Solve for x 3x/2+3|x-2|<=3

$3 \frac{x}{2} + 3 | x - 2 | \leq 3$

Write $3 (\frac{x}{2}) + 3 | x - 2 | \leq 3$ as a piecewise.

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

$x - 2 \geq 0$

Add $2$ to both sides of the inequality.

$x \geq 2$

In the piece where $x - 2$ is non-negative, remove the absolute value.

$3 (\frac{x}{2}) + 3 (x - 2) \leq 3$

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

$x - 2 < 0$

Add $2$ to both sides of the inequality.

$x < 2$

In the piece where $x - 2$ is negative, remove the absolute value and multiply by $- 1$ .

$3 (\frac{x}{2}) + 3 (- (x - 2)) \leq 3$

Write as a piecewise.

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

Simplify $3 (\frac{x}{2}) + 3 (x - 2) \leq 3$ .

Simplify each term.

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

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

Apply the distributive property.

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

Multiply $3$ by $- 2$ .

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

To write $3 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $3 x$ out of $3 x + 3 x \cdot 2$ .

Factor $3 x$ out of $3 x$ .

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

Factor $3 x$ out of $3 x \cdot 2$ .

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

Factor $3 x$ out of $3 x (1) + 3 x (2)$ .

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

Add $1$ and $2$ .

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

Multiply $3$ by $3$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ 3 (\frac{x}{2}) + 3 (- (x - 2)) \leq 3 & x < 2 \end{cases}$

Simplify $3 (\frac{x}{2}) + 3 (- (x - 2)) \leq 3$ .

Simplify each term.

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

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} + 3 (- (x - 2)) \leq 3 & x < 2 \end{cases}$

Apply the distributive property.

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} + 3 (- x - - 2) \leq 3 & x < 2 \end{cases}$

Multiply $- 1$ by $- 2$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} + 3 (- x + 2) \leq 3 & x < 2 \end{cases}$

Apply the distributive property.

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} + 3 (- x) + 3 \cdot 2 \leq 3 & x < 2 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} - 3 x + 3 \cdot 2 \leq 3 & x < 2 \end{cases}$

Multiply $3$ by $2$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} - 3 x + 6 \leq 3 & x < 2 \end{cases}$

To write $- 3 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} - 3 x \cdot \frac{2}{2} + 6 \leq 3 & x < 2 \end{cases}$

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

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x}{2} + \frac{- 3 x \cdot 2}{2} + 6 \leq 3 & x < 2 \end{cases}$

Combine the numerators over the common denominator.

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x - 3 x \cdot 2}{2} + 6 \leq 3 & x < 2 \end{cases}$

Simplify each term.

Simplify the numerator.

Factor $3 x$ out of $3 x - 3 x \cdot 2$ .

Factor $3 x$ out of $3 x$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x (1) - 3 x \cdot 2}{2} + 6 \leq 3 & x < 2 \end{cases}$

Factor $3 x$ out of $- 3 x \cdot 2$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x (1) + 3 x (- 1 \cdot 2)}{2} + 6 \leq 3 & x < 2 \end{cases}$

Factor $3 x$ out of $3 x (1) + 3 x (- 1 \cdot 2)$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x (1 - 1 \cdot 2)}{2} + 6 \leq 3 & x < 2 \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x (1 - 2)}{2} + 6 \leq 3 & x < 2 \end{cases}$

Subtract $2$ from $1$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{3 x \cdot - 1}{2} + 6 \leq 3 & x < 2 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ \frac{- 3 x}{2} + 6 \leq 3 & x < 2 \end{cases}$

Move the negative in front of the fraction.

${\begin{cases} \frac{9 x}{2} - 6 \leq 3 & x \geq 2 \\ - \frac{3 x}{2} + 6 \leq 3 & x < 2 \end{cases}$

Solve $\frac{9 x}{2} - 6 \leq 3$ when $x \geq 2$ .

Solve $\frac{9 x}{2} - 6 \leq 3$ for $x$ .

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

Add $6$ to both sides of the inequality.

$\frac{9 x}{2} \leq 3 + 6$

Add $3$ and $6$ .

$\frac{9 x}{2} \leq 9$

Multiply both sides by $2$ .

$\frac{9 x}{2} \cdot 2 \leq 9 \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{9 x}{2} \cdot 2 \leq 9 \cdot 2$

Rewrite the expression.

$9 x \leq 9 \cdot 2$

Simplify the right side.

Multiply $9$ by $2$ .

$9 x \leq 18$

Divide each term in $9 x \leq 18$ by $9$ and simplify.

Divide each term in $9 x \leq 18$ by $9$ .

$\frac{9 x}{9} \leq \frac{18}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x}{9} \leq \frac{18}{9}$

Divide $x$ by $1$ .

$x \leq \frac{18}{9}$

Simplify the right side.

Divide $18$ by $9$ .

$x \leq 2$

Find the intersection of $x \leq 2$ and $x \geq 2$ .

$x = 2$

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

Solve $- \frac{3 x}{2} + 6 \leq 3$ for $x$ .

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

Subtract $6$ from both sides of the inequality.

$- \frac{3 x}{2} \leq 3 - 6$

Subtract $6$ from $3$ .

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

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

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

$\frac{- \frac{3 x}{2}}{- 1} \geq \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{3 x}{2}}{1} \geq \frac{- 3}{- 1}$

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

$\frac{3 x}{2} \geq \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$\frac{3 x}{2} \geq 3$

Multiply both sides by $2$ .

$\frac{3 x}{2} \cdot 2 \geq 3 \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 x}{2} \cdot 2 \geq 3 \cdot 2$

Rewrite the expression.

$3 x \geq 3 \cdot 2$

Simplify the right side.

Multiply $3$ by $2$ .

$3 x \geq 6$

Divide each term in $3 x \geq 6$ by $3$ and simplify.

Divide each term in $3 x \geq 6$ by $3$ .

$\frac{3 x}{3} \geq \frac{6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{6}{3}$

Divide $x$ by $1$ .

$x \geq \frac{6}{3}$

Simplify the right side.

Divide $6$ by $3$ .

$x \geq 2$

Find the intersection of $x \geq 2$ and $x < 2$ .

No solution

Find the union of the solutions.

$x = 2$

Solve for x 3x/2+3|x-2|<=3

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