Algebra Examples

$\frac{x - | 5 x + 2 |}{2 x + | 5 x + 2 |} > x$

Step 1

Subtract $x$ from both sides of the inequality.

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

Step 2

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

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

Step 3

Simplify terms.

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Step 3.1

Combine $- x$ and $\frac{2 x + | 5 x + 2 |}{2 x + | 5 x + 2 |}$ .

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

Step 3.2

Combine the numerators over the common denominator.

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

Step 4

Simplify the numerator.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Rewrite using the commutative property of multiplication.

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

Step 4.3

Simplify each term.

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Step 4.3.1

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

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Step 4.3.1.1

Move $x$ .

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

Step 4.3.1.2

Multiply $x$ by $x$ .

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

Step 4.3.2

Multiply $- 1$ by $2$ .

$\frac{x - | 5 x + 2 | - 2 x^{2} - x | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 4.4

Reorder terms.

$\frac{- 2 x^{2} - x | 5 x + 2 | + x - | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 5

Simplify with factoring out.

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Step 5.1

Factor $- 1$ out of $- 2 x^{2}$ .

$\frac{- (2 x^{2}) - x | 5 x + 2 | + x - | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 5.2

Factor $- 1$ out of $- x | 5 x + 2 |$ .

$\frac{- (2 x^{2}) - (x | 5 x + 2 |) + x - | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 5.3

Factor $- 1$ out of $- (2 x^{2}) - (x | 5 x + 2 |)$ .

$\frac{- (2 x^{2} + x | 5 x + 2 |) + x - | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 5.4

Factor $- 1$ out of $x$ .

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

Step 5.5

Factor $- 1$ out of $- (2 x^{2} + x | 5 x + 2 |) - 1 (- x)$ .

$\frac{- (2 x^{2} + x | 5 x + 2 | - x) - | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 5.6

Factor $- 1$ out of $- | 5 x + 2 |$ .

$\frac{- (2 x^{2} + x | 5 x + 2 | - x) - (| 5 x + 2 |)}{2 x + | 5 x + 2 |} > 0$

Step 5.7

Factor $- 1$ out of $- (2 x^{2} + x | 5 x + 2 | - x) - (| 5 x + 2 |)$ .

$\frac{- (2 x^{2} + x | 5 x + 2 | - x + | 5 x + 2 |)}{2 x + | 5 x + 2 |} > 0$

Step 5.8

Simplify the expression.

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Step 5.8.1

Rewrite $- (2 x^{2} + x | 5 x + 2 | - x + | 5 x + 2 |)$ as $- 1 (2 x^{2} + x | 5 x + 2 | - x + | 5 x + 2 |)$ .

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

Step 5.8.2

Move the negative in front of the fraction.

$- \frac{2 x^{2} + x | 5 x + 2 | - x + | 5 x + 2 |}{2 x + | 5 x + 2 |} > 0$

Step 6

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

$2 x^{2} + x | 5 x + 2 | - x + | 5 x + 2 | = 0$

$2 x + | 5 x + 2 | = 0$

Step 7

Move all terms not containing $| 5 x + 2 |$ to the right side of the equation.

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Step 7.1

Subtract $2 x^{2}$ from both sides of the equation.

$x | 5 x + 2 | - x + | 5 x + 2 | = - 2 x^{2}$

Step 7.2

Add $x$ to both sides of the equation.

$x | 5 x + 2 | + | 5 x + 2 | = - 2 x^{2} + x$

Step 8

Factor $| 5 x + 2 |$ out of $x | 5 x + 2 | + | 5 x + 2 |$ .

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Step 8.1

Factor $| 5 x + 2 |$ out of $x | 5 x + 2 |$ .

$| 5 x + 2 | x + | 5 x + 2 | = - 2 x^{2} + x$

Step 8.2

Raise $| 5 x + 2 |$ to the power of $1$ .

$| 5 x + 2 | x + | 5 x + 2 | = - 2 x^{2} + x$

Step 8.3

Factor $| 5 x + 2 |$ out of ${| 5 x + 2 |}^{1}$ .

$| 5 x + 2 | x + | 5 x + 2 | \cdot 1 = - 2 x^{2} + x$

Step 8.4

Factor $| 5 x + 2 |$ out of $| 5 x + 2 | x + | 5 x + 2 | \cdot 1$ .

