Question

Help with the following ecuation:

|2|x-a|-bx|=|2b-|x-a||

| | means absolite value
Where b<0 and -2 12 years ago

Answer 1

wow this is a mess

Answer 2

what are you trying to find though

Answer 3

We have to solve for x.
So, seems possible solve for x in terms of b and a.
And then we are suposed to define intervals for x, given the conditions for b and a.
I think.

Answer 4

ok on the right hand side if you take that part of the equation out of the absolute values it becomes|2|x-a|-bx| =2b+|x-a| correct?

Answer 5

now we do that for the left side too so it becomes 2|x-a|+bx=2b+|x-a|

Answer 6

Should't be 2|x-a|+bx=2b+|x-a| ?

Answer 7

2|x-a|-bx=2b+|x-a| *

Answer 8

no because the -bx was in one of the absolute values once anything comes out of an absolute value it instantly becomes positive.

Answer 9

ok

Answer 10

look at your original problem if you want to make sure

Answer 11

now we still have 2 more absolute values to take care of.
so can you tell me what the problem would look like if you took those out.

Answer 12

2|x-a|+bx=2b+|x-a|
|x-a|=2b-bx

That means
x-a=2b-bx
x(1+b)=a+2b
x= (a+2b)/(1+b)

or

a-x=2b-bx
x(b-1)=2b-a
x=(2b-a)/(b-1)

I think its that

Answer 13

not quite

Answer 14

your second answer is close

Answer 15

this is what i got
2|x-a|+bx=2b+|x-a|
2(x+a)+bx=2b+x+a
2x+2a+bx=2b+x+a
2x+bx-x=2b-a
x(2+b-1)=2b-a
\[x=\frac{ 2b-a }{ 2+b-1 }\]

Answer 16

does this make sense?

Answer 17

hmm, i don't know, i thought that since the definition of absolute value is
|x|={ x if x>0
{-x if x ≤0
we should find more than one solution for this ecuation. At least to define x in terms of a and b.
And i dont know why you only take one part of the definition.

Answer 18

let me look at it again i might have overlooked something

Answer 19

hmmm crap i might have read it wrong

Answer 20

i cant tell what exactly went wrong sorry

Answer 21

from my result b cannot = -1 but thats all i can get from it

Answer 22

sorry that i couldnt be much help

Answer 23

There are 4 candidate solutions that will become apparent. Only 1 of the 4 will end up working, but I will demonstrate that. The solution that does work I will show works for the given interval and I will show that by proving the general case: all "a" and "b" for the given intervals for each. The other 3 can be shown not to work by simply selecting convenient values for "a" and "b".

So, by eliminating the outer || :

2|x - a| - bx = 2b - |x - a| or -2b + |x - a|

The first (P) of these will split into cases 1 & 2, the second (Q) will split into cases 3 & 4. I'll show eventually that only case 1 will be a solution.

Taking P: |x - a| = bx - 2b
x - a = bx - 2b (case #1) or -bx + 2b (case #2)

case #1 : x = (2b - a)/(b - 1)
case #2 : x = (2b + a)/(b + 1)

Taking Q: 3|x - a| = 2b + bx
3x - 3a = 2b + bx (case #3) or 3x - 3a = -2b - bx (case #4)

case #3 : x = (2b + 3a)/(3 - b)
case #4 : x = (3a - 2b)/(3 + b)

Now, proving a case DOESN'T work is easier, so I'll continue by selecting suitable, convenient values for "a" and "b" for cases #2, #3, and #4. You can plug them in and verify that the equality will not hold. This will be sufficient for these cases because the equality is supposed to hold over the range of all values for "a" and "b". So for each, we only need one counter-example:

For case #2, select b = -2 and a = 0
For case #3, select b = -1 and a = 0
For case #4, select b = -2 and a = 0

Case #1 is the only one that works, but we need to show that for the general case: substituting back x = (2b - a)/(b - 1) into the original equation, the left side becomes:\[\left| 2\left| \frac{ 2b - a }{ b - 1 } - a \right| - b \left( \frac{ 2b - a }{ b - 1 } \right) \right| = \left| 2\left| \frac{ 2b - ab }{ b - 1 } \right| - \frac{ b }{ b - 1 }(2b - a) \right|\]\[= \left| 2\left( \frac{ b }{ b - 1 } \right)\left| 2 - a \right| - \left( \frac{ b }{ b - 1 } \right)(2b - a) \right|\]We can pull that b/(b-1) out of the || because it is always positive\[= \left( \frac{ b }{ b - 1 } \right)\left| (4 - 2a) + (a - 2b) \right|\]Likewise, "2 - a" is never negative.\[= \left( \frac{ b }{ b - 1 } \right)(4 - a - 2b)\]The right side will become the same as the immediate above:\[\left| 2b - \left| \frac{ 2b - a }{ b - 1 } - a \right| \right| = \left| 2b - \left| \frac{ 2b - ab }{ b - 1 } \right| \right|\]\[= \left| 2b - \left( \frac{ b }{ b - 1 } \right)\left| 2 - a \right| \right| = \left| 2b - \left( \frac{ b }{ b - 1 } \right)(2 - a) \right|\]\[= \left| 2b + \left( \frac{ b }{ b - 1 } \right)(a - 2) \right| = \left| \frac{ 2b ^{2} - 2b + ab - 2b }{ b - 1 } \right|\]\[= \left( \frac{ b }{ b - 1 } \right)\left| 2b - 4 + a \right| = \left( \frac{ b }{ b - 1 } \right)\left| -(4 - a - 2b) \right|\]\[= \left( \frac{ b }{ b - 1 } \right)(4 - a - 2b)\]

Answer 24

I just saw that "taking P" is "taking Q" and vice versa, I just copied it out a little out of order, but I'm sure you'll see that. @juarismi

Answer 25

Great, many thanks. Now i see it!!.

But just one last thing.
if the only solution is that works is x =(2b - a)/(b - 1) and b<0 and -2<a≤2
That means that x will always be ≤2 ¿?

Answer 26

Ok, thank you so much.

Answer 27

uw! Great working with you and I hope this helps a lot!

Answer 28

@juarismi , I think you might be right that x <= 2

Answer 29

Me too.
I can't find a case where x>2.

Answer 30

Yes, you are right because if we take "a" as large as possible (a = 2) and "b" as large as possible (b approaching 0), the largest we get is not even as large as:

[2(0) - (2)] / (0 - 1) = 2

Answer 31

Fine.
Thanks

Answer 32

I have a better answer than what I gave:

Maximize "a" to be "2"

y = (2b - 2) / (b - 1) = 2

So, when "a" = 2, all "b" give x = 2

Answer 33

Do you agree? That way, because the denominator is always negative, we want to maximize the absolute value of the numerator by making it "more negative" and that is by making "a" as large as possible so we can subtract the max. @juarismi

Answer 34

I agree. Absolutely.
So i think the answer can be write as:
is x =(2b - a)/(b - 1)
and x is <=2.

Answer 35

Yes. I think we nailed this one!