Solve for x |3x-2|-|x-3|=4-|x+2|

$| 3 x - 2 | - | x - 3 | = 4 - | x + 2 |$

Rewrite the equation as $4 - | x + 2 | = | 3 x - 2 | - | x - 3 |$ .

$4 - | x + 2 | = | 3 x - 2 | - | x - 3 |$

Subtract $4$ from both sides of the equation.

$- | x + 2 | = | 3 x - 2 | - | x - 3 | - 4$

Divide each term in $- | x + 2 | = | 3 x - 2 | - | x - 3 | - 4$ by $- 1$ and simplify.

Divide each term in $- | x + 2 | = | 3 x - 2 | - | x - 3 | - 4$ by $- 1$ .

$\frac{- | x + 2 |}{- 1} = \frac{| 3 x - 2 |}{- 1} + \frac{- | x - 3 |}{- 1} + \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| x + 2 |}{1} = \frac{| 3 x - 2 |}{- 1} + \frac{- | x - 3 |}{- 1} + \frac{- 4}{- 1}$

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

$| x + 2 | = \frac{| 3 x - 2 |}{- 1} + \frac{- | x - 3 |}{- 1} + \frac{- 4}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{| 3 x - 2 |}{- 1}$ .

$| x + 2 | = - 1 \cdot | 3 x - 2 | + \frac{- | x - 3 |}{- 1} + \frac{- 4}{- 1}$

Rewrite $- 1 \cdot | 3 x - 2 |$ as $- | 3 x - 2 |$ .

$| x + 2 | = - | 3 x - 2 | + \frac{- | x - 3 |}{- 1} + \frac{- 4}{- 1}$

Dividing two negative values results in a positive value.

$| x + 2 | = - | 3 x - 2 | + \frac{| x - 3 |}{1} + \frac{- 4}{- 1}$

Divide $| x - 3 |$ by $1$ .

$| x + 2 | = - | 3 x - 2 | + | x - 3 | + \frac{- 4}{- 1}$

Divide $- 4$ by $- 1$ .

$| x + 2 | = - | 3 x - 2 | + | x - 3 | + 4$

Solve for $| x - 3 |$ .

Rewrite the equation as $- | 3 x - 2 | + | x - 3 | + 4 = | x + 2 |$ .

$- | 3 x - 2 | + | x - 3 | + 4 = | x + 2 |$

Move all terms not containing $| x - 3 |$ to the right side of the equation.

Add $| 3 x - 2 |$ to both sides of the equation.

$| x - 3 | + 4 = | x + 2 | + | 3 x - 2 |$

Subtract $4$ from both sides of the equation.

$| x - 3 | = | x + 2 | + | 3 x - 2 | - 4$

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

$x - 3 = \pm (| x + 2 | + | 3 x - 2 | - 4)$

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

$x - 3 = | x + 2 | + | 3 x - 2 | - 4$

$x - 3 = - (| x + 2 | + | 3 x - 2 | - 4)$

Solve $x - 3 = | x + 2 | + | 3 x - 2 | - 4$ for $x$ .

Solve for $| 3 x - 2 |$ .

Rewrite the equation as $| x + 2 | + | 3 x - 2 | - 4 = x - 3$ .

$| x + 2 | + | 3 x - 2 | - 4 = x - 3$

Move all terms not containing $| 3 x - 2 |$ to the right side of the equation.

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

$| 3 x - 2 | - 4 = x - 3 - | x + 2 |$

Add $4$ to both sides of the equation.

$| 3 x - 2 | = x - 3 - | x + 2 | + 4$

Add $- 3$ and $4$ .

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

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

$3 x - 2 = \pm (x - | x + 2 | + 1)$

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

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

$3 x - 2 = - (x - | x + 2 | + 1)$

Solve $3 x - 2 = x - | x + 2 | + 1$ for $x$ .

Solve for $| x + 2 |$ .

Rewrite the equation as $x - | x + 2 | + 1 = 3 x - 2$ .

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

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

Subtract $x$ from both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

Subtract $x$ from $3 x$ .

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

Subtract $1$ from $- 2$ .

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

Divide each term in $- | x + 2 | = 2 x - 3$ by $- 1$ and simplify.

