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Inequalities with two different fractions which include x in the denominator. 
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#1
Jan1413, 03:33 PM

P: 100

1. The problem statement, all variables and given/known data
(x2)/(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 


#2
Jan1413, 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 largenegative and largepositive in each case again. There are three relevant ranges of x, one of which is a little trickier than the other two. 


#3
Jan1413, 05:29 PM

P: 100




#4
Jan1413, 06:30 PM

P: 361

Inequalities with two different fractions which include x in the denominator.
One of the ranges is x < 3



#5
Jan1413, 07:05 PM

P: 100

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 


#6
Jan1413, 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 welldefined. You might like to explore this range with some values of x (like 4, 5, 10) to see how the functions are behaving. 


#7
Jan1413, 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? 


#8
Jan1413, 09:11 PM

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as the equivalent inequality: [itex]\displaystyle \ \ \frac{x2}{x+3}\frac{x+1}{x}<0\ .[/itex] Then use a common denominator to combine the two fractions into one fraction. 


#9
Jan1413, 09:23 PM

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#10
Jan1413, 09:43 PM

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Yes, it is really the same thing.
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]. 


#11
Jan1413, 10:06 PM

P: 100

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. 


#12
Jan1513, 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. 


#13
Jan1513, 01:52 PM

P: 100

x cannot be 1/2 is that what you're referring to? 


#14
Jan1513, 03:03 PM

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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. 


#15
Jan1513, 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 $$ \frac{x+\frac 12}{x(x+3)}>0 $$ 


#16
Jan1513, 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. 


#17
Jan1613, 08:23 AM

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You want to determine when [itex](x2)/(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. 


#18
Jan1613, 01:57 PM

P: 100

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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