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Inequalities with two different fractions which include x in the denominator.

by TheAbsoluTurk
Tags: denominator, fractions, include, inequalities
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TheAbsoluTurk
#1
Jan14-13, 03:33 PM
P: 100
1. The problem statement, all variables and given/known data

(x-2)/(x+3) less than (x+1)/(x)

3. The attempt at a solution

I broke it up into cases.

When x+3 less than 0 and x less than 0

and then when both are positive, when one is positive and the other is negative and then the other way around.

I'm not sure if that's the right way though. I was thinking that maybe you can say:

(x+3)(x) less than 0

and a second case

(x+3)(x) greater than 0
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Joffan
#2
Jan14-13, 05:06 PM
P: 361
(x+3) and x are not independent. You have to consider the value ranges of x.

Perhaps a good place to start is to consider the value of each expression as they pass the "difficult" value of x in each case, and how those behave as x is large-negative and large-positive in each case again.

There are three relevant ranges of x, one of which is a little trickier than the other two.
TheAbsoluTurk
#3
Jan14-13, 05:29 PM
P: 100
Quote Quote by Joffan View Post
There are three relevant ranges of x, one of which is a little trickier than the other two.
So you mean to say x≠-3 and x≠0 and x≠ some other number?

Joffan
#4
Jan14-13, 06:30 PM
P: 361
Inequalities with two different fractions which include x in the denominator.

One of the ranges is x < -3
TheAbsoluTurk
#5
Jan14-13, 07:05 PM
P: 100
Quote Quote by Joffan View Post
One of the ranges is x < -3
You mean x < -3 is not in the 'solution set' right? x is -4 makes it not a true statement.

After trying the first method again (the textbook's method except they didn't show any examples with two fractions both with x in the denominator) I found that x > -3 , x cannot equal -1/2 , 0
Joffan
#6
Jan14-13, 07:24 PM
P: 361
x can equal anything. Of the two expressions being considered in the question, each one is undefined at one value of x (which you already have indicated you know).

{x<-3} is a range of values where both expressions are well-defined. You might like to explore this range with some values of x (like -4, -5, -10) to see how the functions are behaving.
TheAbsoluTurk
#7
Jan14-13, 09:05 PM
P: 100
When you input -4 for x, you get 6 < 0.75 which means that x < -3 is not a part of the solution right?

When you input -1/2 for x, you get -1 < -1 which again is not true.

Although plugging values is good to check answers, I wanted to know if my method was correct. I don't have the answer to this question though.

What I did was solve for x under all circumstances. When x+3 > 0 and x > 0 and then when x+3 < 0 and x < 0 and then when x+3 > 0 but x < 0 then when x+3 < 0 and x > 0

The idea behind this is that when solving for x, multiplying by -(x+3) or -(x) would switch the inequality sign. So the possibility that one or the other or both are positive or negative becomes an issue. Is this a correct way to solve this inequality?
SammyS
#8
Jan14-13, 09:11 PM
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Quote Quote by TheAbsoluTurk View Post
1. The problem statement, all variables and given/known data

(x-2)/(x+3) less than (x+1)/(x)

3. The attempt at a solution

I broke it up into cases.

When x+3 less than 0 and x less than 0

and then when both are positive, when one is positive and the other is negative and then the other way around.

I'm not sure if that's the right way though. I was thinking that maybe you can say:

(x+3)(x) less than 0

and a second case

(x+3)(x) greater than 0
It seems a lot easier to write [itex]\displaystyle \ \ \frac{x-2}{x+3}<\frac{x+1}{x}[/itex]

as the equivalent inequality: [itex]\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .[/itex]

Then use a common denominator to combine the two fractions into one fraction.
TheAbsoluTurk
#9
Jan14-13, 09:23 PM
P: 100
Quote Quote by SammyS View Post
It seems a lot easier to write [itex]\displaystyle \ \ \frac{x-2}{x+3}<\frac{x+1}{x}[/itex]

as the equivalent inequality: [itex]\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .[/itex]

Then use a common denominator to combine the two fractions into one fraction.
Don't you still have to consider that x + 3 and x could be negative?
HallsofIvy
#10
Jan14-13, 09:43 PM
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Yes, it is really the same thing.

