Solve the Inequality for x (2x+1)/(x-3)<=3

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

Subtract $3$ from both sides of the inequality.

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

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

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

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

Combine $- 3$ and $\frac{x - 3}{x - 3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 3$ by $- 3$ .

$\frac{2 x + 1 - 3 x + 9}{x - 3} \leq 0$

Subtract $3 x$ from $2 x$ .

$\frac{- x + 1 + 9}{x - 3} \leq 0$

Add $1$ and $9$ .

$\frac{- x + 10}{x - 3} \leq 0$

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

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

Rewrite $10$ as $- 1 (- 10)$ .

$\frac{- (x) - 1 (- 10)}{x - 3} \leq 0$

Factor $- 1$ out of $- (x) - 1 (- 10)$ .

$\frac{- (x - 10)}{x - 3} \leq 0$

Rewrite $- (x - 10)$ as $- 1 (x - 10)$ .

$\frac{- 1 (x - 10)}{x - 3} \leq 0$

Move the negative in front of the fraction.

$- \frac{x - 10}{x - 3} \leq 0$

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

$x - 10 = 0$

$x - 3 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 10$

$x = 3$

Consolidate the solutions.

$x = 10, 3$

Find the domain of $- \frac{x - 10}{x - 3}$ .

Set the denominator in $\frac{x - 10}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Use each root to create test intervals.

$x < 3$

$3 < x < 10$

$x > 10$

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.

$x = 0$

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

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

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

True

Test a value on the interval $3 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 10$ and see if this value makes the original inequality true.

$x = 6$

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

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

The left side $4.\overline{3}$ is greater than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{2 (12) + 1}{(12) - 3} \leq 3$

The left side $2.\overline{7}$ is less than the right side $3$ , which means that the given statement is always true.

True

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

$x < 3$ True

$3 < x < 10$ False

$x > 10$ True

$x < 3$ True

$3 < x < 10$ False

$x > 10$ True

The solution consists of all of the true intervals.

$x < 3$ or $x \geq 10$

The result can be shown in multiple forms.

Inequality Form:

$x < 3 or x \geq 10$

Interval Notation:

$(- \infty, 3) \cup [10, \infty)$

Solve the Inequality for x (2x+1)/(x-3)<=3

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