Solve for s |2s+1|+3<=2

$| 2 s + 1 | + 3 \leq 2$

Step 1

Write $| 2 s + 1 | + 3 \leq 2$ as a piecewise.

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

$2 s + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 s \geq - 1$

Divide each term in $2 s \geq - 1$ by $2$ and simplify.

Divide each term in $2 s \geq - 1$ by $2$ .

$\frac{2 s}{2} \geq \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 s}{2} \geq \frac{- 1}{2}$

Divide $s$ by $1$ .

$s \geq \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$s \geq - \frac{1}{2}$

In the piece where $2 s + 1$ is non-negative, remove the absolute value.

$2 s + 1 + 3 \leq 2$

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

$2 s + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 s < - 1$

Divide each term in $2 s < - 1$ by $2$ and simplify.

Divide each term in $2 s < - 1$ by $2$ .

$\frac{2 s}{2} < \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 s}{2} < \frac{- 1}{2}$

Divide $s$ by $1$ .

$s < \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$s < - \frac{1}{2}$

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

$- (2 s + 1) + 3 \leq 2$

Write as a piecewise.

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

Add $1$ and $3$ .

${\begin{cases} 2 s + 4 \leq 2 & s \geq - \frac{1}{2} \\ - (2 s + 1) + 3 \leq 2 & s < - \frac{1}{2} \end{cases}$

Simplify $- (2 s + 1) + 3 \leq 2$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 2 s + 4 \leq 2 & s \geq - \frac{1}{2} \\ - (2 s) - 1 \cdot 1 + 3 \leq 2 & s < - \frac{1}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 s + 4 \leq 2 & s \geq - \frac{1}{2} \\ - 2 s - 1 \cdot 1 + 3 \leq 2 & s < - \frac{1}{2} \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 2 s + 4 \leq 2 & s \geq - \frac{1}{2} \\ - 2 s - 1 + 3 \leq 2 & s < - \frac{1}{2} \end{cases}$

Add $- 1$ and $3$ .

${\begin{cases} 2 s + 4 \leq 2 & s \geq - \frac{1}{2} \\ - 2 s + 2 \leq 2 & s < - \frac{1}{2} \end{cases}$

Step 2

Solve $2 s + 4 \leq 2$ when $s \geq - \frac{1}{2}$ .

Solve $2 s + 4 \leq 2$ for $s$ .

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

Subtract $4$ from both sides of the inequality.

$2 s \leq 2 - 4$

Subtract $4$ from $2$ .

$2 s \leq - 2$

Divide each term in $2 s \leq - 2$ by $2$ and simplify.

Divide each term in $2 s \leq - 2$ by $2$ .

$\frac{2 s}{2} \leq \frac{- 2}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 s}{2} \leq \frac{- 2}{2}$

Divide $s$ by $1$ .

$s \leq \frac{- 2}{2}$

Simplify the right side.

Divide $- 2$ by $2$ .

$s \leq - 1$

Find the intersection of $s \leq - 1$ and $s \geq - \frac{1}{2}$ .

No solution

Step 3

Solve $- 2 s + 2 \leq 2$ when $s < - \frac{1}{2}$ .

Solve $- 2 s + 2 \leq 2$ for $s$ .

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

Subtract $2$ from both sides of the inequality.

$- 2 s \leq 2 - 2$

Subtract $2$ from $2$ .

$- 2 s \leq 0$

Divide each term in $- 2 s \leq 0$ by $- 2$ and simplify.

Divide each term in $- 2 s \leq 0$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 s}{- 2} \geq \frac{0}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 s}{- 2} \geq \frac{0}{- 2}$

Divide $s$ by $1$ .

$s \geq \frac{0}{- 2}$

Simplify the right side.

Divide $0$ by $- 2$ .

$s \geq 0$

Find the intersection of $s \geq 0$ and $s < - \frac{1}{2}$ .

No solution

Step 4

Find the union of the solutions.

No solution

Solve for s |2s+1|+3<=2

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