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

$\frac{4}{2 x - 1} \geq \frac{1}{x + 1}$

Step 1

Subtract $\frac{1}{x + 1}$ from both sides of the inequality.

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

Step 2

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

To write $\frac{4}{2 x - 1}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

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

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

Write each expression with a common denominator of $(2 x - 1) (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{2 x - 1}$ by $\frac{x + 1}{x + 1}$ .

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

Multiply $\frac{1}{x + 1}$ by $\frac{2 x - 1}{2 x - 1}$ .

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

Reorder the factors of $(x + 1) (2 x - 1)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $4$ by $1$ .

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $2 x$ from $4 x$ .

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

Add $4$ and $1$ .

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

Step 3

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

$2 x + 5 = 0$

$2 x - 1 = 0$

$x + 1 = 0$

Step 4

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 5

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 6

Add $1$ to both sides of the equation.

$2 x = 1$

Step 7

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 $1$ .

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

Step 8

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9

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

$x = - \frac{5}{2}$

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

$x = - 1$

Step 10

Consolidate the solutions.

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

Step 11

Find the domain of $\frac{2 x + 5}{(2 x - 1) (x + 1)}$ .

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

$(2 x - 1) (x + 1) = 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 1 = 0$

$x + 1 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ 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 $1$ .

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

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(2 x - 1) (x + 1) = 0$ true.

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

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

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

Step 12

Use each root to create test intervals.

$x < - \frac{5}{2}$

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

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

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

Step 13

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 $x < - \frac{5}{2}$ to see if it makes the inequality true.

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

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$\frac{4}{2 (- 5) - 1} \geq \frac{1}{(- 5) + 1}$

The left side $- 0.\overline{36}$ is less than the right side $- 0.25$ , which means that the given statement is false.

False

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

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

$x = - 2$

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

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

The left side $- 0.8$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $- 4$ is less than the right side $1$ , 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.

$x = 3$

Replace $x$ with $3$ in the original inequality.

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

The left side $0.8$ is greater than the right side $0.25$ , which means that the given statement is always true.

True

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

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < - 1$ True

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

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

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < - 1$ True

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

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

Step 14

The solution consists of all of the true intervals.

$- \frac{5}{2} \leq x < - 1$ or $x > \frac{1}{2}$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$- \frac{5}{2} \leq x < - 1 or x > \frac{1}{2}$

Interval Notation:

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

Step 16

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

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