Solve the Absolute Value Inequality for u -3|2-4u|+5<-13

$- 3 | 2 - 4 u | + 5 < - 13$

Write $- 3 | 2 - 4 u | + 5 < - 13$ as a piecewise.

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

$2 - 4 u \geq 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$- 4 u \geq - 2$

Divide each term in $- 4 u \geq - 2$ by $- 4$ and simplify.

Divide each term in $- 4 u \geq - 2$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 u}{- 4} \leq \frac{- 2}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 u}{- 4} \leq \frac{- 2}{- 4}$

Divide $u$ by $1$ .

$u \leq \frac{- 2}{- 4}$

Simplify the right side.

Cancel the common factor of $- 2$ and $- 4$ .

Factor $- 2$ out of $- 2$ .

$u \leq \frac{- 2 (1)}{- 4}$

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

$u \leq \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Cancel the common factor.

$u \leq \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Rewrite the expression.

$u \leq \frac{1}{2}$

In the piece where $2 - 4 u$ is non-negative, remove the absolute value.

$- 3 (2 - 4 u) + 5 < - 13$

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

$2 - 4 u < 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$- 4 u < - 2$

Divide each term in $- 4 u < - 2$ by $- 4$ and simplify.

Divide each term in $- 4 u < - 2$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 u}{- 4} > \frac{- 2}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 u}{- 4} > \frac{- 2}{- 4}$

Divide $u$ by $1$ .

$u > \frac{- 2}{- 4}$

Simplify the right side.

Cancel the common factor of $- 2$ and $- 4$ .

Factor $- 2$ out of $- 2$ .

$u > \frac{- 2 (1)}{- 4}$

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

$u > \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Cancel the common factor.

$u > \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Rewrite the expression.

$u > \frac{1}{2}$

In the piece where $2 - 4 u$ is negative, remove the absolute value and multiply by $- 1$ .

$- 3 (- (2 - 4 u)) + 5 < - 13$

Write as a piecewise.

${\begin{cases} - 3 (2 - 4 u) + 5 < - 13 & u \leq \frac{1}{2} \\ - 3 (- (2 - 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Simplify $- 3 (2 - 4 u) + 5 < - 13$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - 3 \cdot 2 - 3 (- 4 u) + 5 < - 13 & u \leq \frac{1}{2} \\ - 3 (- (2 - 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $- 3$ by $2$ .

${\begin{cases} - 6 - 3 (- 4 u) + 5 < - 13 & u \leq \frac{1}{2} \\ - 3 (- (2 - 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $- 4$ by $- 3$ .

${\begin{cases} - 6 + 12 u + 5 < - 13 & u \leq \frac{1}{2} \\ - 3 (- (2 - 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Add $- 6$ and $5$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 3 (- (2 - 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Simplify $- 3 (- (2 - 4 u)) + 5 < - 13$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 3 (- 1 \cdot 2 - (- 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 3 (- 2 - (- 4 u)) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $- 4$ by $- 1$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 3 (- 2 + 4 u) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 3 \cdot - 2 - 3 (4 u) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $- 3$ by $- 2$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ 6 - 3 (4 u) + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Multiply $4$ by $- 3$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ 6 - 12 u + 5 < - 13 & u > \frac{1}{2} \end{cases}$

Add $6$ and $5$ .

${\begin{cases} 12 u - 1 < - 13 & u \leq \frac{1}{2} \\ - 12 u + 11 < - 13 & u > \frac{1}{2} \end{cases}$

Solve $12 u - 1 < - 13$ for $u$ .

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

Add $1$ to both sides of the inequality.

$12 u < - 13 + 1$

Add $- 13$ and $1$ .

$12 u < - 12$

Divide each term in $12 u < - 12$ by $12$ and simplify.

Divide each term in $12 u < - 12$ by $12$ .

$\frac{12 u}{12} < \frac{- 12}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 u}{12} < \frac{- 12}{12}$

Divide $u$ by $1$ .

$u < \frac{- 12}{12}$

Simplify the right side.

Divide $- 12$ by $12$ .

$u < - 1$

Solve $- 12 u + 11 < - 13$ for $u$ .

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

Subtract $11$ from both sides of the inequality.

$- 12 u < - 13 - 11$

Subtract $11$ from $- 13$ .

$- 12 u < - 24$

Divide each term in $- 12 u < - 24$ by $- 12$ and simplify.

Divide each term in $- 12 u < - 24$ by $- 12$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 12 u}{- 12} > \frac{- 24}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 u}{- 12} > \frac{- 24}{- 12}$

Divide $u$ by $1$ .

$u > \frac{- 24}{- 12}$

Simplify the right side.

Divide $- 24$ by $- 12$ .

$u > 2$

Find the union of the solutions.

$u < - 1$ or $u > 2$

The result can be shown in multiple forms.

Inequality Form:

$u < - 1 or u > 2$

Interval Notation:

$(- \infty, - 1) \cup (2, \infty)$

Solve the Absolute Value Inequality for u -3|2-4u|+5<-13

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