compound inequalities: |x| > -3, |6 - 7x| > 27

[sak7704]

I have homework problems on subjects we haven't gone over in class. I'm hoping someone here can help.

In class, we went over "|x - 2| < 1", which means that the distance between x and 2 is less than 1. The answer the instructor gave is "S = (1, 3)" (so the "solution" is "all x in between 1 and 3").

Here's where I start to have problems.

1) |x| > -3

What does this even mean? That "the distance between x to 0 is greater than negative 3"?

2) |6 - 7x| > 27

Here's what I did:

6 - 7x > 27 or 6 - 7x < -27

6 - 27 > 7x or 6 + 27 < 7x

-21 > 7x or 33 < 7x

-3 > x or 33/7 < x

Therefore, x < -3 or x > 4 ⁵/₇, so the solution is S = (-infinity. -3) U (4 5/7, infinity).

Am I even close on this solution? I would really appreciate some help with this. Thank you!

 

[soroban]

Hello, sak7704!

In class, we went over \(\displaystyle x\,-\,2|\:<\:1\),
. . which means that the distance between \(\displaystyle x\) and \(\displaystyle 2\) is less than \(\displaystyle 1\).
The answer the instructor gave is: \(\displaystyle S \:= \1,\,3)\)
. . so the solution is "all \(\displaystyle x\) in between \(\displaystyle 1\) and \(\displaystyle 3\).

That is all correct. .Here's another approach . . .

. . \(\displaystyle |x\,-\,2|\:<\:1\) can be written: \(\displaystyle \:-1 \:<\:x\,-\,2\:<\:1\:\) **

Solve for \(\displaystyle x\) (add 2 to all three sides): \(\displaystyle \:1 \:<\:x\:<\:3\)

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
The inequality has "less than".
The function is between \(\displaystyle \,-1\) and \(\displaystyle +1.\)

Here's where I start to have problems.
. . \(\displaystyle 1)\;|x|\:>\:-3\)

This is actually a trick question . . .

The absolute value of any number is always greater than or equal to zero.
. . So, of course, \(\displaystyle |x|\) will be greater than negative-three.

Answer: \(\displaystyle \:S \:=\-\infty,\,\infty)\)

\(\displaystyle 2)\;|6\,-\,7x| \:> \:27\)

What you did was absolutely correct!

This one has "greater than".
. . The function is outside \(\displaystyle \,-27\) and \(\displaystyle +27\).

I write it like this: \(\displaystyle \:6\,-\,7x \:<\:-27\) . . . \(\displaystyle 6\,-\,7x\:>\:27\)

. . and solve for \(\displaystyle x\).

 

[tkhunny]

1) |x| > -3

What does this even mean? That "the distance between x to 0 is greater than negative 3"?

All Real Numbers. Why are you confused? The absolute values is always non-negative.

2) |6 - 7x| > 27

Here's what I did:

6 - 7x > 27 or 6 - 7x < -27

6 - 27 > 7x or 6 + 27 < 7x

-21 > 7x or 33 < 7x

-3 > x or 33/7 < x

Therefore, x < -3 or x > 4 ⁵/₇, so the solution is S = (-infinity. -3) U (4 5/7, infinity).

Am I even close on this solution?

Why do you doubt? Check some things and see.

Try x = 0. That should NOT work.

|6 - 7(0)| = |6 - 0| = |6| = 6 and that is NOT greater than 27. Good.

Try x = -5. That SHOULD work.

|6 - 7(-5)| = |6 + 35| = |41| = 41 and that IS greater than 27. Good.

You can convince yourself and maybe learn something in the doing of it.