Math Is Fun Forum

adam_dsutton

Member

Registered: 2007-09-12

Posts: 10

Inequality Help Please

Solve the following inequality:

lx-1l+lx-2l ≥5

I am struggling how to work this out, I tried case analysis but I don't think i'm doing it correctly, all help is appreciated.
Also how do you write the modulus lines using latex?

Thanks, Adam:)

Last edited by adam_dsutton (2007-10-07 04:08:13)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Inequality Help Please

There are three regions to consider: (i) x < 1, (ii) 1 ≤ x ≤ 2, and (iii) x > 2.

(i): Both x−1 and x−2 are negative, so |x−1| + |x−2| = 1−x + 2−x = 3−2x.

(ii): x−2 is still not positive but x−1 is no longer negative, so |x−1| + |x−2| = x−1 + 2−x = 1. This is not ≥ 5, so the solution doesn’t like in this region.

(iii): Now both x−1 and x−2 are positive, so |x−1| + |x−2| = x−1 + x−2 = 2x−3.

Hence the solution is 3 − 2x ≥ 5 or 2x−3 ≥ 5

i.e. x ≤ −1 or x ≥ 4.

Last edited by JaneFairfax (2007-10-07 04:43:31)

adam_dsutton

Member

Registered: 2007-09-12

Posts: 10

Re: Inequality Help Please

Thanks Jane, but if say lx-1l<0, why must it be equal to -x+1???
Also how do you determine the considered regions, I can understand graphically but not algebraically.

HallsofIvy

Guest

Re: Inequality Help Please

Thanks Jane, but if say lx-1l<0, why must it be equal to -x+1???
Also how do you determine the considered regions, I can understand graphically but not algebraically.

|x-1| is NEVER less than 0!  I think you mean x-1< 0 or what Jane meant by x< 1.  If a<0 then |a|= -a so if x-1< 0, then |x-1|= -(x-1)= -x+ 1.  You determine those regions by looking at each of the absolute values.  Where is the quantity inside the absolute value equal to 0?  That's where the absolute value suddenly changes "formula".

adam_dsutton

Member

Registered: 2007-09-12

Posts: 10

Re: Inequality Help Please

Okay, that makes things alot clearer, thankyou both.