Algebra Examples

$20 > \frac{2}{3} | 6 + x | + 6$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$\frac{2}{3} \cdot | 6 + x | + 6 < 20$

Step 2

Write $\frac{2}{3} \cdot | 6 + x | + 6 < 20$ as a piecewise.

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Step 2.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$6 + x \geq 0$

Step 2.2

Subtract $6$ from both sides of the inequality.

$x \geq - 6$

Step 2.3

In the piece where $6 + x$ is non-negative, remove the absolute value.

$\frac{2}{3} \cdot (6 + x) + 6 < 20$

Step 2.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$6 + x < 0$

Step 2.5

Subtract $6$ from both sides of the inequality.

$x < - 6$

Step 2.6

In the piece where $6 + x$ is negative, remove the absolute value and multiply by $- 1$ .

$\frac{2}{3} \cdot (- (6 + x)) + 6 < 20$

Step 2.7

Write as a piecewise.

${\begin{cases} \frac{2}{3} \cdot (6 + x) + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8

Simplify $\frac{2}{3} \cdot (6 + x) + 6 < 20$ .

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Step 2.8.1

Simplify each term.

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Step 2.8.1.1

Apply the distributive property.

${\begin{cases} \frac{2}{3} \cdot 6 + \frac{2}{3} x + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.1.2

Cancel the common factor of $3$ .

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Step 2.8.1.2.1

Factor $3$ out of $6$ .

${\begin{cases} \frac{2}{3} \cdot (3 (2)) + \frac{2}{3} x + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.1.2.2

Cancel the common factor.

${\begin{cases} \frac{2}{3} \cdot (3 \cdot 2) + \frac{2}{3} x + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.1.2.3

Rewrite the expression.

${\begin{cases} 2 \cdot 2 + \frac{2}{3} x + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.1.3

Multiply $2$ by $2$ .

${\begin{cases} 4 + \frac{2}{3} x + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.1.4

Combine $\frac{2}{3}$ and $x$ .

${\begin{cases} 4 + \frac{2 x}{3} + 6 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.8.2

Add $4$ and $6$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- (6 + x)) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9

Simplify $\frac{2}{3} \cdot (- (6 + x)) + 6 < 20$ .

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Step 2.9.1

Simplify each term.

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Step 2.9.1.1

Apply the distributive property.

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- 1 \cdot 6 - x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.2

Multiply $- 1$ by $6$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (- 6 - x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.3

Apply the distributive property.

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot - 6 + \frac{2}{3} (- x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.4

Cancel the common factor of $3$ .

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Step 2.9.1.4.1

Factor $3$ out of $- 6$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (3 (- 2)) + \frac{2}{3} (- x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.4.2

Cancel the common factor.

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ \frac{2}{3} \cdot (3 \cdot - 2) + \frac{2}{3} (- x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.4.3

Rewrite the expression.

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ 2 \cdot - 2 + \frac{2}{3} (- x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.5

Multiply $2$ by $- 2$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ - 4 + \frac{2}{3} (- x) + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.1.6

Combine $\frac{2}{3}$ and $x$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ - 4 - \frac{2 x}{3} + 6 < 20 & x < - 6 \end{cases}$

Step 2.9.2

Add $- 4$ and $6$ .

${\begin{cases} \frac{2 x}{3} + 10 < 20 & x \geq - 6 \\ - \frac{2 x}{3} + 2 < 20 & x < - 6 \end{cases}$

Step 3

Solve $\frac{2 x}{3} + 10 < 20$ when $x \geq - 6$ .

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Step 3.1

Solve $\frac{2 x}{3} + 10 < 20$ for $x$ .

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Step 3.1.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 3.1.1.1

Subtract $10$ from both sides of the inequality.

$\frac{2 x}{3} < 20 - 10$

Step 3.1.1.2

Subtract $10$ from $20$ .

$\frac{2 x}{3} < 10$

Step 3.1.2

Multiply both sides by $3$ .

$\frac{2 x}{3} \cdot 3 < 10 \cdot 3$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Simplify the left side.

