Solve the absolute value equation:?

Question

3|x+2|-2>7

Please give the final answer in a interval form:

(example,example) U (example,example)

I tried using quickmath.com but it doesn't work for me.

Answer 1

x>1 or x<-5

or as you want it:

x is in (-∞,-5)U(1,∞)

How to see this:

3|x+2|-2>7
3|x+2|>9
|x+2|>3
x+2>3 or -(x+2)>3
x>1 or x+2<-3
x>1 or x<-5

Answer 2

x>1

Answer 3

x>1

Answer 4

Some fo the other answers are correct but the form of the answer needs to be modified to be in interval form.

3|x+2|-2>7 Given
3|x+2|>9 Add 2 to each side
|x+2|>3 Divide each side by 3.
now split into two problems.
(x+2)>3 -(x+2)>3
this is because either a + or - result in parentheseis will be treated as a positive answer within absoulute value.
x>1 Subtract 2 -x-2>3 Distribution of - sign
-x>5 Add 2 to each side.
x<-5 Multiply by -1
(changes Direction of
greater than to less than)
Therefore we have x>1 and x<-5.


Interval notation: (-infinity,-5) U (1,Infinity)


Which is saying in english; Any number from negative infinty to -5 and an nuber from 1 to infinity, will work.

Answer 5

3|x + 2| - 2 > 7
3|x + 2| > 9
|x + 2| > 3
x + 2 > 3 or x + 2 < -3
x > 1 or x < -5

Answer 6

Start with simple algebra:

3|x+2| - 2 > 7
3|x+2| > 9
|x+2| > 3
which means: (x+2) > 3 or (x+2) < -3
x > 1 or x < -5

(-infinity, -5)U(1, infinity)

Answer 7

3|x+2|>9
|x+2|>3
x>1 OR x<-5

Answer 8

3|x+2| - 2 > 7
3|x + 2| > 9
|x + 2| >3
x + 2 > 3 OR x + 2 < -3
x > 1 OR x < -5

(-∞,-5) U (1, ∞)

Answer 9

3|x+2|-2>7 =>
3|x+2|>9 =>
|x+2|>3 =>
x+2>3 0r x+2<-3
so do like this
x>1 U x<-5
and in the end
(+∞,1) U (-5,-∞)

Answer 10

3|x + 2| - 2 > 7
3|x + 2| > 9
|x + 2| > 3

x + 2 > 3
x > 1

-x - 2 > 3
-x > 5
x < -5

x > 1 or x < -5

( - infinity symbol, -5) U (1, + infinity symbol)