$| 5 x + 2 | (x + 1) = - 2 x^{2} + x$

Step 9

Divide each term in $| 5 x + 2 | (x + 1) = - 2 x^{2} + x$ by $x + 1$ and simplify.

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Step 9.1

Divide each term in $| 5 x + 2 | (x + 1) = - 2 x^{2} + x$ by $x + 1$ .

$\frac{| 5 x + 2 | (x + 1)}{x + 1} = \frac{- 2 x^{2}}{x + 1} + \frac{x}{x + 1}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Cancel the common factor of $x + 1$ .

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Step 9.2.1.1

Cancel the common factor.

$\frac{| 5 x + 2 | (x + 1)}{x + 1} = \frac{- 2 x^{2}}{x + 1} + \frac{x}{x + 1}$

Step 9.2.1.2

Divide $| 5 x + 2 |$ by $1$ .

$| 5 x + 2 | = \frac{- 2 x^{2}}{x + 1} + \frac{x}{x + 1}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Combine the numerators over the common denominator.

$| 5 x + 2 | = \frac{- 2 x^{2} + x}{x + 1}$

Step 9.3.2

Factor $x$ out of $- 2 x^{2} + x$ .

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Step 9.3.2.1

Factor $x$ out of $- 2 x^{2}$ .

$| 5 x + 2 | = \frac{x (- 2 x) + x}{x + 1}$

Step 9.3.2.2

Raise $x$ to the power of $1$ .

$| 5 x + 2 | = \frac{x (- 2 x) + x^{1}}{x + 1}$

Step 9.3.2.3

Factor $x$ out of $x^{1}$ .

$| 5 x + 2 | = \frac{x (- 2 x) + x \cdot 1}{x + 1}$

Step 9.3.2.4

Factor $x$ out of $x (- 2 x) + x \cdot 1$ .

$| 5 x + 2 | = \frac{x (- 2 x + 1)}{x + 1}$

Step 9.3.3

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

$| 5 x + 2 | = \frac{x (- (2 x) + 1)}{x + 1}$

Step 9.3.4

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

$| 5 x + 2 | = \frac{x (- (2 x) - 1 (- 1))}{x + 1}$

Step 9.3.5

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

$| 5 x + 2 | = \frac{x (- (2 x - 1))}{x + 1}$

Step 9.3.6

Simplify the expression.

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Step 9.3.6.1

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

$| 5 x + 2 | = \frac{x (- 1 (2 x - 1))}{x + 1}$

Step 9.3.6.2

Move the negative in front of the fraction.

$| 5 x + 2 | = - \frac{x (2 x - 1)}{x + 1}$

Step 10

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$5 x + 2 = \pm (- \frac{x (2 x - 1)}{x + 1})$

Step 11

The complete solution is the result of both the positive and negative portions of the solution.

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Step 11.1

First, use the positive value of the $\pm$ to find the first solution.

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

Step 11.2

Subtract $2$ from both sides of the equation.

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

Step 11.3

Find the LCD of the terms in the equation.

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Step 11.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x + 1, 1$

Step 11.3.2

Remove parentheses.

$1, x + 1, 1$

Step 11.3.3

The LCM of one and any expression is the expression.

$x + 1$

Step 11.4

Multiply each term in $5 x = - \frac{x (2 x - 1)}{x + 1} - 2$ by $x + 1$ to eliminate the fractions.

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Step 11.4.1

Multiply each term in $5 x = - \frac{x (2 x - 1)}{x + 1} - 2$ by $x + 1$ .

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

Step 11.4.2

Simplify the left side.

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Step 11.4.2.1

Apply the distributive property.

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

Step 11.4.2.2

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

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Step 11.4.2.2.1

Move $x$ .

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

Step 11.4.2.2.2

Multiply $x$ by $x$ .

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

Step 11.4.2.3

Multiply $5$ by $1$ .

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

Step 11.4.3

Simplify the right side.

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Step 11.4.3.1

Simplify each term.

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Step 11.4.3.1.1

Cancel the common factor of $x + 1$ .

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Step 11.4.3.1.1.1

Move the leading negative in $- \frac{x (2 x - 1)}{x + 1}$ into the numerator.

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

Step 11.4.3.1.1.2

Cancel the common factor.