When x+3 less than 0 and x less than 0
and then when both are positive, when one is positive and the other is negative and then the other way around.
Well, x+3< 0 is the same as x< -3 and, then, of course, x< 0.
The three cases you want to consider are
x< -3 when x+3< 0 and x< 0 are BOTH true.

[itex]-3\le x< 0[/itex] when [itex]x+ 3\le 0[/itex] and x< 0.

[itex]0\le x[/itex] when both x+3> 0 and [itex]x\ge 0[/itex].
TheAbsoluTurk
#11
Jan14-13, 10:06 PM
P: 100
Quote Quote by HallsofIvy View Post
Yes, it is really the same thing.


Well, x+3< 0 is the same as x< -3 and, then, of course, x< 0.
The three cases you want to consider are
x< -3 when x+3< 0 and x< 0 are BOTH true.

[itex]-3\le x< 0[/itex] when [itex]x+ 3\le 0[/itex] and x< 0.

[itex]0\le x[/itex] when both x+3> 0 and [itex]x\ge 0[/itex].
Yes, I quickly found out that when both are negative and when x+3<0 and x>0 there are no solutions. Maybe in the future I will have the instinct to realize that from the start. But for now I am just trying to get the medthod down.

I got the same answer by putting both terms together and finding the commoj denomintor. Surely the answer must be correct.

I thank you all for your help in rehabilitating my mathematics.
Joffan
#12
Jan15-13, 06:02 AM
P: 361
You're not done yet.

It's possible to establish the answer for the ranges {x<-3} and {x>0} just by considering function values relative to 1 (= x/x).

However, the intermediate range {-3<x<0} is more complicated. Look at both function values at -2.9 and at -0.1 which should tell you there is more to discover.
TheAbsoluTurk
#13
Jan15-13, 01:52 PM
P: 100
Quote Quote by Joffan View Post
You're not done yet.

It's possible to establish the answer for the ranges {x<-3} and {x>0} just by considering function values relative to 1 (= x/x).

However, the intermediate range {-3<x<0} is more complicated. Look at both function values at -2.9 and at -0.1 which should tell you there is more to discover.
Are you supposed to state that there are asymptotes at -3 and 0 ?

x cannot be -1/2 is that what you're referring to?
SammyS
#14
Jan15-13, 03:03 PM
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Quote Quote by SammyS View Post
... the equivalent inequality: [itex]\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .[/itex]

Then use a common denominator to combine the two fractions into one fraction.
Quote Quote by TheAbsoluTurk View Post
Don't you still have to consider that x + 3 and x could be negative?
The resulting inequality is: [itex]\displaystyle \ \ \frac{-6x-3}{x(x+3)}<0\ .[/itex]

This simplifies to: [itex]\displaystyle \ \ \frac{2x+1}{x(x+3)}>0\ .[/itex]

You have three factors, one in the numerator and two in the denominator. Either all three must be positive, or one must be positive and two of them negative.
Joffan
#15
Jan15-13, 06:46 PM
P: 361
There is another range for x which satisfies the condition.

Both original expressions - and the combined expression too - are valid for ##x = -\frac 12##

SammyS's simplified expression can also be expressed as $$ \frac{x+\frac 12}{x(x+3)}>0 $$
TheAbsoluTurk
#16
Jan15-13, 08:02 PM
P: 100
It seems like I kept making careless errors. Thank you all for being patient with me.

-3 < x < -1/2 , x > 0

That should be the right answer.
HallsofIvy
#17
Jan16-13, 08:23 AM
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You want to determine when [itex](x-2)/(x+3)< (x+1)/(x)[/itex].

