Algebra Examples

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

Step 1

Solve $- \frac{1}{2} < x < \frac{1}{2} or x > 3$ .

Tap for more steps...

Step 1.1

Move all the expressions to the left side of the equation.

Tap for more steps...

Step 1.1.1

Subtract $x$ from both sides of the inequality.

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

Step 1.1.2

Add $2$ to both sides of the inequality.

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

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Combine the numerators over the common denominator.

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

Step 1.2.4

Simplify the numerator.

Tap for more steps...

Step 1.2.4.1

Apply the distributive property.

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

Step 1.2.4.2

Rewrite using the commutative property of multiplication.

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

Step 1.2.4.3

Multiply $- x \cdot - 1$ .

Tap for more steps...

Step 1.2.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 1.2.4.3.2

Multiply $x$ by $1$ .

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

Step 1.2.4.4

Simplify each term.

Tap for more steps...

Step 1.2.4.4.1

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.2.4.4.1.1

Move $x$ .

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

Step 1.2.4.4.1.2

Multiply $x$ by $x$ .

$\frac{5 - 1 \cdot 2 x^{2} + x}{2 x - 1} + 2 < 0$

Step 1.2.4.4.2

Multiply $- 1$ by $2$ .

$\frac{5 - 2 x^{2} + x}{2 x - 1} + 2 < 0$

Step 1.2.4.5

Reorder terms.

$\frac{- 2 x^{2} + x + 5}{2 x - 1} + 2 < 0$

Step 1.2.5

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

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

Step 1.2.6

Combine the numerators over the common denominator.

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

Step 1.2.7

Simplify the numerator.

Tap for more steps...

Step 1.2.7.1

Apply the distributive property.

$\frac{- 2 x^{2} + x + 5 + 2 (2 x) + 2 \cdot - 1}{2 x - 1} < 0$

Step 1.2.7.2

Multiply $2$ by $2$ .

$\frac{- 2 x^{2} + x + 5 + 4 x + 2 \cdot - 1}{2 x - 1} < 0$

Step 1.2.7.3

Multiply $2$ by $- 1$ .

$\frac{- 2 x^{2} + x + 5 + 4 x - 2}{2 x - 1} < 0$

Step 1.2.7.4

Add $x$ and $4 x$ .

$\frac{- 2 x^{2} + 5 x + 5 - 2}{2 x - 1} < 0$

Step 1.2.7.5

Subtract $2$ from $5$ .

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

Step 1.2.7.6

Factor by grouping.

Tap for more steps...

Step 1.2.7.6.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 2 \cdot 3 = - 6$ and whose sum is $b = 5$ .

Tap for more steps...

Step 1.2.7.6.1.1

Factor $5$ out of $5 x$ .

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

Step 1.2.7.6.1.2

Rewrite $5$ as $- 1$ plus $6$

$\frac{- 2 x^{2} + (- 1 + 6) x + 3}{2 x - 1} < 0$

Step 1.2.7.6.1.3

Apply the distributive property.

$\frac{- 2 x^{2} - 1 x + 6 x + 3}{2 x - 1} < 0$

Step 1.2.7.6.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.2.7.6.2.1

Group the first two terms and the last two terms.

$\frac{(- 2 x^{2} - 1 x) + 6 x + 3}{2 x - 1} < 0$

Step 1.2.7.6.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.2.7.6.3

Factor the polynomial by factoring out the greatest common factor, $- 2 x - 1$ .

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

Step 1.2.8

Factor $- 1$ out of $- 2 x$ .

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

Step 1.2.9

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

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

Step 1.2.10

Factor $- 1$ out of $- (2 x) - 1 (1)$ .

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

Step 1.2.11

Rewrite $- (2 x + 1)$ as $- 1 (2 x + 1)$ .

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

Step 1.2.12

Move the negative in front of the fraction.

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

Step 1.3

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

$2 x + 1 = 0$

$x - 3 = 0$

$2 x - 1 = 0$

Step 1.4

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 1.5

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

Tap for more steps...

Step 1.5.1

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

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

Step 1.5.2

Simplify the left side.

Tap for more steps...

Step 1.5.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.5.2.1.1

Cancel the common factor.

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

Step 1.5.2.1.2

Divide $x$ by $1$ .

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

Step 1.5.3

Simplify the right side.

Tap for more steps...

Step 1.5.3.1

Move the negative in front of the fraction.

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

Step 1.6

Add $3$ to both sides of the equation.

$x = 3$

Step 1.7

Add $1$ to both sides of the equation.

$2 x = 1$

Step 1.8

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

Tap for more steps...

Step 1.8.1

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

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

Step 1.8.2

Simplify the left side.

Tap for more steps...

Step 1.8.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.8.2.1.1

Cancel the common factor.

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

Step 1.8.2.1.2

Divide $x$ by $1$ .

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

Step 1.9

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

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

$x = 3$

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

Step 1.10

Consolidate the solutions.

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

Step 1.11

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

Tap for more steps...

Step 1.11.1

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

$2 x - 1 = 0$

Step 1.11.2

Solve for $x$ .

Tap for more steps...

Step 1.11.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 1.11.2.2

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

Tap for more steps...

Step 1.11.2.2.1

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

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

Step 1.11.2.2.2

Simplify the left side.

Tap for more steps...

Step 1.11.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.11.2.2.2.1.1

Cancel the common factor.

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

Step 1.11.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.11.3

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

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

Step 1.12

Use each root to create test intervals.

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

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

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

$x > 3$

Step 1.13

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 1.13.1

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

Tap for more steps...

Step 1.13.1.1

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

$x = - 3$

Step 1.13.1.2

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

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

Step 1.13.1.3

The left side $- 0.\overline{714285}$ is not less than the right side $- 5$ , which means that the given statement is false.

False

Step 1.13.2

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

Tap for more steps...

Step 1.13.2.1

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

$x = 0$

Step 1.13.2.2

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

$\frac{5}{2 (0) - 1} < (0) - 2$

Step 1.13.2.3

The left side $- 5$ is less than the right side $- 2$ , which means that the given statement is always true.

True

Step 1.13.3

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

Tap for more steps...

Step 1.13.3.1

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

$x = 2$

Step 1.13.3.2

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

$\frac{5}{2 (2) - 1} < (2) - 2$

Step 1.13.3.3

The left side $1.\overline{6}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.13.4

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

Tap for more steps...

Step 1.13.4.1

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

$x = 6$

Step 1.13.4.2

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

$\frac{5}{2 (6) - 1} < (6) - 2$

Step 1.13.4.3

The left side $0.\overline{45}$ is less than the right side $4$ , which means that the given statement is always true.

True

Step 1.13.5

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

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

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

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

$x > 3$ True

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

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

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

$x > 3$ True

Step 1.14

The solution consists of all of the true intervals.

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

Step 2

Use the inequality $- \frac{1}{2} < x < \frac{1}{2} or x > 3$ to build the set notation.

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

Step 3