# inequality

• Aug 27th 2009, 04:36 AM
thereddevils
inequality
Solve the inequality :

$|x-3|-|x-1|\leq2$

This is what i did :

squarring both sides .

$(x-3)^2-2(x-3)(x-1)+(x-1)^2\leq4$

$x^2-6x+9-2x^2+8x-6+x^2-2x+1-4\leq0$

then

$0\leq0$ ???

What does that mean ? so wat 's the range of x ?
• Aug 27th 2009, 06:03 AM
Krahl
your solution $0 \leq 0$ means that the equation is valid for all x.

notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get $-2\leq2$ or $2\leq2$ because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

The way i would solve these is to solve the equations;

1. $-(x-3)-(x-1)\leq2$
2. $(x-3)-(x-1)\leq2$
3. $-(x-3)+(x-1)\leq2$
4. $(x-3)+(x-1)\leq2$

and take the union of the solutions. because if x solves any one of these then it solves the original inequality.
• Aug 27th 2009, 08:05 AM
red_dog
Use the inequality:

$|a|-|b|\leq |a-b|, \ \forall a,b\in\mathbb{R}$

Then $|x-3|-|x-1|\leq |x-3-x+1|\Rightarrow |x-3|-|x-1|\leq 2$

Therefore the inequality holds for every real x.
• Aug 27th 2009, 08:07 AM
thereddevils
Quote:

Originally Posted by Krahl
your solution $0 \leq 0$ means that the equation is valid for all x.

notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get $-2\leq2$ or $2\leq2$ because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

The way i would solve these is to solve the equations;

1. $-(x-3)-(x-1)\leq2$
2. $(x-3)-(x-1)\leq2$
3. $-(x-3)+(x-1)\leq2$
4. $(x-3)+(x-1)\leq2$

and take the union of the solutions. because if x solves any one of these then it solves the original inequality.

Thanks but how did you get the 4 equations ? Can u explain that to me ?
• Aug 27th 2009, 10:53 AM
nikhil
correction
Quote:

Originally Posted by Krahl
your solution $0 \leq 0$ means that the equation is valid for all x.

notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get $-2\leq2$ or $2\leq2$ because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

The way i would solve these is to solve the equations;

1. $-(x-3)-(x-1)\leq2$
2. $(x-3)-(x-1)\leq2$
3. $-(x-3)+(x-1)\leq2$
4. $(x-3)+(x-1)\leq2$

and take the union of the solutions. because if x solves any one of these then it solves the original inequality.

you have not included the interval of x for which you opened the mod and you said to take the union of solution obtained from the inequations. this can be misleading sometimes for example
|x-1|-|x-3|>=4
let 1<=x<3 so we have
x-1+(x-3)>=4 or
2x-4>=4 or
x>=4 which contradicts with our interval
you can check this by putting x=4.you will get LHS=2 which is not >=4 so such solution should be discarded.
but if you will not write the interval of x for which you opened mod and took union of the intervals obtained then you might get wrong result.
• Aug 27th 2009, 11:02 AM
nikhil
hi thereddevils!!!
|x|=x if x>=0
=-x if x<0
|x-3|-|x-1|>=2
consider two points 1 and 3
take case i.e
1) x>=3
2) 1<=x<3
3)x<1
you will get 3 inequations. solve them and take the union of solution.
discard the solution if its interval does not match your interval i.e you took x>1 but your corresponding inequation gave x<3.(though you will not have this situation in this problem.)
• Aug 27th 2009, 11:07 AM
nikhil
TIP
whenever equation is like |x-a|+|x-b|+|x-c|....and so on >=n
then draw points a,b,c.... on number line.
now start taking interval i.e x<a then a<=x<b then b<=x<c and so on
this will help to solve inequality in an organised way