Solve for r (|-10+4r|)/10>1

$\frac{| - 10 + 4 r |}{10} > 1$

Step 1

Multiply both sides by $10$ .

$\frac{| - 10 + 4 r |}{10} \cdot 10 > 1 \cdot 10$

Step 2

Simplify.

Simplify the left side.

Simplify $\frac{| - 10 + 4 r |}{10} \cdot 10$ .

Simplify the numerator.

Factor $2$ out of $- 10 + 4 r$ .

Factor $2$ out of $- 10$ .

$\frac{| 2 (- 5) + 4 r |}{10} \cdot 10 > 1 \cdot 10$

Factor $2$ out of $4 r$ .

$\frac{| 2 (- 5) + 2 (2 r) |}{10} \cdot 10 > 1 \cdot 10$

Factor $2$ out of $2 (- 5) + 2 (2 r)$ .

$\frac{| 2 (- 5 + 2 r) |}{10} \cdot 10 > 1 \cdot 10$

Apply the distributive property.

$\frac{| 2 \cdot - 5 + 2 (2 r) |}{10} \cdot 10 > 1 \cdot 10$

Multiply $2$ by $- 5$ .

$\frac{| - 10 + 2 (2 r) |}{10} \cdot 10 > 1 \cdot 10$

Multiply $2$ by $2$ .

$\frac{| - 10 + 4 r |}{10} \cdot 10 > 1 \cdot 10$

Factor $2$ out of $- 10 + 4 r$ .

Factor $2$ out of $- 10$ .

$\frac{| 2 (- 5) + 4 r |}{10} \cdot 10 > 1 \cdot 10$

Factor $2$ out of $4 r$ .

$\frac{| 2 (- 5) + 2 (2 r) |}{10} \cdot 10 > 1 \cdot 10$

Factor $2$ out of $2 (- 5) + 2 (2 r)$ .

$\frac{| 2 (- 5 + 2 r) |}{10} \cdot 10 > 1 \cdot 10$

Remove non-negative terms from the absolute value.

$\frac{2 | - 5 + 2 r |}{10} \cdot 10 > 1 \cdot 10$

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{2 | - 5 + 2 r |}{10} \cdot 10 > 1 \cdot 10$

Rewrite the expression.

$2 | - 5 + 2 r | > 1 \cdot 10$

Simplify the right side.

Multiply $10$ by $1$ .

$2 | - 5 + 2 r | > 10$

Step 3

Solve for $r$ .

Write $2 | - 5 + 2 r | > 10$ as a piecewise.

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

$- 5 + 2 r \geq 0$

Solve the inequality.

Add $5$ to both sides of the inequality.

$2 r \geq 5$

Divide each term in $2 r \geq 5$ by $2$ and simplify.

Divide each term in $2 r \geq 5$ by $2$ .

$\frac{2 r}{2} \geq \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 r}{2} \geq \frac{5}{2}$

Divide $r$ by $1$ .

$r \geq \frac{5}{2}$

In the piece where $- 5 + 2 r$ is non-negative, remove the absolute value.

$2 (- 5 + 2 r) > 10$

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

$- 5 + 2 r < 0$

Solve the inequality.

Add $5$ to both sides of the inequality.

$2 r < 5$

Divide each term in $2 r < 5$ by $2$ and simplify.

Divide each term in $2 r < 5$ by $2$ .

$\frac{2 r}{2} < \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 r}{2} < \frac{5}{2}$

Divide $r$ by $1$ .

$r < \frac{5}{2}$

In the piece where $- 5 + 2 r$ is negative, remove the absolute value and multiply by $- 1$ .

$2 (- (- 5 + 2 r)) > 10$

Write as a piecewise.

${\begin{cases} 2 (- 5 + 2 r) > 10 & r \geq \frac{5}{2} \\ 2 (- (- 5 + 2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Simplify $2 (- 5 + 2 r) > 10$ .

Apply the distributive property.

${\begin{cases} 2 \cdot - 5 + 2 (2 r) > 10 & r \geq \frac{5}{2} \\ 2 (- (- 5 + 2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $2$ by $- 5$ .

${\begin{cases} - 10 + 2 (2 r) > 10 & r \geq \frac{5}{2} \\ 2 (- (- 5 + 2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $2$ by $2$ .

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 2 (- (- 5 + 2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Simplify $2 (- (- 5 + 2 r)) > 10$ .

Apply the distributive property.

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 2 (- - 5 - (2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $- 1$ by $- 5$ .

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 2 (5 - (2 r)) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 2 (5 - 2 r) > 10 & r < \frac{5}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 2 \cdot 5 + 2 (- 2 r) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $2$ by $5$ .

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 10 + 2 (- 2 r) > 10 & r < \frac{5}{2} \end{cases}$

Multiply $- 2$ by $2$ .

${\begin{cases} - 10 + 4 r > 10 & r \geq \frac{5}{2} \\ 10 - 4 r > 10 & r < \frac{5}{2} \end{cases}$

Solve $- 10 + 4 r > 10$ for $r$ .

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

Add $10$ to both sides of the inequality.

$4 r > 10 + 10$

Add $10$ and $10$ .

$4 r > 20$

Divide each term in $4 r > 20$ by $4$ and simplify.

Divide each term in $4 r > 20$ by $4$ .

$\frac{4 r}{4} > \frac{20}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 r}{4} > \frac{20}{4}$

Divide $r$ by $1$ .

$r > \frac{20}{4}$

Simplify the right side.

Divide $20$ by $4$ .

$r > 5$

Solve $10 - 4 r > 10$ for $r$ .

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

Subtract $10$ from both sides of the inequality.

$- 4 r > 10 - 10$

Subtract $10$ from $10$ .

$- 4 r > 0$

Divide each term in $- 4 r > 0$ by $- 4$ and simplify.

Divide each term in $- 4 r > 0$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 r}{- 4} < \frac{0}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 r}{- 4} < \frac{0}{- 4}$

Divide $r$ by $1$ .

$r < \frac{0}{- 4}$

Simplify the right side.

Divide $0$ by $- 4$ .

$r < 0$

Find the union of the solutions.

$r < 0$ or $r > 5$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$r < 0 or r > 5$

Interval Notation:

$(- \infty, 0) \cup (5, \infty)$

Step 5

Solve for r (|-10+4r|)/10>1

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