Solve for x |x+4|+|2x-1|>12

$| x + 4 | + | 2 x - 1 | > 12$

Step 1

Replace $>$ with $=$ in $| x + 4 | + | 2 x - 1 | > 12$ .

$| x + 4 | + | 2 x - 1 | = 12$

Step 2

Subtract $| x + 4 |$ from both sides of the equation.

$| 2 x - 1 | = 12 - | x + 4 |$

Step 3

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

$2 x - 1 = \pm (12 - | x + 4 |)$

Step 4

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

$2 x - 1 = 12 - | x + 4 |$

$2 x - 1 = - (12 - | x + 4 |)$

Step 5

Solve $2 x - 1 = 12 - | x + 4 |$ for $x$ .

Solve for $| x + 4 |$ .

Rewrite the equation as $12 - | x + 4 | = 2 x - 1$ .

$12 - | x + 4 | = 2 x - 1$

Move all terms not containing $| x + 4 |$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$- | x + 4 | = 2 x - 1 - 12$

Subtract $12$ from $- 1$ .

$- | x + 4 | = 2 x - 13$

Divide each term in $- | x + 4 | = 2 x - 13$ by $- 1$ and simplify.

Divide each term in $- | x + 4 | = 2 x - 13$ by $- 1$ .

$\frac{- | x + 4 |}{- 1} = \frac{2 x}{- 1} + \frac{- 13}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| x + 4 |}{1} = \frac{2 x}{- 1} + \frac{- 13}{- 1}$

Divide $| x + 4 |$ by $1$ .

$| x + 4 | = \frac{2 x}{- 1} + \frac{- 13}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

$| x + 4 | = - 1 \cdot (2 x) + \frac{- 13}{- 1}$

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

$| x + 4 | = - (2 x) + \frac{- 13}{- 1}$

Multiply $2$ by $- 1$ .

$| x + 4 | = - 2 x + \frac{- 13}{- 1}$

Divide $- 13$ by $- 1$ .

$| x + 4 | = - 2 x + 13$

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

$x + 4 = \pm (- 2 x + 13)$

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

$x + 4 = - 2 x + 13$

$x + 4 = - (- 2 x + 13)$

Solve $x + 4 = - 2 x + 13$ for $x$ .

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

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

$x + 4 + 2 x = 13$

Add $x$ and $2 x$ .

$3 x + 4 = 13$

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

Subtract $4$ from both sides of the equation.

$3 x = 13 - 4$

Subtract $4$ from $13$ .

$3 x = 9$

Divide each term in $3 x = 9$ by $3$ and simplify.

Divide each term in $3 x = 9$ by $3$ .

$\frac{3 x}{3} = \frac{9}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{9}{3}$

Divide $x$ by $1$ .

$x = \frac{9}{3}$

Simplify the right side.

Divide $9$ by $3$ .

$x = 3$

Solve $x + 4 = - (- 2 x + 13)$ for $x$ .

Simplify $- (- 2 x + 13)$ .

Rewrite.

$x + 4 = 0 + 0 - (- 2 x + 13)$

Simplify by adding zeros.

$x + 4 = - (- 2 x + 13)$

Apply the distributive property.

$x + 4 = - (- 2 x) - 1 \cdot 13$

Multiply.

Multiply $- 2$ by $- 1$ .

$x + 4 = 2 x - 1 \cdot 13$

Multiply $- 1$ by $13$ .

$x + 4 = 2 x - 13$

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

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

$x + 4 - 2 x = - 13$

Subtract $2 x$ from $x$ .

$- x + 4 = - 13$

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

Subtract $4$ from both sides of the equation.

$- x = - 13 - 4$

Subtract $4$ from $- 13$ .

$- x = - 17$

Divide each term in $- x = - 17$ by $- 1$ and simplify.

Divide each term in $- x = - 17$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 17}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 17}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 17}{- 1}$

Simplify the right side.

Divide $- 17$ by $- 1$ .

$x = 17$

Consolidate the solutions.

$x = 3, 17$

Step 6

Solve $2 x - 1 = - (12 - | x + 4 |)$ for $x$ .

Solve for $| x + 4 |$ .

Rewrite the equation as $- (12 - | x + 4 |) = 2 x - 1$ .

$- (12 - | x + 4 |) = 2 x - 1$

Simplify $- (12 - | x + 4 |)$ .

Apply the distributive property.