As I said before, since we clearly want to multiply both sides by x+ 3 and x, to clear the fractions, we need to consider when they are positive of negative. So consider 3 cases:
1) x< -3. Then both x+ 3 and x are negative. Multiplying the above inequality by x(x+ 3) is multiplying by a positive number (the product of two negatives) so the direction of inequality does not change. [itex](x- 2)(x)< (x+ 3)(x+ 1)[/itex]. [itex]x^2- 2x< x^2+ 4x+ 3[/itex]. Subtract [itex]x^2[/itex] from both sides to get [itex]-2x< 4x+ 3[/itex] so that [itex]0< 6x+ 3= 3(x+ 1)[/itex]. Dividing both sides by the positive number 3, we have [itex]0< x+ 1[/itex] or [itex]x< -1[/itex] which can't happen when x< -3.

2) -3< x< 0. Now x+ 3 is positive but x is still negative. Multiplying both sides by x(x+ 3) is now multiplying by a negative number and changes the direction of the inequality: [itex](x- 2)(x)> (x+ 3)(x+ 1)[/itex]. The same calculations as before go through with the changed inequality sign: [itex]x< -2[/itex]. That tells us that the orignal inequality is true for [itex]-3< x< -2[/itex].

3) x> 0. Now both x+ 3 and x are positive so j=multiplying both sides of the inequality by x(x+ 3) does not change the sign: [itex](x- 2)(x)< (x+ 3)(x+ 1)[/itex] and again we get [itex]x> -1[/itex]. Of course that is true for all x> 0 so we have the inequality true for all x> 0.

We have, so far, that the inequality is true for -3< x< -2 and x> 0. You should also check to see if it is true at x= -3, x= -2, and x= 0.
TheAbsoluTurk
#18
Jan16-13, 01:57 PM
P: 100
Quote Quote by HallsofIvy View Post
You want to determine when [itex](x-2)/(x+3)\le (x+1)/(x)[/itex].

As I said before, since we clearly want to multiply both sides by x+ 3 and x, to clear the fractions, we need to consider when they are positive of negative. So consider 3 cases:
1) x< -3. Then both x+ 3 and x are negative. Multiplying the above inequality by x(x+ 3) is multiplying by a positive number (the product of two negatives) so the direction of inequality does not change. [itex](x- 2)(x)\le (x+ 3)(x+ 1)[/itex]. [itex]x^2- 2x\le x^2+ 4x+ 3[/itex]. Subtract [itex]x^2[/itex] from both sides to get [itex]-2x\le 4x+ 3[/itex] so that [itex]0\le 6x+ 3= 3(x+ 2)[/itex]. Dividing both sides by the positive number 3, we have [itex]0\le x+ 2[/itex] or [itex]x\ge -2[/itex] which can't happen when x< -3.

2) -3< x< 0. Now x+ 3 is positive but x is still negative. Multiplying both sides by x(x+ 3) is now multiplying by a negative number and changes the direction of the inequality: [itex](x- 2)(x)\ge (x+ 3)(x+ 1)[/itex]. The same calculations as before go through with the changed inequality sign: [itex]x\le -2[/itex]. That tells us that the orignal inequality is true for [itex]-3< x< -2[/itex].

3) x> 0. Now both x+ 3 and x are positive so j=multiplying both sides of the inequality by x(x+ 3) does not change the sign: [itex](x- 2)(x)\le (x+ 3)(x+ 1)[itex] and again we get [itex]x\ge -2[/itex]. Of course that is true for all x> 0 so we have the inequality true for all x> 0.

We have, so far, that the inequality is true for -3< x< -2 and x> 0. You should also check to see if it is true at x= -3, x= -2, and x= 0.
I think you made a mistake in the firsf case where you ended up with 3(x+2)

Also the question didn't include any inequality symbols with equal signs. (No line under the inequality is what I mean.


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