{x - 3} > 0$

Step 2.2

To write $- \frac{x - 5}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

$\frac{x - 1}{x + 2} \cdot \frac{x - 3}{x - 3} - \frac{x - 5}{x - 3} \cdot \frac{x + 2}{x + 2} > 0$

Step 2.3

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

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Step 2.3.1

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

$\frac{(x - 1) (x - 3)}{(x + 2) (x - 3)} - \frac{x - 5}{x - 3} \cdot \frac{x + 2}{x + 2} > 0$

Step 2.3.2

Multiply $\frac{x - 5}{x - 3}$ by $\frac{x + 2}{x + 2}$ .

$\frac{(x - 1) (x - 3)}{(x + 2) (x - 3)} - \frac{(x - 5) (x + 2)}{(x - 3) (x + 2)} > 0$

Step 2.3.3

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

$\frac{(x - 1) (x - 3)}{(x + 2) (x - 3)} - \frac{(x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{(x - 1) (x - 3) - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5

Simplify the numerator.

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Step 2.5.1

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

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Step 2.5.1.1

Apply the distributive property.

$\frac{x (x - 3) - 1 (x - 3) - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.1.2

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 3 - 1 (x - 3) - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.1.3

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 3 - 1 x - 1 \cdot - 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

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

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 3 - 1 x - 1 \cdot - 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.2.1.2

Move $- 3$ to the left of $x$ .

$\frac{x^{2} - 3 \cdot x - 1 x - 1 \cdot - 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.2.1.3

Rewrite $- 1 x$ as $- x$ .

$\frac{x^{2} - 3 x - x - 1 \cdot - 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.2.1.4

Multiply $- 1$ by $- 3$ .

$\frac{x^{2} - 3 x - x + 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.2.2

Subtract $x$ from $- 3 x$ .

$\frac{x^{2} - 4 x + 3 - (x - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.3

Apply the distributive property.

$\frac{x^{2} - 4 x + 3 + (- x - - 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.4

Multiply $- 1$ by $- 5$ .

$\frac{x^{2} - 4 x + 3 + (- x + 5) (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.5

Expand $(- x + 5) (x + 2)$ using the FOIL Method.

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Step 2.5.5.1

Apply the distributive property.

$\frac{x^{2} - 4 x + 3 - x (x + 2) + 5 (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.5.2

Apply the distributive property.

$\frac{x^{2} - 4 x + 3 - x \cdot x - x \cdot 2 + 5 (x + 2)}{(x + 2) (x - 3)} > 0$

Step 2.5.5.3

Apply the distributive property.

$\frac{x^{2} - 4 x + 3 - x \cdot x - x \cdot 2 + 5 x + 5 \cdot 2}{(x + 2) (x - 3)} > 0$

Step 2.5.6

Simplify and combine like terms.

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Step 2.5.6.1

Simplify each term.

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Step 2.5.6.1.1

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

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Step 2.5.6.1.1.1

Move $x$ .

$\frac{x^{2} - 4 x + 3 - (x \cdot x) - x \cdot 2 + 5 x + 5 \cdot 2}{(x + 2) (x - 3)} > 0$

Step 2.5.6.1.1.2

Multiply $x$ by $x$ .

$\frac{x^{2} - 4 x + 3 - x^{2} - x \cdot 2 + 5 x + 5 \cdot 2}{(x + 2) (x - 3)} > 0$

Step 2.5.6.1.2

Multiply $2$ by $- 1$ .

$\frac{x^{2} - 4 x + 3 - x^{2} - 2 x + 5 x + 5 \cdot 2}{(x + 2) (x - 3)} > 0$

Step 2.5.6.1.3

Multiply $5$ by $2$ .

$\frac{x^{2} - 4 x + 3 - x^{2} - 2 x + 5 x + 10}{(x + 2) (x - 3)} > 0$

Step 2.5.6.2

Add $- 2 x$ and $5 x$ .

$\frac{x^{2} - 4 x + 3 - x^{2} + 3 x + 10}{(x + 2) (x - 3)} > 0$

Step 2.5.7

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

$\frac{- 4 x + 3 + 0 + 3 x + 10}{(x + 2) (x - 3)} > 0$

Step 2.5.8

Add $- 4 x$ and $0$ .