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

Step 11.4.3.1.1.3

Rewrite the expression.

$5 x^{2} + 5 x = - x (2 x - 1) - 2 (x + 1)$

Step 11.4.3.1.2

Apply the distributive property.

$5 x^{2} + 5 x = - x (2 x) - x \cdot - 1 - 2 (x + 1)$

Step 11.4.3.1.3

Rewrite using the commutative property of multiplication.

$5 x^{2} + 5 x = - 1 \cdot 2 x \cdot x - x \cdot - 1 - 2 (x + 1)$

Step 11.4.3.1.4

Multiply $- x \cdot - 1$ .

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Step 11.4.3.1.4.1

Multiply $- 1$ by $- 1$ .

$5 x^{2} + 5 x = - 1 \cdot 2 x \cdot x + 1 x - 2 (x + 1)$

Step 11.4.3.1.4.2

Multiply $x$ by $1$ .

$5 x^{2} + 5 x = - 1 \cdot 2 x \cdot x + x - 2 (x + 1)$

Step 11.4.3.1.5

Simplify each term.

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Step 11.4.3.1.5.1

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

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Step 11.4.3.1.5.1.1

Move $x$ .

$5 x^{2} + 5 x = - 1 \cdot 2 (x \cdot x) + x - 2 (x + 1)$

Step 11.4.3.1.5.1.2

Multiply $x$ by $x$ .

$5 x^{2} + 5 x = - 1 \cdot 2 x^{2} + x - 2 (x + 1)$

Step 11.4.3.1.5.2

Multiply $- 1$ by $2$ .

$5 x^{2} + 5 x = - 2 x^{2} + x - 2 (x + 1)$

Step 11.4.3.1.6

Apply the distributive property.

$5 x^{2} + 5 x = - 2 x^{2} + x - 2 x - 2 \cdot 1$

Step 11.4.3.1.7

Multiply $- 2$ by $1$ .

$5 x^{2} + 5 x = - 2 x^{2} + x - 2 x - 2$

Step 11.4.3.2

Subtract $2 x$ from $x$ .

$5 x^{2} + 5 x = - 2 x^{2} - x - 2$

Step 11.5

Solve the equation.

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Step 11.5.1

Move all terms containing $x$ to the left side of the equation.

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Step 11.5.1.1

Add $2 x^{2}$ to both sides of the equation.

$5 x^{2} + 5 x + 2 x^{2} = - x - 2$

Step 11.5.1.2

Add $x$ to both sides of the equation.

$5 x^{2} + 5 x + 2 x^{2} + x = - 2$

Step 11.5.1.3

Add $5 x^{2}$ and $2 x^{2}$ .

$7 x^{2} + 5 x + x = - 2$

Step 11.5.1.4

Add $5 x$ and $x$ .

$7 x^{2} + 6 x = - 2$

Step 11.5.2

Add $2$ to both sides of the equation.

$7 x^{2} + 6 x + 2 = 0$

Step 11.5.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 11.5.4

Substitute the values $a = 7$ , $b = 6$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (7 \cdot 2)}}{2 \cdot 7}$

Step 11.5.5

Simplify.

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Step 11.5.5.1

Simplify the numerator.

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Step 11.5.5.1.1

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 7 \cdot 2}}{2 \cdot 7}$

Step 11.5.5.1.2

Multiply $- 4 \cdot 7 \cdot 2$ .

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Step 11.5.5.1.2.1

Multiply $- 4$ by $7$ .

$x = \frac{- 6 \pm \sqrt{36 - 28 \cdot 2}}{2 \cdot 7}$

Step 11.5.5.1.2.2

Multiply $- 28$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 56}}{2 \cdot 7}$

Step 11.5.5.1.3

Subtract $56$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 20}}{2 \cdot 7}$

Step 11.5.5.1.4

Rewrite $- 20$ as $- 1 (20)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 20}}{2 \cdot 7}$

Step 11.5.5.1.5

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{20}}{2 \cdot 7}$

Step 11.5.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 6 \pm i \cdot \sqrt{20}}{2 \cdot 7}$

Step 11.5.5.1.7

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 11.5.5.1.7.1

Factor $4$ out of $20$ .

