# Absolute value inequality?

This is a question of algebra II.

Solve the compound inequality below.

|x-3|<4 and |x+2|>8.

I think this answer is 6<x<7. Is this right?Please teach me.

### 4 Answers

- ted sLv 71 month ago
| x - a | < b is all points within b units of a.....| x - 3 | < 4 is all values within 4 units of 3...( - 1, 7).......| x + 2 | > 8 are all values at least 8 units AWAY from - 2.... ( -∞ , - 10 ) U ( 6 , ∞)

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- 1 month ago
Yes, I'm pretty sure that this is right. The first one, |x-3|<4 is the same as solving ±(x-3)<4 which is -x+3<4 and x-3<4 which comes out to -1<x<7. The next one uses the same concept, ±(x+2)>8 which is x+2>8 and -x-2>8 which comes out to -10>x>6. X can't be less than 10 since that doesn't correlate with the other one, and it can't be greater than -1 up until it is greater than 6. It can't be greater than 7 so 6<x<7.

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- mizooLv 71 month ago
|x-3|<4 and |x+2|>8

1. |x-3|<4

2. |x+2|>8

1.

|x-3|<4

-4 < x - 3 < 4

-4 + 3 < x + < 4 + 3

-1 < x < 7

2. |x+2|>8

x + 2 > 8 or x + 2 < -8

x > 6 or x < -10

x < -10 or 6 < x

The question is to find "1 and 2":

{-1 < x < 7} AND {x < -10 or 6 < x} => 6 < x < 7

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- 1 month ago
|x - 3| < 4 translates to x - 3 > -4 and x - 3 < 4

-4 < x - 3 < 4

-4 + 3 < x < 4 + 3

-1 < x < 7

|x + 2| > 8 translates to x + 2 < -8 and x + 2 > 8

x + 2 < -8

x < -10

x + 2 > 8

x > 6

The only solution set that is included in both is 6 < x < 7

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