Algebra Examples

$| 2 x - 6 | + | 3 - x | > 12$

Step 1

Replace $>$ with $=$ in $| 2 x - 6 | + | 3 - x | > 12$ .

$| 2 x - 6 | + | 3 - x | = 12$

Step 2

Subtract $| 2 x - 6 |$ from both sides of the equation.

$| 3 - x | = 12 - | 2 x - 6 |$

Step 3

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

$3 - x = \pm (12 - | 2 x - 6 |)$

Step 4

The result consists of both the positive and negative portions of the $\pm$ .

$3 - x = 12 - | 2 x - 6 |$

$3 - x = - (12 - | 2 x - 6 |)$

Step 5

Solve $3 - x = 12 - | 2 x - 6 |$ for $x$ .

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Step 5.1

Solve for $| 2 x - 6 |$ .

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Step 5.1.1

Rewrite the equation as $12 - | 2 x - 6 | = 3 - x$ .

$12 - | 2 x - 6 | = 3 - x$

Step 5.1.2

Move all terms not containing $| 2 x - 6 |$ to the right side of the equation.

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Step 5.1.2.1

Subtract $12$ from both sides of the equation.

$- | 2 x - 6 | = 3 - x - 12$

Step 5.1.2.2

Subtract $12$ from $3$ .

$- | 2 x - 6 | = - x - 9$

Step 5.1.3

Divide each term in $- | 2 x - 6 | = - x - 9$ by $- 1$ and simplify.

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Step 5.1.3.1

Divide each term in $- | 2 x - 6 | = - x - 9$ by $- 1$ .

$\frac{- | 2 x - 6 |}{- 1} = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Step 5.1.3.2

Simplify the left side.

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Step 5.1.3.2.1

Dividing two negative values results in a positive value.

$\frac{| 2 x - 6 |}{1} = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Step 5.1.3.2.2

Divide $| 2 x - 6 |$ by $1$ .

$| 2 x - 6 | = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Step 5.1.3.3

Simplify the right side.

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Step 5.1.3.3.1

Simplify each term.

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Step 5.1.3.3.1.1

Dividing two negative values results in a positive value.

$| 2 x - 6 | = \frac{x}{1} + \frac{- 9}{- 1}$

Step 5.1.3.3.1.2

Divide $x$ by $1$ .

$| 2 x - 6 | = x + \frac{- 9}{- 1}$

Step 5.1.3.3.1.3

Divide $- 9$ by $- 1$ .

$| 2 x - 6 | = x + 9$

Step 5.2

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

$2 x - 6 = \pm (x + 9)$

Step 5.3

The result consists of both the positive and negative portions of the $\pm$ .

$2 x - 6 = x + 9$

$2 x - 6 = - (x + 9)$

Step 5.4

Solve $2 x - 6 = x + 9$ for $x$ .

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Step 5.4.1

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

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Step 5.4.1.1

Subtract $x$ from both sides of the equation.

$2 x - 6 - x = 9$

Step 5.4.1.2

Subtract $x$ from $2 x$ .

$x - 6 = 9$

Step 5.4.2

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

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Step 5.4.2.1

Add $6$ to both sides of the equation.

$x = 9 + 6$

Step 5.4.2.2

Add $9$ and $6$ .

$x = 15$

Step 5.5

Solve $2 x - 6 = - (x + 9)$ for $x$ .

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Step 5.5.1

Simplify $- (x + 9)$ .

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Step 5.5.1.1

Rewrite.

$2 x - 6 = 0 + 0 - (x + 9)$

Step 5.5.1.2

Simplify by adding zeros.

$2 x - 6 = - (x + 9)$

Step 5.5.1.3

Apply the distributive property.

$2 x - 6 = - x - 1 \cdot 9$

Step 5.5.1.4

Multiply $- 1$ by $9$ .

$2 x - 6 = - x - 9$

Step 5.5.2

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

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Step 5.5.2.1

Add $x$ to both sides of the equation.

$2 x - 6 + x = - 9$

Step 5.5.2.2

Add $2 x$ and $x$ .

$3 x - 6 = - 9$

Step 5.5.3

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

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Step 5.5.3.1

Add $6$ to both sides of the equation.

$3 x = - 9 + 6$

Step 5.5.3.2

Add $- 9$ and $6$ .

$3 x = - 3$

Step 5.5.4

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

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Step 5.5.4.1

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

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

Step 5.5.4.2

Simplify the left side.

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Step 5.5.4.2.1

Cancel the common factor of $3$ .

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Step 5.5.4.2.1.1

Cancel the common factor.

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

Step 5.5.4.2.1.2

Divide $x$ by $1$ .

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

Step 5.5.4.3

Simplify the right side.

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Step 5.5.4.3.1

Divide $- 3$ by $3$ .

$x = - 1$

Step 5.6

Consolidate the solutions.

$x = 15, - 1$

Step 6

Solve $3 - x = - (12 - | 2 x - 6 |)$ for $x$ .

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Step 6.1

Solve for $| 2 x - 6 |$ .

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Step 6.1.1

Rewrite the equation as $- (12 - | 2 x - 6 |) = 3 - x$ .

$- (12 - | 2 x - 6 |) = 3 - x$

Step 6.1.2

Simplify $- (12 - | 2 x - 6 |)$ .

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Step 6.1.2.1

Apply the distributive property.

$- 1 \cdot 12 - - | 2 x - 6 | = 3 - x$

Step 6.1.2.2

Multiply $- 1$ by $12$ .

$- 12 - - | 2 x - 6 | = 3 - x$

Step 6.1.2.3

Multiply $- - | 2 x - 6 |$ .

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Step 6.1.2.3.1

Multiply $- 1$ by $- 1$ .