$x = \frac{- 6 \pm i \cdot \sqrt{4 (5)}}{2 \cdot 7}$

Step 11.5.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 5}}{2 \cdot 7}$

Step 11.5.5.1.8

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (2 \sqrt{5})}{2 \cdot 7}$

Step 11.5.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i \sqrt{5}}{2 \cdot 7}$

Step 11.5.5.2

Multiply $2$ by $7$ .

$x = \frac{- 6 \pm 2 i \sqrt{5}}{14}$

Step 11.5.5.3

Simplify $\frac{- 6 \pm 2 i \sqrt{5}}{14}$ .

$x = \frac{- 3 \pm i \sqrt{5}}{7}$

Step 11.5.6

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{5}}{7}, - \frac{3 + i \sqrt{5}}{7}$

Step 11.6

Next, use the negative value of the $\pm$ to find the second solution.

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

Step 11.7

Subtract $2$ from both sides of the equation.

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

Step 11.8

Find the LCD of the terms in the equation.

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Step 11.8.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x + 1, 1$

Step 11.8.2

Remove parentheses.

$1, x + 1, 1$

Step 11.8.3

The LCM of one and any expression is the expression.

$x + 1$

Step 11.9

Multiply each term in $5 x = \frac{x (2 x - 1)}{x + 1} - 2$ by $x + 1$ to eliminate the fractions.

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Step 11.9.1

Multiply each term in $5 x = \frac{x (2 x - 1)}{x + 1} - 2$ by $x + 1$ .

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

Step 11.9.2

Simplify the left side.

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Step 11.9.2.1

Apply the distributive property.

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

Step 11.9.2.2

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

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Step 11.9.2.2.1

Move $x$ .

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

Step 11.9.2.2.2

Multiply $x$ by $x$ .

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

Step 11.9.2.3

Multiply $5$ by $1$ .

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

Step 11.9.3

Simplify the right side.

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Step 11.9.3.1

Simplify each term.

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Step 11.9.3.1.1

Cancel the common factor of $x + 1$ .

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Step 11.9.3.1.1.1

Cancel the common factor.

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

Step 11.9.3.1.1.2

Rewrite the expression.

$5 x^{2} + 5 x = x (2 x - 1) - 2 (x + 1)$

Step 11.9.3.1.2

Apply the distributive property.

$5 x^{2} + 5 x = x (2 x) + x \cdot - 1 - 2 (x + 1)$

Step 11.9.3.1.3

Rewrite using the commutative property of multiplication.

$5 x^{2} + 5 x = 2 x \cdot x + x \cdot - 1 - 2 (x + 1)$

Step 11.9.3.1.4

Move $- 1$ to the left of $x$ .

$5 x^{2} + 5 x = 2 x \cdot x - 1 \cdot x - 2 (x + 1)$

Step 11.9.3.1.5

Simplify each term.

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Step 11.9.3.1.5.1

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

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Step 11.9.3.1.5.1.1

Move $x$ .

$5 x^{2} + 5 x = 2 (x \cdot x) - 1 \cdot x - 2 (x + 1)$

Step 11.9.3.1.5.1.2

Multiply $x$ by $x$ .

$5 x^{2} + 5 x = 2 x^{2} - 1 \cdot x - 2 (x + 1)$

Step 11.9.3.1.5.2

Rewrite $- 1 x$ as $- x$ .

$5 x^{2} + 5 x = 2 x^{2} - x - 2 (x + 1)$

Step 11.9.3.1.6

Apply the distributive property.

$5 x^{2} + 5 x = 2 x^{2} - x - 2 x - 2 \cdot 1$

Step 11.9.3.1.7

Multiply $- 2$ by $1$ .

$5 x^{2} + 5 x = 2 x^{2} - x - 2 x - 2$

Step 11.9.3.2

Subtract $2 x$ from $- x$ .

$5 x^{2} + 5 x = 2 x^{2} - 3 x - 2$

Step 11.10

Solve the equation.

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Step 11.10.1

Move all terms containing $x$ to the left side of the equation.

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Step 11.10.1.1

Subtract $2 x^{2}$ from both sides of the equation.

$5 x^{2} + 5 x - 2 x^{2} = - 3 x - 2$

Step 11.10.1.2

Add $3 x$ to both sides of the equation.

$5 x^{2} + 5 x - 2 x^{2} + 3 x = - 2$

Step 11.10.1.3

Subtract $2 x^{2}$ from $5 x^{2}$ .

