Question

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

Answer 1

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

`"Equivalently, "x !in (-1,12).`

Explanation:

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

reference :

`|X|=X, if X ge 0`

`=-X, if X lt 0`.

Accordingly, we have to consider `2" Mutually Exclusive Cases :"`

`"Case 1 : "(11-2x) ge 0.`

`(11-2x) ge 0 rArr |11-2x|=11-2x ("by Defn.")`

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

`rArr-2ge2x.` Multiplying this last inequality by `1/2`,

&, keeping in mind that as `1/2 gt0`,

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

`-1 ge x.............(1)`.

`"Also, "11-2x ge 0 rArr 11 ge 2x rArr 11/2 ge x, i.e., x le 11/2.`

`(1)` is in accordanace with this.

`"Case 2 : "(11-2x) lt 0`.

`:. |11-2x|=-(11-2x)=2x-11..."[Defn.]"`

`:. |11-2x| ge 13 rArr 2x-11 ge 13 rArr 2x ge 11+13=24`

`rArr x ge 12............(2)`

As `11-2x lt 0 rArr 11 lt 2x rArr 11/2 lt x.`

#(2) is in accordance with this.

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

`"Equivalently, "x !in (-1,12).`

Enjoy Maths.!

Answer 2

`|(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,

`(ast)...|x-a| lt delta iff a-delta lt x lt a+delta...[x,a in RR, delta in RR^+]`.

Now, `|11-2x| ge 13 iff |2(11/2-x)| ge 13`

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

Hence, `|11-2x| ge 13 iff |(x-11/2)| ge 13/2, "or, equivalently,"`

`|(11-2x)| cancel< 13/2 iff |(x-11/2)| cancel< 13/2.`

Using `(ast),` we have,

`|(x-11/2)| lt 13/2 iff (11/2-13/2)lt x lt (11/2+13/2)`

`iff -1 lt x lt 12.`

`:. |(11-2x)| ge 13 iff -1 ge x, or, xge 12," ie., equivalently, "x !in (-1,12).`