Solve the Inequality for x 3/(2x-1)>-5/x

$\frac{3}{2 x - 1} > \frac{- 5}{x}$

Subtract $\frac{- 5}{x}$ from both sides of the inequality.

$\frac{3}{2 x - 1} - \frac{- 5}{x} > 0$

Simplify $\frac{3}{2 x - 1} - \frac{- 5}{x}$ .

Simplify each term.

Move the negative in front of the fraction.

$\frac{3}{2 x - 1} - - \frac{5}{x} > 0$

Multiply $- - \frac{5}{x}$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{5}{x}$ by $1$ .

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

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

$\frac{3}{2 x - 1} \cdot \frac{x}{x} + \frac{5}{x} > 0$

To write $\frac{5}{x}$ as a fraction with a common denominator, multiply by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{3}{2 x - 1} \cdot \frac{x}{x} + \frac{5}{x} \cdot \frac{2 x - 1}{2 x - 1} > 0$

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

Multiply $\frac{3}{2 x - 1}$ by $\frac{x}{x}$ .

$\frac{3 x}{(2 x - 1) x} + \frac{5}{x} \cdot \frac{2 x - 1}{2 x - 1} > 0$

Multiply $\frac{5}{x}$ by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{3 x}{(2 x - 1) x} + \frac{5 (2 x - 1)}{x (2 x - 1)} > 0$

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

$\frac{3 x}{x (2 x - 1)} + \frac{5 (2 x - 1)}{x (2 x - 1)} > 0$

Combine the numerators over the common denominator.

$\frac{3 x + 5 (2 x - 1)}{x (2 x - 1)} > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 x + 5 (2 x) + 5 \cdot - 1}{x (2 x - 1)} > 0$

Multiply $2$ by $5$ .

$\frac{3 x + 10 x + 5 \cdot - 1}{x (2 x - 1)} > 0$

Multiply $5$ by $- 1$ .

$\frac{3 x + 10 x - 5}{x (2 x - 1)} > 0$

Add $3 x$ and $10 x$ .

$\frac{13 x - 5}{x (2 x - 1)} > 0$

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

$x = 0$

$13 x - 5 = 0$

$2 x - 1 = 0$

Add $5$ to both sides of the equation.

$13 x = 5$

Divide each term in $13 x = 5$ by $13$ and simplify.

Divide each term in $13 x = 5$ by $13$ .

$\frac{13 x}{13} = \frac{5}{13}$

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 x}{13} = \frac{5}{13}$

Divide $x$ by $1$ .

$x = \frac{5}{13}$

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.

$x = 0$

$x = \frac{5}{13}$

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

Consolidate the solutions.

$x = 0, \frac{5}{13}, \frac{1}{2}$

Find the domain of $\frac{13 x - 5}{x (2 x - 1)}$ .

Set the denominator in $\frac{13 x - 5}{x (2 x - 1)}$ equal to $0$ to find where the expression is undefined.

$x (2 x - 1) = 0$

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 $0$ .

$x = 0$

$2 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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 $1$ .

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

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

$x = 0, \frac{1}{2}$

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

$(- \infty, 0) \cup (0, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{5}{13}$

$\frac{5}{13} < x < \frac{1}{2}$

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

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 $x < 0$ to see if it makes the inequality true.

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

$x = - 2$

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

$\frac{3}{2 (- 2) - 1} > \frac{- 5}{- 2}$

The left side $- 0.6$ is not greater than the right side $2.5$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < \frac{5}{13}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{5}{13}$ and see if this value makes the original inequality true.

$x = 0.19$

Replace $x$ with $0.19$ in the original inequality.

$\frac{3}{2 (0.19) - 1} > \frac{- 5}{0.19}$

The left side $- 4.83870967$ is greater than the right side $- 26.31578947$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{5}{13} < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{5}{13} < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0.44$

Replace $x$ with $0.44$ in the original inequality.

$\frac{3}{2 (0.44) - 1} > \frac{- 5}{0.44}$

The left side $- 25$ is not greater than the right side $- 11.\overline{36}$ , 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.

$\frac{3}{2 (3) - 1} > \frac{- 5}{3}$

The left side $0.6$ is greater than the right side $- 1.\overline{6}$ , which means that the given statement is always true.

True

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

$x < 0$ False

$0 < x < \frac{5}{13}$ True

$\frac{5}{13} < x < \frac{1}{2}$ False

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

$x < 0$ False

$0 < x < \frac{5}{13}$ True

$\frac{5}{13} < x < \frac{1}{2}$ False

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

The solution consists of all of the true intervals.

$0 < x < \frac{5}{13}$ or $x > \frac{1}{2}$

The result can be shown in multiple forms.

Inequality Form:

$0 < x < \frac{5}{13} or x > \frac{1}{2}$

Interval Notation:

$(0, \frac{5}{13}) \cup (\frac{1}{2}, \infty)$

Solve the Inequality for x 3/(2x-1)>-5/x

Related Questions

Support LCMGCF.com by disabling your adblocker.