# How do you rewrite the inequality abs(11-2x)>=13 as a compound inequality?

Jan 31, 2017

x le -1, or, x ge 12".

$\text{Equivalently, } x \notin \left(- 1 , 12\right) .$

#### Explanation:

Recall the Defn. of Absolute Value , reproduced below for ready

reference :

$| X | = X , \mathmr{if} X \ge 0$

$= - X , \mathmr{if} X < 0$.

Accordingly, we have to consider $2 \text{ Mutually Exclusive Cases :}$

$\text{Case 1 : } \left(11 - 2 x\right) \ge 0.$

$\left(11 - 2 x\right) \ge 0 \Rightarrow | 11 - 2 x | = 11 - 2 x \left(\text{by Defn.}\right)$

:. |11-2x|ge13 rArr 11-2xge13rArr11-13ge2x

$\Rightarrow - 2 \ge 2 x .$ Multiplying this last inequality by $\frac{1}{2}$,

&, keeping in mind that as $\frac{1}{2} > 0$,

the sign of the inequality will not be reversed, we get,

$- 1 \ge x \ldots \ldots \ldots \ldots . \left(1\right)$.

$\text{Also, } 11 - 2 x \ge 0 \Rightarrow 11 \ge 2 x \Rightarrow \frac{11}{2} \ge x , i . e . , x \le \frac{11}{2.}$

$\left(1\right)$ is in accordanace with this.

$\text{Case 2 : } \left(11 - 2 x\right) < 0$.

$\therefore | 11 - 2 x | = - \left(11 - 2 x\right) = 2 x - 11. . . \text{[Defn.]}$

$\therefore | 11 - 2 x | \ge 13 \Rightarrow 2 x - 11 \ge 13 \Rightarrow 2 x \ge 11 + 13 = 24$

$\Rightarrow x \ge 12. \ldots \ldots \ldots . . \left(2\right)$

As $11 - 2 x < 0 \Rightarrow 11 < 2 x \Rightarrow \frac{11}{2} < x .$

(2) is in accordance with this.

Combining (1), &, (2)" we have, "x le -1, or, x ge 12".

$\text{Equivalently, } x \notin \left(- 1 , 12\right) .$

Enjoy Maths.!

Jan 31, 2017

|(11-2x)| ge 13 iff -1 ge x, or, xge 12;" i.e., equivalently, "x !in (-1,12).

#### Explanation:

Let us have a Second Method to solve this Problem.

We know that,

$\left(\ast\right) \ldots | x - a | < \delta \iff a - \delta < x < a + \delta \ldots \left[x , a \in \mathbb{R} , \delta \in {\mathbb{R}}^{+}\right]$.

Now, $| 11 - 2 x | \ge 13 \iff | 2 \left(\frac{11}{2} - x\right) | \ge 13$

iff |2||(11/2-x)\ ge 13 iff |(x-11/2)| ge 13/2,.............................[because, |x|=|-x|

Hence, $| 11 - 2 x | \ge 13 \iff | \left(x - \frac{11}{2}\right) | \ge \frac{13}{2} , \text{or, equivalently,}$

$| \left(11 - 2 x\right) | \cancel{<} \frac{13}{2} \iff | \left(x - \frac{11}{2}\right) | \cancel{<} \frac{13}{2.}$

Using $\left(\ast\right) ,$ we have,

$| \left(x - \frac{11}{2}\right) | < \frac{13}{2} \iff \left(\frac{11}{2} - \frac{13}{2}\right) < x < \left(\frac{11}{2} + \frac{13}{2}\right)$

$\iff - 1 < x < 12.$

$\therefore | \left(11 - 2 x\right) | \ge 13 \iff - 1 \ge x , \mathmr{and} , x \ge 12 , \text{ ie., equivalently, } x \notin \left(- 1 , 12\right) .$