Divide each term in $- | x + 2 | = 2 x - 3$ by $- 1$ .

$\frac{- | x + 2 |}{- 1} = \frac{2 x}{- 1} + \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| x + 2 |}{1} = \frac{2 x}{- 1} + \frac{- 3}{- 1}$

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

$| x + 2 | = \frac{2 x}{- 1} + \frac{- 3}{- 1}$

Simplify the right side.

Simplify each term.

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

$| x + 2 | = - 1 \cdot (2 x) + \frac{- 3}{- 1}$

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

$| x + 2 | = - (2 x) + \frac{- 3}{- 1}$

Multiply $2$ by $- 1$ .

$| x + 2 | = - 2 x + \frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

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

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

$x + 2 = \pm (- 2 x + 3)$

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

$x + 2 = - 2 x + 3$

$x + 2 = - (- 2 x + 3)$

Solve $x + 2 = - 2 x + 3$ for $x$ .

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

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

$x + 2 + 2 x = 3$

Add $x$ and $2 x$ .

$3 x + 2 = 3$

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

Subtract $2$ from both sides of the equation.

$3 x = 3 - 2$

Subtract $2$ from $3$ .

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Solve $x + 2 = - (- 2 x + 3)$ for $x$ .

Simplify $- (- 2 x + 3)$ .

Rewrite.

$x + 2 = 0 + 0 - (- 2 x + 3)$

Simplify by adding zeros.

$x + 2 = - (- 2 x + 3)$

Apply the distributive property.

$x + 2 = - (- 2 x) - 1 \cdot 3$

Multiply.

Multiply $- 2$ by $- 1$ .

$x + 2 = 2 x - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x + 2 = 2 x - 3$

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

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

$x + 2 - 2 x = - 3$

Subtract $2 x$ from $x$ .

$- x + 2 = - 3$

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

Subtract $2$ from both sides of the equation.

$- x = - 3 - 2$

Subtract $2$ from $- 3$ .

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

Consolidate the solutions.

$x = \frac{1}{3}, 5$

Solve $3 x - 2 = - (x - | x + 2 | + 1)$ for $x$ .

Solve for $| x + 2 |$ .

Rewrite the equation as $- (x - | x + 2 | + 1) = 3 x - 2$ .

$- (x - | x + 2 | + 1) = 3 x - 2$

Simplify $- (x - | x + 2 | + 1)$ .

Apply the distributive property.

$- x - - | x + 2 | - 1 \cdot 1 = 3 x - 2$

Simplify.

Multiply $- - | x + 2 |$ .

Multiply $- 1$ by $- 1$ .

$- x + 1 | x + 2 | - 1 \cdot 1 = 3 x - 2$

Multiply $| x + 2 |$ by $1$ .

$- x + | x + 2 | - 1 \cdot 1 = 3 x - 2$

Multiply $- 1$ by $1$ .

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

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

Add $x$ to both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Add $3 x$ and $x$ .

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

Add $- 2$ and $1$ .

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

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

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

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

$x + 2 = 4 x - 1$

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

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

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

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

$x + 2 - 4 x = - 1$

Subtract $4 x$ from $x$ .

$- 3 x + 2 = - 1$

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

Subtract $2$ from both sides of the equation.

$- 3 x = - 1 - 2$

Subtract $2$ from $- 1$ .

$- 3 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 3$ .

$x = 1$

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

Simplify $- (4 x - 1)$ .

Rewrite.

$x + 2 = 0 + 0 - (4 x - 1)$

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $4$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

$x + 2 = - 4 x + 1$

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

Add $4 x$ to both sides of the equation.

$x + 2 + 4 x = 1$

Add $x$ and $4 x$ .

$5 x + 2 = 1$

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

Subtract $2$ from both sides of the equation.

$5 x = 1 - 2$

Subtract $2$ from $1$ .

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Consolidate the solutions.

$x = 1, - \frac{1}{5}$

Consolidate the solutions.

$x = \frac{1}{3}, 5, - \frac{1}{5}$

Solve $x - 3 = - (| x + 2 | + | 3 x - 2 | - 4)$ for $x$ .

Solve for $| 3 x - 2 |$ .