$3 x^{2} + 5 x + 3 x = - 2$

Step 11.10.1.4

Add $5 x$ and $3 x$ .

$3 x^{2} + 8 x = - 2$

Step 11.10.2

Add $2$ to both sides of the equation.

$3 x^{2} + 8 x + 2 = 0$

Step 11.10.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 11.10.4

Substitute the values $a = 3$ , $b = 8$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (3 \cdot 2)}}{2 \cdot 3}$

Step 11.10.5

Simplify.

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Step 11.10.5.1

Simplify the numerator.

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Step 11.10.5.1.1

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}$

Step 11.10.5.1.2

Multiply $- 4 \cdot 3 \cdot 2$ .

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Step 11.10.5.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 12 \cdot 2}}{2 \cdot 3}$

Step 11.10.5.1.2.2

Multiply $- 12$ by $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 24}}{2 \cdot 3}$

Step 11.10.5.1.3

Subtract $24$ from $64$ .

$x = \frac{- 8 \pm \sqrt{40}}{2 \cdot 3}$

Step 11.10.5.1.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 11.10.5.1.4.1

Factor $4$ out of $40$ .

$x = \frac{- 8 \pm \sqrt{4 (10)}}{2 \cdot 3}$

Step 11.10.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 8 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 3}$

Step 11.10.5.1.5

Pull terms out from under the radical.

$x = \frac{- 8 \pm 2 \sqrt{10}}{2 \cdot 3}$

Step 11.10.5.2

Multiply $2$ by $3$ .

$x = \frac{- 8 \pm 2 \sqrt{10}}{6}$

Step 11.10.5.3

Simplify $\frac{- 8 \pm 2 \sqrt{10}}{6}$ .

$x = \frac{- 4 \pm \sqrt{10}}{3}$

Step 11.10.6

The final answer is the combination of both solutions.

$x = - \frac{4 - \sqrt{10}}{3}, - \frac{4 + \sqrt{10}}{3}$

Step 11.11

The complete solution is the result of both the positive and negative portions of the solution.

$x = - \frac{3 - i \sqrt{5}}{7}, - \frac{3 + i \sqrt{5}}{7}, - \frac{4 - \sqrt{10}}{3}, - \frac{4 + \sqrt{10}}{3}$

Step 12

Subtract $2 x$ from both sides of the equation.

$| 5 x + 2 | = - 2 x$

Step 13

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$5 x + 2 = \pm (- 2 x)$

Step 14

The complete solution is the result of both the positive and negative portions of the solution.

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Step 14.1

First, use the positive value of the $\pm$ to find the first solution.

$5 x + 2 = - 2 x$

Step 14.2

Move all terms containing $x$ to the left side of the equation.

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Step 14.2.1

Add $2 x$ to both sides of the equation.

$5 x + 2 + 2 x = 0$

Step 14.2.2

Add $5 x$ and $2 x$ .

$7 x + 2 = 0$

Step 14.3

Subtract $2$ from both sides of the equation.

$7 x = - 2$

Step 14.4

Divide each term in $7 x = - 2$ by $7$ and simplify.

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Step 14.4.1

Divide each term in $7 x = - 2$ by $7$ .

$\frac{7 x}{7} = \frac{- 2}{7}$

Step 14.4.2

Simplify the left side.

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Step 14.4.2.1

Cancel the common factor of $7$ .

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Step 14.4.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 2}{7}$

Step 14.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2}{7}$

Step 14.4.3

Simplify the right side.

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Step 14.4.3.1

Move the negative in front of the fraction.

$x = - \frac{2}{7}$

Step 14.5

Next, use the negative value of the $\pm$ to find the second solution.

$5 x + 2 = 2 x$

Step 14.6

Move all terms containing $x$ to the left side of the equation.

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Step 14.6.1

Subtract $2 x$ from both sides of the equation.

$5 x + 2 - 2 x = 0$

Step 14.6.2

Subtract $2 x$ from $5 x$ .

$3 x + 2 = 0$

Step 14.7

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 14.8

Divide each term in $3 x = - 2$ by $3$ and simplify.

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Step 14.8.1

Divide each term in $3 x = - 2$ by $3$ .

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

Step 14.8.2

Simplify the left side.

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Step 14.8.2.1

Cancel the common factor of $3$ .

