Absolute values help?

Question

what is the answer to this question and how do you show ur work

-2 I x-15 I + I x-15 I < 27

Thanks for the help guys but where does the one come from

-2 I x-15 I + I x-15 I < 27
factor out the |x-15|
|x - 15|(-2 + 1) < 27

the one?

Answer 1

-2 I x-15 I + I x-15 I < 27
factor out the |x-15|
|x - 15|(-2 + 1) < 27
|x - 15|(-1) < 27
when you divide both sides by (-1) remember to flip the sign.
|x - 15| > -27
That translates to:
-27 < x - 15 < 27
-27 + 15 < x < 27 + 15
-12 < x < 42

============
"Where does the one come from?"

There is always an "invisible" one in front of variables if there isn't a different number there.

Ex:
-2 |x-15| + (1) |x-15| < 27

Answer 2

For any x, I x-15 I>=0
so
I x-15 I>- 27
- I x-15 I<27
(-2+1) I x-15 I< 27
-2 I x-15 I + I x-15 I < 27

Answer 3

Ques: -2 I x-15 I + I x-15 I < 27 ans => -2 I x-15 I + I x-15 I < 27
factor out the |x-15|
|x - 15|(-2 + 1) < 27 -|x-15| < 27
Now, multiply through out by -1, then |x-15| > -27
This should be equal to (x-15) = -27 and -(x-15) = -27
Which gives, x = 12 and x = 42

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Answer 4

You can treat | x-15 | as just some number y. Then

-2y + y < 27
=> -y < 27
=> y > - 27
=> | x-15 | > -27

| x - 15| is always >= 0 and therefore > -27 for all x. So any number is a solution.

Answer 5

Before we get started, we're going to get one of two cases:
1) |y| > c, for some positive number c.

This is an inequality that translates to an OR. This means
(y > c) OR (y < -c)

2) |y| < c, for some positive number c.

This translates to an AND. This means
(y < c) AND (y > -c). This can also be expressed as
-c < y < c.

-2 |x - 15| + |x - 15| < 27

Group together the -2 |x - 15| + |x - 15| the same way you would group -2y + y.

What is -2y + y? The answer is -y. But in our case, our "y" is equal to |x - 15|, so we get

- |x - 15| < 27

Divide both sides by (-1). This eliminates the negative, and flips the inequality sign.

|x - 15| > -27

Hmmm... we did not get either case #1 or 2. The absolute value of a number is *always* greater than or equal to 0. This makes this inequality always true (because the absolute value of *any* real number, including negative ones, is always going to be greater than a negative number).

Unless I'm mistaken, the answer is all real numbers.

Answer 6

meaning 2x-5 must be decrease than 9 and additionally extra suitable than -9. So clean up 2x - 5 < 9 AND 2x - 5 > -9, with the help of the techniques we've shown you in previous solutions. you're able to finally end up with x between -2 and 7, i.e. -2 < x < 7 word emy is the only different answerer who's right suited (up till at last now)

Answer 7

-2(|x-15|) + (|x-15|) <27

The modulus take only positive values.

-2(x-15)+(x-15)<27

Take out (x-15) as a common factor.

(x-15)(-1) <27
-x+15 <27

subtract both sides by 15 and get,

-x+15-15<27-15
-x<12

Multiply both sides by negative.

x>-12


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Answer 8

when x>15
-2 I x-15 I + I x-15 I < 27 is equal to
-2 ( x-15 ) + ( x-15 ) < 27
-( x-15 ) < 27
-x +15 < 27
-x<12
or x>-12
thus x>15

when x<15
-2 I x-15 I + I x-15 I < 27 is equal to
2 ( x-15 ) - ( x-15 ) < 27
x - 15 < 27
or x<12
thus x<12

to answer sych questions break the abs value function and then solve

Answer 9

The answer is: x E {(-inf, -12) U (-12,+inf)}
in other words x needs to be different to -12


Procedure:

-I x-15 I < 27 (factorized)
I x-15 I >- 27
x-15<-27 then x<-27 +15, x<-12
and
x-15>-27 then x>-27+15, x>-12

x E {(-inf, -12) U (-12,+inf)}

Answer 10

-2x - 30 + x + 15 < 27
-x - 15 < 27
x > - 42

-2x - 30 - x - 15 < 27
3x + 45 > - 27
3x > - 72
x > - 24

-2 (- x + 15) + x - 15 < 27
2x - 30 + x - 15 < 27
3x < 72
x < 24

2x - 30 - x + 15 < 27
x - 15 < 27
x < 42

- 42 < x < 42