Rewrite the equation as $- (| x + 2 | + | 3 x - 2 | - 4) = x - 3$ .

$- (| x + 2 | + | 3 x - 2 | - 4) = x - 3$

Simplify $- (| x + 2 | + | 3 x - 2 | - 4)$ .

Apply the distributive property.

$- | x + 2 | - | 3 x - 2 | - - 4 = x - 3$

Multiply $- 1$ by $- 4$ .

$- | x + 2 | - | 3 x - 2 | + 4 = x - 3$

Move all terms not containing $| 3 x - 2 |$ to the right side of the equation.

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

$- | 3 x - 2 | + 4 = x - 3 + | x + 2 |$

Subtract $4$ from both sides of the equation.

$- | 3 x - 2 | = x - 3 + | x + 2 | - 4$

Subtract $4$ from $- 3$ .

$- | 3 x - 2 | = x + | x + 2 | - 7$

Divide each term in $- | 3 x - 2 | = x + | x + 2 | - 7$ by $- 1$ and simplify.

Divide each term in $- | 3 x - 2 | = x + | x + 2 | - 7$ by $- 1$ .

$\frac{- | 3 x - 2 |}{- 1} = \frac{x}{- 1} + \frac{| x + 2 |}{- 1} + \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| 3 x - 2 |}{1} = \frac{x}{- 1} + \frac{| x + 2 |}{- 1} + \frac{- 7}{- 1}$

Divide $| 3 x - 2 |$ by $1$ .

$| 3 x - 2 | = \frac{x}{- 1} + \frac{| x + 2 |}{- 1} + \frac{- 7}{- 1}$

Simplify the right side.

Simplify each term.

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

$| 3 x - 2 | = - 1 \cdot x + \frac{| x + 2 |}{- 1} + \frac{- 7}{- 1}$

Rewrite $- 1 \cdot x$ as $- x$ .

$| 3 x - 2 | = - x + \frac{| x + 2 |}{- 1} + \frac{- 7}{- 1}$

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

$| 3 x - 2 | = - x - 1 \cdot | x + 2 | + \frac{- 7}{- 1}$

Rewrite $- 1 \cdot | x + 2 |$ as $- | x + 2 |$ .

$| 3 x - 2 | = - x - | x + 2 | + \frac{- 7}{- 1}$

Divide $- 7$ by $- 1$ .

$| 3 x - 2 | = - x - | x + 2 | + 7$

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

$3 x - 2 = \pm (- x - | x + 2 | + 7)$

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

$3 x - 2 = - x - | x + 2 | + 7$

$3 x - 2 = - (- x - | x + 2 | + 7)$

Solve $3 x - 2 = - x - | x + 2 | + 7$ for $x$ .

Solve for $| x + 2 |$ .

Rewrite the equation as $- x - | x + 2 | + 7 = 3 x - 2$ .

$- x - | x + 2 | + 7 = 3 x - 2$

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

Add $x$ to both sides of the equation.

$- | x + 2 | + 7 = 3 x - 2 + x$

Subtract $7$ from both sides of the equation.

$- | x + 2 | = 3 x - 2 + x - 7$

Add $3 x$ and $x$ .

$- | x + 2 | = 4 x - 2 - 7$

Subtract $7$ from $- 2$ .

$- | x + 2 | = 4 x - 9$

Divide each term in $- | x + 2 | = 4 x - 9$ by $- 1$ and simplify.

Divide each term in $- | x + 2 | = 4 x - 9$ by $- 1$ .

$\frac{- | x + 2 |}{- 1} = \frac{4 x}{- 1} + \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| x + 2 |}{1} = \frac{4 x}{- 1} + \frac{- 9}{- 1}$

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

$| x + 2 | = \frac{4 x}{- 1} + \frac{- 9}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{4 x}{- 1}$ .

$| x + 2 | = - 1 \cdot (4 x) + \frac{- 9}{- 1}$

Rewrite $- 1 \cdot (4 x)$ as $- (4 x)$ .

$| x + 2 | = - (4 x) + \frac{- 9}{- 1}$

Multiply $4$ by $- 1$ .

$| x + 2 | = - 4 x + \frac{- 9}{- 1}$

Divide $- 9$ by $- 1$ .