$- 1 \cdot 12 - - | x + 4 | = 2 x - 1$

Multiply $- 1$ by $12$ .

$- 12 - - | x + 4 | = 2 x - 1$

Multiply $- - | x + 4 |$ .

Multiply $- 1$ by $- 1$ .

$- 12 + 1 | x + 4 | = 2 x - 1$

Multiply $| x + 4 |$ by $1$ .

$- 12 + | x + 4 | = 2 x - 1$

Move all terms not containing $| x + 4 |$ to the right side of the equation.

Add $12$ to both sides of the equation.

$| x + 4 | = 2 x - 1 + 12$

Add $- 1$ and $12$ .

$| x + 4 | = 2 x + 11$

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

$x + 4 = \pm (2 x + 11)$

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

$x + 4 = 2 x + 11$

$x + 4 = - (2 x + 11)$

Solve $x + 4 = 2 x + 11$ for $x$ .

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

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

$x + 4 - 2 x = 11$

Subtract $2 x$ from $x$ .

$- x + 4 = 11$

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

Subtract $4$ from both sides of the equation.

$- x = 11 - 4$

Subtract $4$ from $11$ .

$- x = 7$

Divide each term in $- x = 7$ by $- 1$ and simplify.

Divide each term in $- x = 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x = - 7$

Solve $x + 4 = - (2 x + 11)$ for $x$ .

Simplify $- (2 x + 11)$ .

Rewrite.

$x + 4 = 0 + 0 - (2 x + 11)$

Simplify by adding zeros.

$x + 4 = - (2 x + 11)$

Apply the distributive property.

$x + 4 = - (2 x) - 1 \cdot 11$

Multiply.

Multiply $2$ by $- 1$ .

$x + 4 = - 2 x - 1 \cdot 11$

Multiply $- 1$ by $11$ .

$x + 4 = - 2 x - 11$

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

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

$x + 4 + 2 x = - 11$

Add $x$ and $2 x$ .

$3 x + 4 = - 11$

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

Subtract $4$ from both sides of the equation.

$3 x = - 11 - 4$

Subtract $4$ from $- 11$ .

$3 x = - 15$

Divide each term in $3 x = - 15$ by $3$ and simplify.

Divide each term in $3 x = - 15$ by $3$ .

$\frac{3 x}{3} = \frac{- 15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 15}{3}$

Divide $x$ by $1$ .

$x = \frac{- 15}{3}$

Simplify the right side.

Divide $- 15$ by $3$ .

$x = - 5$

Consolidate the solutions.

$x = - 7, - 5$

Step 7

Consolidate the solutions.

$x = 3, 17, - 7, - 5$

Step 8

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - 5$

$- 5 < x < 3$

$3 < x < 17$

$x > 17$

Step 9

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 $x < - 7$ to see if it makes the inequality true.

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

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$| (- 10) + 4 | + | 2 (- 10) - 1 | > 12$

The left side $27$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < - 5$ and see if this value makes the original inequality true.

$x = - 6$

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

$| (- 6) + 4 | + | 2 (- 6) - 1 | > 12$

The left side $15$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$| (0) + 4 | + | 2 (0) - 1 | > 12$

The left side $5$ is not greater than the right side $12$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 17$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 17$ and see if this value makes the original inequality true.

$x = 6$

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

$| (6) + 4 | + | 2 (6) - 1 | > 12$

The left side $21$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $x > 17$ to see if it makes the inequality true.

Choose a value on the interval $x > 17$ and see if this value makes the original inequality true.

$x = 20$

Replace $x$ with $20$ in the original inequality.

$| (20) + 4 | + | 2 (20) - 1 | > 12$

The left side $63$ is greater than the right side $12$ , which means that the given statement is always true.

True

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

$x < - 7$ True

$- 7 < x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 17$ True

$x > 17$ True

$x < - 7$ True

$- 7 < x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 17$ True

$x > 17$ True

Step 10

The solution consists of all of the true intervals.

$x < - 7$ or $- 7 < x < - 5$ or $3 < x < 17$ or $x > 17$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < - 7 or - 7 < x < - 5 or 3 < x < 17 or x > 17$

Interval Notation:

$(- \infty, - 7) \cup (- 7, - 5) \cup (3, 17) \cup (17, \infty)$

Step 12

Solve for x |x+4|+|2x-1|>12

Related Questions

Support LCMGCF.com by disabling your adblocker.