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Step 3.1.3.1.1

Cancel the common factor of $3$ .

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Step 3.1.3.1.1.1

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 < 10 \cdot 3$

Step 3.1.3.1.1.2

Rewrite the expression.

$2 x < 10 \cdot 3$

Step 3.1.3.2

Simplify the right side.

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Step 3.1.3.2.1

Multiply $10$ by $3$ .

$2 x < 30$

Step 3.1.4

Divide each term in $2 x < 30$ by $2$ and simplify.

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Step 3.1.4.1

Divide each term in $2 x < 30$ by $2$ .

$\frac{2 x}{2} < \frac{30}{2}$

Step 3.1.4.2

Simplify the left side.

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Step 3.1.4.2.1

Cancel the common factor of $2$ .

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Step 3.1.4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} < \frac{30}{2}$

Step 3.1.4.2.1.2

Divide $x$ by $1$ .

$x < \frac{30}{2}$

Step 3.1.4.3

Simplify the right side.

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Step 3.1.4.3.1

Divide $30$ by $2$ .

$x < 15$

Step 3.2

Find the intersection of $x < 15$ and $x \geq - 6$ .

$- 6 \leq x < 15$

Step 4

Solve $- \frac{2 x}{3} + 2 < 20$ when $x < - 6$ .

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Step 4.1

Solve $- \frac{2 x}{3} + 2 < 20$ for $x$ .

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Step 4.1.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 4.1.1.1

Subtract $2$ from both sides of the inequality.

$- \frac{2 x}{3} < 20 - 2$

Step 4.1.1.2

Subtract $2$ from $20$ .

$- \frac{2 x}{3} < 18$

Step 4.1.2

Divide each term in $- \frac{2 x}{3} < 18$ by $- 1$ and simplify.

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Step 4.1.2.1

Divide each term in $- \frac{2 x}{3} < 18$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{2 x}{3}}{- 1} > \frac{18}{- 1}$

Step 4.1.2.2

Simplify the left side.

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Step 4.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{2 x}{3}}{1} > \frac{18}{- 1}$

Step 4.1.2.2.2

Divide $\frac{2 x}{3}$ by $1$ .

$\frac{2 x}{3} > \frac{18}{- 1}$

Step 4.1.2.3

Simplify the right side.

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Step 4.1.2.3.1

Divide $18$ by $- 1$ .

$\frac{2 x}{3} > - 18$

Step 4.1.3

Multiply both sides by $3$ .

$\frac{2 x}{3} \cdot 3 > - 18 \cdot 3$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Simplify the left side.

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Step 4.1.4.1.1

Cancel the common factor of $3$ .

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Step 4.1.4.1.1.1

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 > - 18 \cdot 3$

Step 4.1.4.1.1.2

Rewrite the expression.

$2 x > - 18 \cdot 3$

Step 4.1.4.2

Simplify the right side.

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Step 4.1.4.2.1

Multiply $- 18$ by $3$ .

$2 x > - 54$

Step 4.1.5

Divide each term in $2 x > - 54$ by $2$ and simplify.

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Step 4.1.5.1

Divide each term in $2 x > - 54$ by $2$ .

$\frac{2 x}{2} > \frac{- 54}{2}$

Step 4.1.5.2

Simplify the left side.

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Step 4.1.5.2.1

Cancel the common factor of $2$ .

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Step 4.1.5.2.1.1

Cancel the common factor.

$\frac{2 x}{2} > \frac{- 54}{2}$

Step 4.1.5.2.1.2

Divide $x$ by $1$ .

$x > \frac{- 54}{2}$

Step 4.1.5.3

Simplify the right side.

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Step 4.1.5.3.1

Divide $- 54$ by $2$ .

$x > - 27$

Step 4.2

Find the intersection of $x > - 27$ and $x < - 6$ .

$- 27 < x < - 6$

Step 5

Find the union of the solutions.

$- 27 < x < 15$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$- 27 < x < 15$

Interval Notation:

$(- 27, 15)$

Step 7