$| x + 2 | = - 4 x + 9$

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

$x + 2 = \pm (- 4 x + 9)$

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

$x + 2 = - 4 x + 9$

$x + 2 = - (- 4 x + 9)$

Solve $x + 2 = - 4 x + 9$ for $x$ .

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

Add $4 x$ to both sides of the equation.

$x + 2 + 4 x = 9$

Add $x$ and $4 x$ .

$5 x + 2 = 9$

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

Subtract $2$ from both sides of the equation.

$5 x = 9 - 2$

Subtract $2$ from $9$ .

$5 x = 7$

Divide each term in $5 x = 7$ by $5$ and simplify.

Divide each term in $5 x = 7$ by $5$ .

$\frac{5 x}{5} = \frac{7}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{7}{5}$

Divide $x$ by $1$ .

$x = \frac{7}{5}$

Solve $x + 2 = - (- 4 x + 9)$ for $x$ .

Simplify $- (- 4 x + 9)$ .

Rewrite.

$x + 2 = 0 + 0 - (- 4 x + 9)$

Simplify by adding zeros.

$x + 2 = - (- 4 x + 9)$

Apply the distributive property.

$x + 2 = - (- 4 x) - 1 \cdot 9$

Multiply.

Multiply $- 4$ by $- 1$ .

$x + 2 = 4 x - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$x + 2 = 4 x - 9$

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

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

$x + 2 - 4 x = - 9$

Subtract $4 x$ from $x$ .

$- 3 x + 2 = - 9$

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

Subtract $2$ from both sides of the equation.

$- 3 x = - 9 - 2$

Subtract $2$ from $- 9$ .

$- 3 x = - 11$

Divide each term in $- 3 x = - 11$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 11$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 11}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 11}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 11}{- 3}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{11}{3}$

Consolidate the solutions.

$x = \frac{7}{5}, \frac{11}{3}$

Solve $3 x - 2 = - (- x - | x + 2 | + 7)$ for $x$ .

Solve for $| x + 2 |$ .

Rewrite the equation as $- (- x - | x + 2 | + 7) = 3 x - 2$ .

$- (- x - | x + 2 | + 7) = 3 x - 2$

Simplify $- (- x - | x + 2 | + 7)$ .

Apply the distributive property.

$- - x - - | x + 2 | - 1 \cdot 7 = 3 x - 2$

Simplify.

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$1 x - - | x + 2 | - 1 \cdot 7 = 3 x - 2$

Multiply $x$ by $1$ .

$x - - | x + 2 | - 1 \cdot 7 = 3 x - 2$

Multiply $- - | x + 2 |$ .

Multiply $- 1$ by $- 1$ .

$x + 1 | x + 2 | - 1 \cdot 7 = 3 x - 2$

Multiply $| x + 2 |$ by $1$ .

$x + | x + 2 | - 1 \cdot 7 = 3 x - 2$

Multiply $- 1$ by $7$ .

$x + | x + 2 | - 7 = 3 x - 2$

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

Subtract $x$ from both sides of the equation.

$| x + 2 | - 7 = 3 x - 2 - x$

Add $7$ to both sides of the equation.

$| x + 2 | = 3 x - 2 - x + 7$

Subtract $x$ from $3 x$ .

$| x + 2 | = 2 x - 2 + 7$

Add $- 2$ and $7$ .

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

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

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

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

$x + 2 = 2 x + 5$

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

Solve $x + 2 = 2 x + 5$ for $x$ .

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

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

$x + 2 - 2 x = 5$

Subtract $2 x$ from $x$ .

$- x + 2 = 5$

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

Subtract $2$ from both sides of the equation.

$- x = 5 - 2$

Subtract $2$ from $5$ .

$- x = 3$

Divide each term in $- x = 3$ by $- 1$ and simplify.

Divide each term in $- x = 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $3$ by $- 1$ .

$x = - 3$

Solve $x + 2 = - (2 x + 5)$ for $x$ .

Simplify $- (2 x + 5)$ .

Rewrite.

$x + 2 = 0 + 0 - (2 x + 5)$

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $2$ by $- 1$ .

$x + 2 = - 2 x - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x + 2 = - 2 x - 5$

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

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

$x + 2 + 2 x = - 5$

Add $x$ and $2 x$ .