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Step 14.8.2.1.1

Cancel the common factor.

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

Step 14.8.2.1.2

Divide $x$ by $1$ .

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

Step 14.8.3

Simplify the right side.

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Step 14.8.3.1

Move the negative in front of the fraction.

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

Step 14.9

The complete solution is the result of both the positive and negative portions of the solution.

$x = - \frac{2}{7}, - \frac{2}{3}$

Step 15

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

$- \frac{4 - \sqrt{10}}{3}, - \frac{4 + \sqrt{10}}{3}$

$x = - \frac{2}{7}, - \frac{2}{3}$

Step 16

Consolidate the solutions.

$x = - \frac{4 - \sqrt{10}}{3}, - \frac{4 + \sqrt{10}}{3}, - \frac{2}{7}, - \frac{2}{3}$

Step 17

Find the domain of $\frac{x - | 5 x + 2 |}{2 x + | 5 x + 2 |} - x$ .

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Step 17.1

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

$2 x + | 5 x + 2 | = 0$

Step 17.2

Solve for $x$ .

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Step 17.2.1

Subtract $2 x$ from both sides of the equation.

$| 5 x + 2 | = - 2 x$

Step 17.2.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$5 x + 2 = \pm (- 2 x)$

Step 17.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 17.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$5 x + 2 = - 2 x$

Step 17.2.3.2

Move all terms containing $x$ to the left side of the equation.

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Step 17.2.3.2.1

Add $2 x$ to both sides of the equation.

$5 x + 2 + 2 x = 0$

Step 17.2.3.2.2

Add $5 x$ and $2 x$ .

$7 x + 2 = 0$

Step 17.2.3.3

Subtract $2$ from both sides of the equation.

$7 x = - 2$

Step 17.2.3.4

Divide each term in $7 x = - 2$ by $7$ and simplify.

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Step 17.2.3.4.1

Divide each term in $7 x = - 2$ by $7$ .

$\frac{7 x}{7} = \frac{- 2}{7}$

Step 17.2.3.4.2

Simplify the left side.

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Step 17.2.3.4.2.1

Cancel the common factor of $7$ .

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Step 17.2.3.4.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 2}{7}$

Step 17.2.3.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2}{7}$

Step 17.2.3.4.3

Simplify the right side.

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Step 17.2.3.4.3.1

Move the negative in front of the fraction.

$x = - \frac{2}{7}$

Step 17.2.3.5

Next, use the negative value of the $\pm$ to find the second solution.

$5 x + 2 = 2 x$

Step 17.2.3.6

Move all terms containing $x$ to the left side of the equation.

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Step 17.2.3.6.1

Subtract $2 x$ from both sides of the equation.

$5 x + 2 - 2 x = 0$

Step 17.2.3.6.2

Subtract $2 x$ from $5 x$ .

$3 x + 2 = 0$

Step 17.2.3.7

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 17.2.3.8

Divide each term in $3 x = - 2$ by $3$ and simplify.

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Step 17.2.3.8.1

Divide each term in $3 x = - 2$ by $3$ .

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

Step 17.2.3.8.2

Simplify the left side.

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Step 17.2.3.8.2.1

Cancel the common factor of $3$ .

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Step 17.2.3.8.2.1.1

Cancel the common factor.

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

Step 17.2.3.8.2.1.2

Divide $x$ by $1$ .

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

Step 17.2.3.8.3

Simplify the right side.

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Step 17.2.3.8.3.1

Move the negative in front of the fraction.

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

Step 17.2.3.9

The complete solution is the result of both the positive and negative portions of the solution.

$x = - \frac{2}{7}, - \frac{2}{3}$

Step 17.3

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

$(- \infty, - \frac{2}{3}) \cup (- \frac{2}{3}, - \frac{2}{7}) \cup (- \frac{2}{7}, \infty)$

Step 18

Use each root to create test intervals.

$x < - \frac{4 + \sqrt{10}}{3}$

$- \frac{4 + \sqrt{10}}{3} < x < - \frac{2}{3}$

$- \frac{2}{3} < x < - \frac{2}{7}$

$- \frac{2}{7} < x < - \frac{4 - \sqrt{10}}{3}$

$x > - \frac{4 - \sqrt{10}}{3}$

Step 19

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

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Step 19.1

Test a value on the interval $x < - \frac{4 + \sqrt{10}}{3}$ to see if it makes the inequality true.