$\frac{- 4 x + 3 + 3 x + 10}{(x + 2) (x - 3)} > 0$

Step 2.5.9

Add $- 4 x$ and $3 x$ .

$\frac{- x + 3 + 10}{(x + 2) (x - 3)} > 0$

Step 2.5.10

Add $3$ and $10$ .

$\frac{- x + 13}{(x + 2) (x - 3)} > 0$

Step 2.6

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

$\frac{- (x) + 13}{(x + 2) (x - 3)} > 0$

Step 2.7

Rewrite $13$ as $- 1 (- 13)$ .

$\frac{- (x) - 1 (- 13)}{(x + 2) (x - 3)} > 0$

Step 2.8

Factor $- 1$ out of $- (x) - 1 (- 13)$ .

$\frac{- (x - 13)}{(x + 2) (x - 3)} > 0$

Step 2.9

Rewrite $- (x - 13)$ as $- 1 (x - 13)$ .

$\frac{- 1 (x - 13)}{(x + 2) (x - 3)} > 0$

Step 2.10

Move the negative in front of the fraction.

$- \frac{x - 13}{(x + 2) (x - 3)} > 0$

Step 3

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

$x - 13 = 0$

$x + 2 = 0$

$x - 3 = 0$

Step 4

Add $13$ to both sides of the equation.

$x = 13$

Step 5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

Add $3$ to both sides of the equation.

$x = 3$

Step 7

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

$x = 13$

$x = - 2$

$x = 3$

Step 8

Consolidate the solutions.

$x = 13, - 2, 3$

Step 9

Find the domain of $- \frac{x - 13}{(x + 2) (x - 3)}$ .

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Step 9.1

Set the denominator in $\frac{x - 13}{(x + 2) (x - 3)}$ equal to $0$ to find where the expression is undefined.

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

Step 9.2

Solve for $x$ .

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Step 9.2.1

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

$x + 2 = 0$

$x - 3 = 0$

Step 9.2.2

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

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Step 9.2.2.1

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

$x + 2 = 0$

Step 9.2.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 9.2.3

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

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Step 9.2.3.1

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

$x - 3 = 0$

Step 9.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 9.2.4

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

$x = - 2, 3$

Step 9.3

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

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

Step 10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 3$

$3 < x < 13$

$x > 13$

Step 11

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 11.1

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

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Step 11.1.1

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

$x = - 4$

Step 11.1.2

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

$\frac{(- 4) - 1}{(- 4) + 2} > \frac{(- 4) - 5}{(- 4) - 3}$

Step 11.1.3

The left side $2.5$ is greater than the right side $1.\overline{285714}$ , which means that the given statement is always true.

True

Step 11.2

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

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Step 11.2.1

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

$x = 0$

Step 11.2.2

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

$\frac{(0) - 1}{(0) + 2} > \frac{(0) - 5}{(0) - 3}$

Step 11.2.3

The left side $- 0.5$ is not greater than the right side $1.\overline{6}$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $3 < x < 13$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $3 < x < 13$ and see if this value makes the original inequality true.

$x = 8$

Step 11.3.2

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

$\frac{(8) - 1}{(8) + 2} > \frac{(8) - 5}{(8) - 3}$

Step 11.3.3

The left side $0.7$ is greater than the right side $0.6$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 13$ to see if it makes the inequality true.

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Step 11.4.1

Choose a value on the interval $x > 13$ and see if this value makes the original inequality true.

$x = 16$

Step 11.4.2

Replace $x$ with $16$ in the original inequality.

$\frac{(16) - 1}{(16) + 2} > \frac{(16) - 5}{(16) - 3}$

Step 11.4.3

The left side $0.8\overline{3}$ is not greater than the right side $0.\overline{846153}$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 2$ True

$- 2 < x < 3$ False

$3 < x < 13$ True

$x > 13$ False

$x < - 2$ True

$- 2 < x < 3$ False

$3 < x < 13$ True

$x > 13$ False

Step 12

The solution consists of all of the true intervals.

$x < - 2$ or $3 < x < 13$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 2) \cup (3, 13)$

Step 14