$3 x + 2 = - 5$

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

Subtract $2$ from both sides of the equation.

$3 x = - 5 - 2$

Subtract $2$ from $- 5$ .

$3 x = - 7$

Divide each term in $3 x = - 7$ by $3$ and simplify.

Divide each term in $3 x = - 7$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Consolidate the solutions.

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

Consolidate the solutions.

$x = \frac{7}{5}, \frac{11}{3}, - 3, - \frac{7}{3}$

Consolidate the solutions.

$x = \frac{1}{3}, 5, - \frac{1}{5}, \frac{7}{5}, \frac{11}{3}, - 3, - \frac{7}{3}$

Use each root to create test intervals.

$x < - 3$

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

$- \frac{7}{3} < x < - \frac{1}{5}$

$- \frac{1}{5} < x < \frac{1}{3}$

$\frac{1}{3} < x < \frac{7}{5}$

$\frac{7}{5} < x < \frac{11}{3}$

$\frac{11}{3} < x < 5$

$x > 5$

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$| 3 (- 6) - 2 | - | (- 6) - 3 | = 4 - | (- 6) + 2 |$

The left side $11$ does not equal to the right side $0$ , which means that the given statement is false.

False

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

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

$x = - 2.67$

Replace $x$ with $- 2.67$ in the original inequality.

$| 3 (- 2.67) - 2 | - | (- 2.67) - 3 | = 4 - | (- 2.67) + 2 |$

The left side $4.34$ does not equal to the right side $3.33$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{7}{3} < x < - \frac{1}{5}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{7}{3} < x < - \frac{1}{5}$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$| 3 (- 1) - 2 | - | (- 1) - 3 | = 4 - | (- 1) + 2 |$

The left side $1$ does not equal to the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{1}{5} < x < \frac{1}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{5} < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$| 3 (0) - 2 | - | (0) - 3 | = 4 - | (0) + 2 |$

The left side $- 1$ does not equal to the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $\frac{1}{3} < x < \frac{7}{5}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{3} < x < \frac{7}{5}$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$| 3 (1) - 2 | - | (1) - 3 | = 4 - | (1) + 2 |$

The left side $- 1$ does not equal to the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $\frac{7}{5} < x < \frac{11}{3}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{7}{5} < x < \frac{11}{3}$ and see if this value makes the original inequality true.

$x = 3$

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

$| 3 (3) - 2 | - | (3) - 3 | = 4 - | (3) + 2 |$

The left side $7$ does not equal to the right side $- 1$ , which means that the given statement is false.

False

Test a value on the interval $\frac{11}{3} < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $\frac{11}{3} < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$| 3 (4) - 2 | - | (4) - 3 | = 4 - | (4) + 2 |$

The left side $9$ does not equal to the right side $- 2$ , which means that the given statement is false.

False

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

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

$x = 8$

Replace $x$ with $8$ in the original inequality.

$| 3 (8) - 2 | - | (8) - 3 | = 4 - | (8) + 2 |$

The left side $17$ does not equal to the right side $- 6$ , which means that the given statement is false.

False

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

$x < - 3$ False

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

$- \frac{7}{3} < x < - \frac{1}{5}$ False

$- \frac{1}{5} < x < \frac{1}{3}$ False

$\frac{1}{3} < x < \frac{7}{5}$ False

$\frac{7}{5} < x < \frac{11}{3}$ False

$\frac{11}{3} < x < 5$ False

$x > 5$ False

$x < - 3$ False

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

$- \frac{7}{3} < x < - \frac{1}{5}$ False

$- \frac{1}{5} < x < \frac{1}{3}$ False

$\frac{1}{3} < x < \frac{7}{5}$ False

$\frac{7}{5} < x < \frac{11}{3}$ False

$\frac{11}{3} < x < 5$ False

$x > 5$ False

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Exclude the solutions that do not make $| 3 x - 2 | - | x - 3 | = 4 - | x + 2 |$ true.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.4, - 2.\overline{3}$

Mixed Number Form:

$x = 1 \frac{2}{5}, - 2 \frac{1}{3}$

Solve for x |3x-2|-|x-3|=4-|x+2|

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