What is the number of solutions of x = |x-|30-2x||?

|x-|30-2x|| = x if the blue portion is equal to x or -x.

Case 1: x-|30-2x| = x
0 = |30-2x|
The equation above is valid if 30-2x=0:
30-2x = 0
30 = 2x
15 = x

Case 2: x-|30-2x| = -x
2x = |30-2x|

Case 2a:
2x = 30-2x
4x = 30
x = 30/4 = 7.5

Case 2b:
-2x = 30-2x
0 = 30
Not valid.

Two solutions:
x=15 and x=7.5


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solve for expression ; \(x = |x-|30-2x||\)
we get x= 15 and 15/2
total solutions ; 2
IMO C

Since x = |y| => x>0

Let us divide x>0 in 2 regions

Region 1 : 0<x<15
x = |x- (30-2x|
x= |3x - 30| = 3 |x-10|
Suppose 15>x>10
x = 3x -30
x=15 Solution 1.
Now suppose 0<x<10
x = 30 -3x
4x = 30
x = 7.5 Solution 2

Region 2: x>15
x=|x-(2x-30)|
x= |30 -x|
Suppose x>30
x = x-30 No Solution
Now suppose 15<x<30
x = 30 -x
x=15 Which is same as Solution 1.

We see that there are only 2 solutions.

IMO C
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Can this question be solved without dividing number line into seperate regions?
If yes, please provide way to solve such questions.
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Case 1
if x=<10, x<|30-2x| and 30-2x>0

x=30-2x-x
x=7.5

Case 2
if 10<x=<15, x>|30-2x| and 30-2x>0

x=x-30+2x
x=15

Case 3
if 15<x=<30, x>|30-2x| and 30-2x<0

x= x- 2x+30
x= 15 (not possible)

Case 4
if x>30, x<|30-2x| and 30-2x<0

x=2x-30-x
Not possible

There are 2 values of x possible

How do you decide exact regions in a nested modulus equation without opening some inner modulus first?
e.g In x<10, how do you identify 10?
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I didn't explicitly mention it.
You gotta do couple more steps to divide the regions.

When x>|30-2x|
i) for x>15
x>2x-30
x<30

ii) for x<15
x>30-2x
x>10

Similarly you can find other cases.

hi.. can you please elaborate how you directly decided the the values of the two regions??

I tried to open |2x-30| first

Posted from my mobile device
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=>

The equation \(x = |x-|30-2x||\) is equivalent to\(x = |x-2|x-15||\)

If \(x ≥ 15\), then \(x = |x-2|x-15||\) or \(x = | x – 2(x-15) | = | x – 2x + 30 | = | -x + 30 | = | x – 30 |\)

If \(x ≥ 30\), then \(x = | x – 30 | = x – 30\) or \(0 = -30\), which doesn’t make sense.

If \(15 ≤ x < 30,\) then \(x = - ( x – 30 ) = -x + 30\) or \(2x = 30.\) It follows that \(x = 15.\)

If \(x < 15\), then \(x = |x-2|x-15||\) is equivalent to \(x = | x + 2(x-15) | = | x + 2x - 30 | = | 3x - 30 | = 3| x – 10 |\)

If \(10 ≤ x < 15\), then \(x = 3| x – 10 | = 3(x-10) = 3x -30,\) so, \(2x – 30 = 0.\) It follows that \(x = 15,\) which is not a solution since \(10 ≤ x < 15.\)

If \(x < 10,\) then \(x = 3| x – 10 | = -3(x-10) = -3x + 30\) and \(4x = 30.\)

So, \(x = 7.5.\)

Thus, there are two solutions: \(7.5\) and \(15.\)

Therefore, the answer is C.
Answer: C
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