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by Frankenstein » Sun Jun 12, 2011 5:13 am
Hi,
Let A(x,y) be a point on the coordinate plane.
|x-1| is the normal distance of A from from the vertical line x=1
|x+2| is the normal distance of A from from the vertical line x=-2
So, for any -2<x<1, |x-1|+|x+2| will be the distance between lines x=1 and x+2=0. This distance is 3 units.
So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

(or)
|x-1|+|x+2|=5. Let a = x-1. So, |a|+|a+3| = 5 => |a+3| = 5 -|a| =>|a+3|^2 = (5-|a|)^2
i.e. a^2+6a+9 = a^2 -10|a| +25 =>6a+10|a| = 16.
case-1: If a>0, |a| = a. So, 16a=16 =>a=1
case-2: If a<0, |a| = -a. So, 6a-10a = 16 =>a = -4
So, a=1 or -4
x = a+1. So, x is 2 or -3.
Last edited by Frankenstein on Sun Jun 12, 2011 5:17 am, edited 1 time in total.
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by vinayreguri » Sun Jun 12, 2011 5:17 am
I solved this other way around


x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.

Will this work for others too.
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by Anurag@Gurome » Sun Jun 12, 2011 5:20 am
vinayreguri wrote:solve for x where |x-1|+|x+2|=5
For this problem the expression |x - 1| or |x + 2| changes their sign at x = -2 and x = 1. Hence, we have three ranges of values of x for which either of the expression changes sign.

1. x < -2 :
  • |x - 1| = -(x - 1) and |x + 2| = -(x + 2)
    --> |x - 1| + |x + 2| = 5
    --> -(x - 1) - (x + 2) = 5
    --> x = -3
1. -2 ≤ x < 1 :
  • |x - 1| = -(x - 1) and |x + 2| = (x + 2)
    --> |x - 1| + |x + 2| = 5
    --> -(x - 1) + (x + 2) = 5
    --> No Solution
1. 1 ≤ x:
  • |x - 1| = (x - 1) and |x + 2| = (x + 2)
    --> |x - 1| + |x + 2| = 5
    --> (x - 1) + (x + 2) = 5
    --> x = 2
Hence, the solution for the equation is x = -3 and x = 2.
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by Frankenstein » Sun Jun 12, 2011 5:24 am
vinayreguri wrote:I solved this other way around


x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.

Will this work for others too.
Hi,
This method is actually incorrect because you have effectively taken |(x-1)+(x+2)| = 5, which is incorrect.
|x-1|+|x+2| and |(x-1)+(x+2)| will be equal only if (x-1)&(x+2) are of the same sign.
Consider : |x-1|+|x+2| = 3.
Using you method: x-1+x+2=3 or x-1+x+2=-3.
Solving you get x = 1 or x = -2. But actually many more values of x(any value between -2 and 1) satisfy this equation.
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by Anurag@Gurome » Sun Jun 12, 2011 5:32 am
vinayreguri wrote:x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.
Will this work for others too.
This is neither correct nor applicable to other problems.

Removing absolute values from the equation will result in four equations.
  • 1. (x - 1) + (x + 2) = 5
    2. -(x - 1) + (x + 2) = 5
    3. (x - 1) - (x + 2) = 5
    4. -(x - 1) - (x + 2) = 5 --> (x - 1) + (x + 2) = -5
Though 1 and 4 only will result in some feasible solution, 2 is also an interpretation of the given equation. But 2 has no solution. And from mechanical view, 3 is also an interpretation of the given equation but it is impractical.

For other problems, we have to consider all four of them and check which one is possible and which one is giving some feasible solution.
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by [email protected] » Sun Jun 12, 2011 6:50 am
Anurag could you please explain me the 1st explanation, especially the following line:

So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

I did not get this point...
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by Frankenstein » Sun Jun 12, 2011 7:25 am
[email protected] wrote:Anurag could you please explain me the 1st explanation, especially the following line:

So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

I did not get this point...
Hi,
I think you have understood up to the part that for any value of x between -2 and 1 , the sum of distances is zero.
Now, consider a point with x>1. Let the distance of this point from x=1 be a, The distance of this point from x=-2 is a + (distance between x=1 and x=-2) = a+3
So, total is a+(a+3) = 2a+3. We need that sum to be equal to 5. So, 2a+3=5 =>a=1.
So, x should be at a distance from x=1 to the right of it. So, x=2.
Similarly we will solve for x to the left of x=-2.
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