$- 12 + 1 | 2 x - 6 | = 3 - x$

Step 6.1.2.3.2

Multiply $| 2 x - 6 |$ by $1$ .

$- 12 + | 2 x - 6 | = 3 - x$

Step 6.1.3

Move all terms not containing $| 2 x - 6 |$ to the right side of the equation.

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Step 6.1.3.1

Add $12$ to both sides of the equation.

$| 2 x - 6 | = 3 - x + 12$

Step 6.1.3.2

Add $3$ and $12$ .

$| 2 x - 6 | = - x + 15$

Step 6.2

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

$2 x - 6 = \pm (- x + 15)$

Step 6.3

The result consists of both the positive and negative portions of the $\pm$ .

$2 x - 6 = - x + 15$

$2 x - 6 = - (- x + 15)$

Step 6.4

Solve $2 x - 6 = - x + 15$ for $x$ .

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Step 6.4.1

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

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Step 6.4.1.1

Add $x$ to both sides of the equation.

$2 x - 6 + x = 15$

Step 6.4.1.2

Add $2 x$ and $x$ .

$3 x - 6 = 15$

Step 6.4.2

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

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Step 6.4.2.1

Add $6$ to both sides of the equation.

$3 x = 15 + 6$

Step 6.4.2.2

Add $15$ and $6$ .

$3 x = 21$

Step 6.4.3

Divide each term in $3 x = 21$ by $3$ and simplify.

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Step 6.4.3.1

Divide each term in $3 x = 21$ by $3$ .

$\frac{3 x}{3} = \frac{21}{3}$

Step 6.4.3.2

Simplify the left side.

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Step 6.4.3.2.1

Cancel the common factor of $3$ .

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Step 6.4.3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{21}{3}$

Step 6.4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{21}{3}$

Step 6.4.3.3

Simplify the right side.

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Step 6.4.3.3.1

Divide $21$ by $3$ .

$x = 7$

Step 6.5

Solve $2 x - 6 = - (- x + 15)$ for $x$ .

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Step 6.5.1

Simplify $- (- x + 15)$ .

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Step 6.5.1.1

Rewrite.

$2 x - 6 = 0 + 0 - (- x + 15)$

Step 6.5.1.2

Simplify by adding zeros.

$2 x - 6 = - (- x + 15)$

Step 6.5.1.3

Apply the distributive property.

$2 x - 6 = - - x - 1 \cdot 15$

Step 6.5.1.4

Multiply $- - x$ .

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Step 6.5.1.4.1

Multiply $- 1$ by $- 1$ .

$2 x - 6 = 1 x - 1 \cdot 15$

Step 6.5.1.4.2

Multiply $x$ by $1$ .

$2 x - 6 = x - 1 \cdot 15$

Step 6.5.1.5

Multiply $- 1$ by $15$ .

$2 x - 6 = x - 15$

Step 6.5.2

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

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Step 6.5.2.1

Subtract $x$ from both sides of the equation.

$2 x - 6 - x = - 15$

Step 6.5.2.2

Subtract $x$ from $2 x$ .

$x - 6 = - 15$

Step 6.5.3

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

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Step 6.5.3.1

Add $6$ to both sides of the equation.

$x = - 15 + 6$

Step 6.5.3.2

Add $- 15$ and $6$ .

$x = - 9$

Step 6.6

Consolidate the solutions.

$x = 7, - 9$

Step 7

Consolidate the solutions.

$x = 15, - 1, 7, - 9$

Step 8

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 1$

$- 1 < x < 7$

$7 < x < 15$

$x > 15$

Step 9

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 9.1

Test a value on the interval $x < - 9$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Step 9.1.2

Replace $x$ with $- 12$ in the original inequality.

$| 2 (- 12) - 6 | + | 3 - (- 12) | > 12$

Step 9.1.3

The left side $45$ is greater than the right side $12$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- 9 < x < - 1$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 9 < x < - 1$ and see if this value makes the original inequality true.

$x = - 5$

Step 9.2.2

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

$| 2 (- 5) - 6 | + | 3 - (- 5) | > 12$

Step 9.2.3

The left side $24$ is greater than the right side $12$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $- 1 < x < 7$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $- 1 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Step 9.3.2

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

$| 2 (0) - 6 | + | 3 - (0) | > 12$

Step 9.3.3

The left side $9$ is not greater than the right side $12$ , which means that the given statement is false.

False

Step 9.4

Test a value on the interval $7 < x < 15$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $7 < x < 15$ and see if this value makes the original inequality true.

$x = 11$

Step 9.4.2

Replace $x$ with $11$ in the original inequality.

$| 2 (11) - 6 | + | 3 - (11) | > 12$

Step 9.4.3

The left side $24$ is greater than the right side $12$ , which means that the given statement is always true.

True

Step 9.5

Test a value on the interval $x > 15$ to see if it makes the inequality true.

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Step 9.5.1

Choose a value on the interval $x > 15$ and see if this value makes the original inequality true.

$x = 18$

Step 9.5.2

Replace $x$ with $18$ in the original inequality.

$| 2 (18) - 6 | + | 3 - (18) | > 12$

Step 9.5.3

The left side $45$ is greater than the right side $12$ , which means that the given statement is always true.

True

Step 9.6

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

$x < - 9$ True

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$7 < x < 15$ True

$x > 15$ True

$x < - 9$ True

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$7 < x < 15$ True

$x > 15$ True

Step 10

The solution consists of all of the true intervals.

$x < - 9$ or $- 9 < x < - 1$ or $7 < x < 15$ or $x > 15$

Step 11

Convert the inequality to interval notation.

$(- \infty, - 9) \cup (- 9, - 1) \cup (7, 15) \cup (15, \infty)$

Step 12