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

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

Subtract $3$ from both sides of the inequality.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $- 3$ .

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

Multiply $- 3$ by $1$ .

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

Subtract $6 x$ from $4 x$ .

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

Subtract $3$ from $- 3$ .

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

Factor $2$ out of $- 2 x - 6$ .

Factor $2$ out of $- 2 x$ .

$\frac{2 (- x) - 6}{2 x + 1} \leq 0$

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $2 (- x) + 2 (- 3)$ .

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

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

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Factor $- 1$ out of $- (x) - 1 (3)$ .

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

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

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

Move the negative in front of the fraction.

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

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

$x + 3 = 0$

$2 x + 1 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$x = - 3$

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

Consolidate the solutions.

$x = - 3, - \frac{1}{2}$

Find the domain of $- \frac{2 (x + 3)}{2 x + 1}$ .

Set the denominator in $\frac{2 (x + 3)}{2 x + 1}$ equal to $0$ to find where the expression is undefined.

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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.

$x < - 3$

$- 3 < x < - \frac{1}{2}$

$x > - \frac{1}{2}$

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.

$\frac{4 (- 6) - 3}{2 (- 6) + 1} \leq 3$

The left side $2.\overline{45}$ is less than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < - \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = - 2$

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

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

The left side $3.\overline{6}$ is greater than the right side $3$ , which means that the given statement is false.

False

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.

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

The left side $- 3$ 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

$- 3 < x < - \frac{1}{2}$ False

$x > - \frac{1}{2}$ True

$x < - 3$ True

$- 3 < x < - \frac{1}{2}$ False

$x > - \frac{1}{2}$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $x > - \frac{1}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x > - \frac{1}{2}$

Interval Notation:

$(- \infty, - 3] \cup (- \frac{1}{2}, \infty)$

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

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