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Step 19.1.1

Choose a value on the interval $x < - \frac{4 + \sqrt{10}}{3}$ and see if this value makes the original inequality true.

$x = - 5$

Step 19.1.2

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

$\frac{(- 5) - | 5 (- 5) + 2 |}{2 (- 5) + | 5 (- 5) + 2 |} > - 5$

Step 19.1.3

The left side $- 2.\overline{153846}$ is greater than the right side $- 5$ , which means that the given statement is always true.

True

Step 19.2

Test a value on the interval $- \frac{4 + \sqrt{10}}{3} < x < - \frac{2}{3}$ to see if it makes the inequality true.

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Step 19.2.1

Choose a value on the interval $- \frac{4 + \sqrt{10}}{3} < x < - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Step 19.2.2

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

$\frac{(- 2) - | 5 (- 2) + 2 |}{2 (- 2) + | 5 (- 2) + 2 |} > - 2$

Step 19.2.3

The left side $- 2.5$ is not greater than the right side $- 2$ , which means that the given statement is false.

False

Step 19.3

Test a value on the interval $- \frac{2}{3} < x < - \frac{2}{7}$ to see if it makes the inequality true.

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Step 19.3.1

Choose a value on the interval $- \frac{2}{3} < x < - \frac{2}{7}$ and see if this value makes the original inequality true.

$x = - 0.48$

Step 19.3.2

Replace $x$ with $- 0.48$ in the original inequality.

$\frac{(- 0.48) - | 5 (- 0.48) + 2 |}{2 (- 0.48) + | 5 (- 0.48) + 2 |} > - 0.48$

Step 19.3.3

The left side $1.\overline{571428}$ is greater than the right side $- 0.48$ , which means that the given statement is always true.

True

Step 19.4

Test a value on the interval $- \frac{2}{7} < x < - \frac{4 - \sqrt{10}}{3}$ to see if it makes the inequality true.

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Step 19.4.1

Choose a value on the interval $- \frac{2}{7} < x < - \frac{4 - \sqrt{10}}{3}$ and see if this value makes the original inequality true.

$x = - 0.28$

Step 19.4.2

Replace $x$ with $- 0.28$ in the original inequality.

$\frac{(- 0.28) - | 5 (- 0.28) + 2 |}{2 (- 0.28) + | 5 (- 0.28) + 2 |} > - 0.28$

Step 19.4.3

The left side $- 22$ is not greater than the right side $- 0.28$ , which means that the given statement is false.

False

Step 19.5

Test a value on the interval $x > - \frac{4 - \sqrt{10}}{3}$ to see if it makes the inequality true.

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Step 19.5.1

Choose a value on the interval $x > - \frac{4 - \sqrt{10}}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 19.5.2

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

$\frac{(0) - | 5 (0) + 2 |}{2 (0) + | 5 (0) + 2 |} > 0$

Step 19.5.3

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 19.6

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

$x < - \frac{4 + \sqrt{10}}{3}$ True

$- \frac{4 + \sqrt{10}}{3} < x < - \frac{2}{3}$ False

$- \frac{2}{3} < x < - \frac{2}{7}$ True

$- \frac{2}{7} < x < - \frac{4 - \sqrt{10}}{3}$ False

$x > - \frac{4 - \sqrt{10}}{3}$ False

$x < - \frac{4 + \sqrt{10}}{3}$ True

$- \frac{4 + \sqrt{10}}{3} < x < - \frac{2}{3}$ False

$- \frac{2}{3} < x < - \frac{2}{7}$ True

$- \frac{2}{7} < x < - \frac{4 - \sqrt{10}}{3}$ False

$x > - \frac{4 - \sqrt{10}}{3}$ False

Step 20

The solution consists of all of the true intervals.

$x < - \frac{4 + \sqrt{10}}{3}$ or $- \frac{2}{3} < x < - \frac{2}{7}$

Step 21

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{4 + \sqrt{10}}{3} or - \frac{2}{3} < x < - \frac{2}{7}$

Interval Notation:

$(- \infty, - \frac{4 + \sqrt{10}}{3}) \cup (- \frac{2}{3}, - \frac{2}{7})$

Step 22