Solve the Inequality for x (x+4)/(2x-1)<=3

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

Subtract $3$ from both sides of the inequality.

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

Simplify $\frac{x + 4}{2 x - 1} - 3$ .

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

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

Combine $- 3$ and $\frac{2 x - 1}{2 x - 1}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $- 3$ .

$\frac{x + 4 - 6 x - 3 \cdot - 1}{2 x - 1} \leq 0$

Multiply $- 3$ by $- 1$ .

$\frac{x + 4 - 6 x + 3}{2 x - 1} \leq 0$

Subtract $6 x$ from $x$ .

$\frac{- 5 x + 4 + 3}{2 x - 1} \leq 0$

Add $4$ and $3$ .

$\frac{- 5 x + 7}{2 x - 1} \leq 0$

Factor $- 1$ out of $- 5 x$ .

$\frac{- (5 x) + 7}{2 x - 1} \leq 0$

Rewrite $7$ as $- 1 (- 7)$ .

$\frac{- (5 x) - 1 (- 7)}{2 x - 1} \leq 0$

Factor $- 1$ out of $- (5 x) - 1 (- 7)$ .

$\frac{- (5 x - 7)}{2 x - 1} \leq 0$

Rewrite $- (5 x - 7)$ as $- 1 (5 x - 7)$ .

$\frac{- 1 (5 x - 7)}{2 x - 1} \leq 0$

Move the negative in front of the fraction.

$- \frac{5 x - 7}{2 x - 1} \leq 0$

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

$5 x - 7 = 0$

$2 x - 1 = 0$

Add $7$ to both sides of the equation.

$5 x = 7$

Divide each term in $5 x = 7$ by $5$ and simplify.

Divide each term in $5 x = 7$ by $5$ .

$\frac{5 x}{5} = \frac{7}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{7}{5}$

Divide $x$ by $1$ .

$x = \frac{7}{5}$

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

$x = \frac{7}{5}$

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

Consolidate the solutions.

$x = \frac{7}{5}, \frac{1}{2}$

Find the domain of $- \frac{5 x - 7}{2 x - 1}$ .

Set the denominator in $\frac{5 x - 7}{2 x - 1}$ equal to $0$ to find where the expression is undefined.

$2 x - 1 = 0$

Solve for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Use each root to create test intervals.

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

$\frac{1}{2} < x < \frac{7}{5}$

$x > \frac{7}{5}$

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 $x < \frac{1}{2}$ to see if it makes the inequality true.

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

$x = 0$

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

$\frac{(0) + 4}{2 (0) - 1} \leq 3$

The left side $- 4$ is less than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{1}{2} < x < \frac{7}{5}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{2} < x < \frac{7}{5}$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

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

The left side $5$ is greater than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{7}{5}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{7}{5}$ and see if this value makes the original inequality true.

$x = 4$

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

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

The left side $1.\overline{142857}$ 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.

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

$\frac{1}{2} < x < \frac{7}{5}$ False

$x > \frac{7}{5}$ True

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

$\frac{1}{2} < x < \frac{7}{5}$ False

$x > \frac{7}{5}$ True

The solution consists of all of the true intervals.

$x < \frac{1}{2}$ or $x \geq \frac{7}{5}$

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{1}{2} or x \geq \frac{7}{5}$

Interval Notation:

$(- \infty, \frac{1}{2}) \cup [\frac{7}{5}, \infty)$

Solve the Inequality for x (x+4)/(2x-1)<=3

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