Solve the Inequality for x (3x-15)/x<=5-x

$\frac{3 x - 15}{x} \leq 5 - x$

Step 1

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

Subtract $5$ from both sides of the inequality.

$\frac{3 x - 15}{x} - 5 \leq - x$

Add $x$ to both sides of the inequality.

$\frac{3 x - 15}{x} - 5 + x \leq 0$

Step 2

Simplify $\frac{3 x - 15}{x} - 5 + x$ .

Factor $3$ out of $3 x - 15$ .

Factor $3$ out of $3 x$ .

$\frac{3 (x) - 15}{x} - 5 + x \leq 0$

Factor $3$ out of $- 15$ .

$\frac{3 x + 3 \cdot - 5}{x} - 5 + x \leq 0$

Factor $3$ out of $3 x + 3 \cdot - 5$ .

$\frac{3 (x - 5)}{x} - 5 + x \leq 0$

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

$\frac{3 (x - 5)}{x} - 5 \cdot \frac{x}{x} + x \leq 0$

Combine $- 5$ and $\frac{x}{x}$ .

$\frac{3 (x - 5)}{x} + \frac{- 5 x}{x} + x \leq 0$

Combine the numerators over the common denominator.

$\frac{3 (x - 5) - 5 x}{x} + x \leq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 x + 3 \cdot - 5 - 5 x}{x} + x \leq 0$

Multiply $3$ by $- 5$ .

$\frac{3 x - 15 - 5 x}{x} + x \leq 0$

Subtract $5 x$ from $3 x$ .

$\frac{- 2 x - 15}{x} + x \leq 0$

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

$\frac{- 2 x - 15}{x} + \frac{x \cdot x}{x} \leq 0$

Combine the numerators over the common denominator.

$\frac{- 2 x - 15 + x \cdot x}{x} \leq 0$

Simplify the numerator.

Multiply $x$ by $x$ .

$\frac{- 2 x - 15 + x^{2}}{x} \leq 0$

Reorder terms.

$\frac{x^{2} - 2 x - 15}{x} \leq 0$

Factor $x^{2} - 2 x - 15$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$\frac{(x - 5) (x + 3)}{x} \leq 0$

Step 3

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

$x = 0$

$x - 5 = 0$

$x + 3 = 0$

Step 4

Add $5$ to both sides of the equation.

$x = 5$

Step 5

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

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

$x = 0$

$x = 5$

$x = - 3$

Step 7

Consolidate the solutions.

$x = 0, 5, - 3$

Step 8

Find the domain of $\frac{(x - 5) (x + 3)}{x}$ .

Set the denominator in $\frac{(x - 5) (x + 3)}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

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

$(- \infty, 0) \cup (0, \infty)$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$0 < x < 5$

$x > 5$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 6$

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

$\frac{3 (- 6) - 15}{- 6} \leq 5 - (- 6)$

The left side $5.5$ is less than the right side $11$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

$\frac{3 (- 2) - 15}{- 2} \leq 5 - (- 2)$

The left side $10.5$ is greater than the right side $7$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 5$ and see if this value makes the original inequality true.

$x = 2$

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

$\frac{3 (2) - 15}{2} \leq 5 - (2)$

The left side $- 4.5$ is less than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

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

$\frac{3 (8) - 15}{8} \leq 5 - (8)$

The left side $1.125$ is greater than the right side $- 3$ , which means that the given statement is false.

False

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

$x < - 3$ True

$- 3 < x < 0$ False

$0 < x < 5$ True

$x > 5$ False

$x < - 3$ True

$- 3 < x < 0$ False

$0 < x < 5$ True

$x > 5$ False

Step 11

The solution consists of all of the true intervals.

$x \leq - 3$ or $0 < x \leq 5$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or 0 < x \leq 5$

Interval Notation:

$(- \infty, - 3] \cup (0, 5]$

Step 13

Solve the Inequality for x (3x-15)/x<=5-x

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