Algebra Examples

$\frac{3 x + 4}{x - 4} \leq x - 1$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $x$ from both sides of the inequality.

$\frac{3 x + 4}{x - 4} - x \leq - 1$

Step 1.2

Add $1$ to both sides of the inequality.

$\frac{3 x + 4}{x - 4} - x + 1 \leq 0$

Step 2

Simplify $\frac{3 x + 4}{x - 4} - x + 1$ .

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Step 2.1

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

$\frac{3 x + 4}{x - 4} - x \cdot \frac{x - 4}{x - 4} + 1 \leq 0$

Step 2.2

Combine $- x$ and $\frac{x - 4}{x - 4}$ .

$\frac{3 x + 4}{x - 4} + \frac{- x (x - 4)}{x - 4} + 1 \leq 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{3 x + 4 - x (x - 4)}{x - 4} + 1 \leq 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{3 x + 4 - x \cdot x - x \cdot - 4}{x - 4} + 1 \leq 0$

Step 2.4.2

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

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Step 2.4.2.1

Move $x$ .

$\frac{3 x + 4 - (x \cdot x) - x \cdot - 4}{x - 4} + 1 \leq 0$

Step 2.4.2.2

Multiply $x$ by $x$ .

$\frac{3 x + 4 - x^{2} - x \cdot - 4}{x - 4} + 1 \leq 0$

Step 2.4.3

Multiply $- 4$ by $- 1$ .

$\frac{3 x + 4 - x^{2} + 4 x}{x - 4} + 1 \leq 0$

Step 2.4.4

Add $3 x$ and $4 x$ .

$\frac{7 x + 4 - x^{2}}{x - 4} + 1 \leq 0$

Step 2.4.5

Reorder terms.

$\frac{- x^{2} + 7 x + 4}{x - 4} + 1 \leq 0$

Step 2.5

Write $1$ as a fraction with a common denominator.

$\frac{- x^{2} + 7 x + 4}{x - 4} + \frac{x - 4}{x - 4} \leq 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{- x^{2} + 7 x + 4 + x - 4}{x - 4} \leq 0$

Step 2.7

Simplify the numerator.

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Step 2.7.1

Add $7 x$ and $x$ .

$\frac{- x^{2} + 8 x + 4 - 4}{x - 4} \leq 0$

Step 2.7.2

Subtract $4$ from $4$ .

$\frac{- x^{2} + 8 x + 0}{x - 4} \leq 0$

Step 2.7.3

Add $- x^{2} + 8 x$ and $0$ .

$\frac{- x^{2} + 8 x}{x - 4} \leq 0$

Step 2.7.4

Factor $x$ out of $- x^{2} + 8 x$ .

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Step 2.7.4.1

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

$\frac{x (- x) + 8 x}{x - 4} \leq 0$

Step 2.7.4.2

Factor $x$ out of $8 x$ .

$\frac{x (- x) + x \cdot 8}{x - 4} \leq 0$

Step 2.7.4.3

Factor $x$ out of $x (- x) + x \cdot 8$ .

$\frac{x (- x + 8)}{x - 4} \leq 0$

Step 2.8

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

$\frac{x (- (x) + 8)}{x - 4} \leq 0$

Step 2.9

Rewrite $8$ as $- 1 (- 8)$ .

$\frac{x (- (x) - 1 (- 8))}{x - 4} \leq 0$

Step 2.10

Factor $- 1$ out of $- (x) - 1 (- 8)$ .

$\frac{x (- (x - 8))}{x - 4} \leq 0$

Step 2.11

Rewrite $- (x - 8)$ as $- 1 (x - 8)$ .

$\frac{x (- 1 (x - 8))}{x - 4} \leq 0$

Step 2.12

Move the negative in front of the fraction.

$- \frac{x (x - 8)}{x - 4} \leq 0$

Step 3

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

$- x = 0$

$x - 8 = 0$

$x - 4 = 0$

Step 4

Divide each term in $- x = 0$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- x = 0$ by $- 1$ .

$\frac{- x}{- 1} = \frac{0}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{0}{- 1}$

Step 4.2.2

Divide $x$ by $1$ .

$x = \frac{0}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 1$ .

$x = 0$

Step 5

Add $8$ to both sides of the equation.

$x = 8$

Step 6

Add $4$ to both sides of the equation.

$x = 4$

Step 7

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

$x = 0$

$x = 8$

$x = 4$

Step 8

Consolidate the solutions.

$x = 0, 8, 4$

Step 9

Find the domain of $- \frac{x (x - 8)}{x - 4}$ .

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Step 9.1

Set the denominator in $\frac{x (x - 8)}{x - 4}$ equal to $0$ to find where the expression is undefined.

$x - 4 = 0$

Step 9.2

Add $4$ to both sides of the equation.

$x = 4$

Step 9.3

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

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

Step 10

Use each root to create test intervals.

$x < 0$

$0 < x < 4$

$4 < x < 8$

$x > 8$

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 $x < 0$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 11.1.2

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

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

Step 11.1.3

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

False

Step 11.2

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 11.2.2

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

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

Step 11.2.3

The left side $- 5$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $4 < x < 8$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $4 < x < 8$ and see if this value makes the original inequality true.

$x = 6$

Step 11.3.2

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

$\frac{3 (6) + 4}{(6) - 4} \leq (6) - 1$

Step 11.3.3

The left side $11$ is greater than the right side $5$ , which means that the given statement is false.

False

Step 11.4

Test a value on the interval $x > 8$ to see if it makes the inequality true.

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Step 11.4.1

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Step 11.4.2

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

$\frac{3 (10) + 4}{(10) - 4} \leq (10) - 1$

Step 11.4.3

The left side $5.\overline{6}$ is less than the right side $9$ , which means that the given statement is always true.

True

Step 11.5

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

$x < 0$ False

$0 < x < 4$ True

$4 < x < 8$ False

$x > 8$ True

$x < 0$ False

$0 < x < 4$ True

$4 < x < 8$ False

$x > 8$ True

Step 12

The solution consists of all of the true intervals.

$0 \leq x < 4$ or $x \geq 8$

Step 13

Convert the inequality to interval notation.

$[0, 4) \cup [8, \infty)$

Step 14