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Futuristic
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Another number line problem... 25 Jun 2006, 17:44
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|x-1| + |x-3| + |x-5| >=9

How do I solve for x?



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Another number line problem... 25 Jun 2006, 17:58
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0 > x > 5
By substituting values.
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Futuristic
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Another number line problem... 25 Jun 2006, 19:22
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sgrover
0 > x > 5
By substituting values.
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Sorry, this sounds completely incorrect.

I know that if you were to expand all of these you would get 2^3 = 8 inequalities to solve, I want to know what the shortcut is.
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Another number line problem... 25 Jun 2006, 21:11
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Take the following ranges:

x < 1; 1 <= x < 3; 3 <= x < 5 and x >= 5

You should get x >= 6 in the fourth range and x >= 2 in the second range. You should also get x >= 0 in the first range.

So a combination could be 0 <=x < 1; 2 <= x <3 and x >=6.
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Another number line problem... 30 Jun 2006, 11:23
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Hong Hu answered this beautifully for me, and I will solve it again here since I believe it may help us all:-

|x-1| + |x-3| + |x-5| >=9

1,3,5 are the turning points.[ Turning points are the points at which the actual sign inside the absolute value changes. So for |x-1|, the turning point is 1, since anything below this value means x-1 is -ve and anything above this value causes x-1 to be +ve.]

Based on the turning points we can calculate the ranges we must look at:-

1. If x <1 then

-(x-1) -(x-3) - (x-5) >=9
-x+1 -x+3-x+5 >=9
-3x >=0
x <=0

[What we've done above is to evaluate the actual sign of the expression within each absolute value expression for x <1. We've then removed the abs. value and solved it as a linear inequality]

We need to make sure that the solution is consistent with the value of x chosen. Basically x<=0 must belong in the range x<1. In this case it does so x<=0 is one of the solutions.

2. 1<=x<3

(x-1) -(x-3) -(x-5) >=9
x-1-x+3-x+5>=9
-2x+8>=9
-2x>=1
x<= -1/2 DOES NOT FALL INTO THE RANGE

3. 3<=x<5

(x-1) +(x-3) -(x-5) >=9
x+1 >=9
x>=8 DOES NOT FALL INTO THE RANGE

4. 5<=x

x-1 + x-3 + x-5 >=9
3x - 9 >=9
x>=6 VALID FOR THE RANGE

Therefore the range is x>=6, x<=0
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Another number line problem... 30 Jun 2006, 17:39
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I agree, it is beautiful. I did it the 2^3-way, which is too long...
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Another number line problem... 30 Jun 2006, 20:05
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Futuristic
Hong Hu answered this beautifully for me, and I will solve it again here since I believe it may help us all:-

|x-1| + |x-3| + |x-5| >=9

1,3,5 are the turning points.[ Turning points are the points at which the actual sign inside the absolute value changes. So for |x-1|, the turning point is 1, since anything below this value means x-1 is -ve and anything above this value causes x-1 to be +ve.]

Based on the turning points we can calculate the ranges we must look at:-

1. If x <1 then

-(x-1) -(x-3) - (x-5) >=9
-x+1 -x+3-x+5 >=9
-3x >=0
x <=0

[What we've done above is to evaluate the actual sign of the expression within each absolute value expression for x <1. We've then removed the abs. value and solved it as a linear inequality]

We need to make sure that the solution is consistent with the value of x chosen. Basically x<=0 must belong in the range x<1. In this case it does so x<=0 is one of the solutions.

2. 1<=x<3

(x-1) -(x-3) -(x-5) >=9
x-1-x+3-x+5>=9
-2x+8>=9
-2x>=1
x<= -1/2 DOES NOT FALL INTO THE RANGE

3. 3<=x<5

(x-1) +(x-3) -(x-5) >=9
x+1 >=9
x>=8 DOES NOT FALL INTO THE RANGE

4. 5<=x

x-1 + x-3 + x-5 >=9
3x - 9 >=9
x>=6 VALID FOR THE RANGE

Therefore the range is x>=6, x<=0
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Very nicely done, but one question..

2. 1<=x<3

(x-1) -(x-3) -(x-5) >=9
x-1-x+3-x+5>=9
-2x+8>=9
-2x>=1
x<= -1/2 DOES NOT FALL INTO THE RANGE

shouldn't this be?
-x+7>=9
x<=-2 DOES NOT FALL INTO THE RANGE
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Another number line problem... 07 Jul 2006, 08:00
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Yes, it's a small error, but it does not influence the final result, because -2 also does not fall into the range.

It should be like this:
(-@, 1> .......... x<=0
(1,3> ............. x<=-2
(3,5> ............. x>=8
(5,+@> .......... x>=6

Result x is from interval <-@,0> u <6,+@)
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Another number line problem... 07 Jul 2006, 08:13
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|x-1| + |x-3| + |x-5|

Not as beautiful, but effective:

We're told that the distance from x to 1 added to the distance from x to 3 added to the distance from x to 5 must be greater than 9. For any x between 1 and 5, this will not happen!

So for x less than 1 we get 9-3x>=9 i.e 0>=x
For x greater than 5 we get 3x-9>=9 i.e x>=6

So x>=6 or 0>=x



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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