Trouble with very basic algebra question

[mhazelm]

Homework Statement

Determine all x in R such that the following hold:

1) (x+4/x-2) < x
2) |x+4/x-2| < x
3) |2x| > |5-2x|

Homework Equations

We have the triangle inequality, |a + b| $$\leq$$ |a| + |b|, which also implies the relation ||a| - |b|| $$\leq$$ |a-b|.

Also, relations such as: |x| < a implies -a < x < a, etc.
|ab| = |a|*|b|

The Attempt at a Solution

I feel very bad asking about these, I'm sure I should have learned this sort of thing earlier than analysis I (call it my weak spot). Anyway, here is what is bothering me:

1) (x+4/x-2) < x --> 0 < x - (x+4/x-2) --> 0 < [x(x-2) - (x+4)]/(x-2)
--> 0 < (x^2 - 3x + 4)/2. Now if we use the quadratic formula to find the roots of the numerator, we get (3 +/- $$\sqrt{9-16}$$ )/2, so the roots are complex. So am I to assume that there are no x in R such that this inequality is true? I can accept that, but then the next one is the same, except the absolute value. So for that one:

2) |x+4/x-2| < x --> -x < (x+4/x-2) < x --> -x(x-2) < x + 4 < x(x-2)
--> -x^2 + x -4 < 0 < x^2 - 3x - 4. Now the lhs doesn't factor, but the quadratic formula results in complex roots again, and the rhs roots, which would be x=4 and x=-1, pose problems: if you substitute x = 4 or x = -1 into the inequality you get 4 < 4 and -1 < -1 which is obviously never true! So - do I have to assume again that there are no real x in R such that this inequality holds? I'm getting worried that all my problems (there are more very similar to these) seem to have no real x possible! It makes me think I am doing something wrong...

3) |2x| > |5-2x| --> 2|x| > |5-2x| >= ||5| + |-2x|| --> I'm not sure what to do next! There is obviously some very simple answer that I should know. Maybe I am just having a slow night (lack of sleep does that!). I thought I'd be using the triangle inequality, but then, it's a difference and not a sum... so I'm just not sure what I'm supposed to do.

Any tips for those of us who are SLOW SLOW SLOW with this type of inequality? I feel like I can handle proofs pretty well, with sup/inf concepts, proving other similar sorts of things, and have always been fine at abstract and linear algebra. But concrete math problems with numbers, whoa! No good at that... :grumpy:

 

[Chemistry101]

Are you trying to graph them on a number line and put them into inequalities?

 

[mhazelm]

I'm just supposed to determine which real x's would satisfy them. I'm not sure what level of rigor is expected - I'm not sure if he just wanted to review a few computational examples of the basic real number theorems or if we are supposed to be very rigorous in showing how we make each step. I guess we are trying to put them into inequalities. I just haven't seen this sort of problem for about 7 years!

 

[Chemistry101]

I am trying to help out lol, let me read up a little more and I will get back to you.

 

[mhazelm]

thanks! :-D

 

[Math Man900]

Homework Statement

Determine all x in R such that the following hold:

1) (x+4/x-2) < x
2) |x+4/x-2| < x
3) |2x| > |5-2x|

The first one is pretty interesting. first u simply subtract "x" from each side to get
(x+4)/(x-2)-x<0
now, to find a common denominator
((x+4)-x(x-2))/(x-2)<0
now, you CANT multiply each side by "x-2" because u don't know if x>2 or x<2 yet.
Instead, just simplify some more
(x+4-x^2+2x)/(x-2)<0
(-x^2+3x+4)/(x-2)<0
now, it is easier (but not necessary) to multiply each side by -1 (and flipping the < to a >)
(x^2-3x-4)/(x-2)>0
and, now comes the interesting part!
(x+1)(x-4)/(x-2)>0
alright, notice how the left hand side is only true, when the product of (x+1) and (x-4) is divided by (x-2).
Notice, however, that the same "greater than zero" rules apply for products as they do for quotients. what i mean by this is: the previous equation is also true when
(x+1)(x-4)(x-2)>0
Now, all you have to do is find "for what values of x is (x+1) and (x-4) and (x-2) all positive?"
Once you find that, you have to go through the different cases when 2 of them are negative, and the last other one is positive.
The solution ends up being:

-1<x<2 or x>4

this solution can be easily verified, if you check out each case of "x" from x=-2 all the way to x=5

 

[Math Man900]

2) |x+4/x-2| < x --> -x < (x+4/x-2) < x --> -x(x-2) < x + 4 < x(x-2)
--> -x^2 + x -4 < 0 < x^2 - 3x - 4. Now the lhs doesn't factor, but the quadratic formula results in complex roots again, and the rhs roots, which would be x=4 and x=-1, pose problems: if you substitute x = 4 or x = -1 into the inequality you get 4 < 4 and -1 < -1 which is obviously never true!

You were very close in this one! when u plugged in x=4 and x=-1, you found the bounds of what x can or cannot be. All you need to do is try 3 situations. x<-1 , -1<x<4 , 4<x.
If you try the numbers x=-2,1,5 you find the the statement becomes False,False,True
that means that the correct answer is simply

x>4

you can also prove this one very easily by using any sort of graphing device.

 

[Math Man900]

For the 3rd one, an educated guess and check will suffice.
|2x|>|5-2x|
2|x|>2|(5-2x)/2|

|x|>|5/2-x|

If we guess x<0, you can see the equation becomes
-x>5/2-x
now add x to each side
0>5/2

This is NEVER true, which means that x must be greater than 0.

Now if we try x>5/2 our equation simplifies to:
x>-(5/2-x)
x>x-5/2
subtracting x from each side gives
0>-5/2

Now this is ALWAYS true, so that means that x could be greater than 5/2.

Now for our last attempt, we will have 0<x<5/2
This leaves us
x>5/2-x
adding x to each side
2x>5/2
x>5/4

Now combining out previous results, we obtain
This means that (x>5/4 AND x>0) OR x>5/2
the "(x>5/4 AND x>0)" simplifies to "x>5/4". Thus:
(x>5/4) OR x>5/2
and this is only true when x>5/4. Thus, the final answer to this simple yet hard problem is just

x>5/4

 

[HallsofIvy]

The simplest way to solve complicated inequalities is to first solve the corresponding equality.

For example, to solve |2x|> |5-2x| first solve |2x|= |x-2x| which then can be written as either 2x= 5- 2x or 2x= -(5- 2x).

If 2x= 5- 2x then 4x= 5 so x= 5/4. If 2x= -(x- 2x)= -5+ 2x then 0= -5 which is not true. The equality is true only for x= 5/4

Those two points separate intervals where the inequality is true from intervals whereit is false. It is only necessary to look at a single point in each interval.

For example, 0< 5/4 and |2(0)|= 0 which is NOT greater than |5- 2(0)|= 5.
The inequality is false for all x< 5/4.

5/4< 2 and |2(2)|= 4 which is greater than |4- 2(2)|= 1. The inequality is true for all x> 5/4.