Convert to Interval Notation |10x+20|<=10

$| 10 x + 20 | \leq 10$

Step 1

Write $| 10 x + 20 | \leq 10$ as a piecewise.

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

$10 x + 20 \geq 0$

Solve the inequality.

Subtract $20$ from both sides of the inequality.

$10 x \geq - 20$

Divide each term in $10 x \geq - 20$ by $10$ and simplify.

Divide each term in $10 x \geq - 20$ by $10$ .

$\frac{10 x}{10} \geq \frac{- 20}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} \geq \frac{- 20}{10}$

Divide $x$ by $1$ .

$x \geq \frac{- 20}{10}$

Simplify the right side.

Divide $- 20$ by $10$ .

$x \geq - 2$

In the piece where $10 x + 20$ is non-negative, remove the absolute value.

$10 x + 20 \leq 10$

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

$10 x + 20 < 0$

Solve the inequality.

Subtract $20$ from both sides of the inequality.

$10 x < - 20$

Divide each term in $10 x < - 20$ by $10$ and simplify.

Divide each term in $10 x < - 20$ by $10$ .

$\frac{10 x}{10} < \frac{- 20}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} < \frac{- 20}{10}$

Divide $x$ by $1$ .

$x < \frac{- 20}{10}$

Simplify the right side.

Divide $- 20$ by $10$ .

$x < - 2$

In the piece where $10 x + 20$ is negative, remove the absolute value and multiply by $- 1$ .

$- (10 x + 20) \leq 10$

Write as a piecewise.

${\begin{cases} 10 x + 20 \leq 10 & x \geq - 2 \\ - (10 x + 20) \leq 10 & x < - 2 \end{cases}$

Simplify $- (10 x + 20) \leq 10$ .

Apply the distributive property.

${\begin{cases} 10 x + 20 \leq 10 & x \geq - 2 \\ - (10 x) - 1 \cdot 20 \leq 10 & x < - 2 \end{cases}$

Multiply $10$ by $- 1$ .

${\begin{cases} 10 x + 20 \leq 10 & x \geq - 2 \\ - 10 x - 1 \cdot 20 \leq 10 & x < - 2 \end{cases}$

Multiply $- 1$ by $20$ .

${\begin{cases} 10 x + 20 \leq 10 & x \geq - 2 \\ - 10 x - 20 \leq 10 & x < - 2 \end{cases}$

Step 2

Solve $10 x + 20 \leq 10$ when $x \geq - 2$ .

Solve $10 x + 20 \leq 10$ for $x$ .

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

Subtract $20$ from both sides of the inequality.

$10 x \leq 10 - 20$

Subtract $20$ from $10$ .

$10 x \leq - 10$

Divide each term in $10 x \leq - 10$ by $10$ and simplify.

Divide each term in $10 x \leq - 10$ by $10$ .

$\frac{10 x}{10} \leq \frac{- 10}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} \leq \frac{- 10}{10}$

Divide $x$ by $1$ .

$x \leq \frac{- 10}{10}$

Simplify the right side.

Divide $- 10$ by $10$ .

$x \leq - 1$

Find the intersection of $x \leq - 1$ and $x \geq - 2$ .

$- 2 \leq x \leq - 1$

Step 3

Solve $- 10 x - 20 \leq 10$ when $x < - 2$ .

Solve $- 10 x - 20 \leq 10$ for $x$ .

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

Add $20$ to both sides of the inequality.

$- 10 x \leq 10 + 20$

Add $10$ and $20$ .

$- 10 x \leq 30$

Divide each term in $- 10 x \leq 30$ by $- 10$ and simplify.

Divide each term in $- 10 x \leq 30$ by $- 10$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 10 x}{- 10} \geq \frac{30}{- 10}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 x}{- 10} \geq \frac{30}{- 10}$

Divide $x$ by $1$ .

$x \geq \frac{30}{- 10}$

Simplify the right side.

Divide $30$ by $- 10$ .

$x \geq - 3$

Find the intersection of $x \geq - 3$ and $x < - 2$ .

$- 3 \leq x < - 2$

Step 4

Find the union of the solutions.

$- 3 \leq x \leq - 1$

Step 5

Convert the inequality to interval notation.

$[- 3, - 1]$

Step 6

Convert to Interval Notation |10x+20|<=10

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