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

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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 (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.

 Quote by Joffan 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?

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

One of the ranges is x < -3

 Quote by Joffan 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

 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.
 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?

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 Quote by TheAbsoluTurk 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 $\displaystyle \ \ \frac{x-2}{x+3}<\frac{x+1}{x}$

as the equivalent inequality: $\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .$

Then use a common denominator to combine the two fractions into one fraction.

 Quote by SammyS It seems a lot easier to write $\displaystyle \ \ \frac{x-2}{x+3}<\frac{x+1}{x}$ as the equivalent inequality: $\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .$ 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?

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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.

$-3\le x< 0$ when $x+ 3\le 0$ and x< 0.

$0\le x$ when both x+3> 0 and $x\ge 0$.

 Quote by HallsofIvy 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. $-3\le x< 0$ when $x+ 3\le 0$ and x< 0. $0\le x$ when both x+3> 0 and $x\ge 0$.
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.

 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

 Quote by Joffan 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
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?

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 Quote by SammyS ... the equivalent inequality: $\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .$ Then use a common denominator to combine the two fractions into one fraction.
 Quote by TheAbsoluTurk Don't you still have to consider that x + 3 and x could be negative?
The resulting inequality is: $\displaystyle \ \ \frac{-6x-3}{x(x+3)}<0\ .$

This simplifies to: $\displaystyle \ \ \frac{2x+1}{x(x+3)}>0\ .$

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.

 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 $$\frac{x+\frac 12}{x(x+3)}>0$$
 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus You want to determine when $(x-2)/(x+3)< (x+1)/(x)$. 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. $(x- 2)(x)< (x+ 3)(x+ 1)$. $x^2- 2x< x^2+ 4x+ 3$. Subtract $x^2$ from both sides to get $-2x< 4x+ 3$ so that $0< 6x+ 3= 3(x+ 1)$. Dividing both sides by the positive number 3, we have $0< x+ 1$ or $x< -1$ 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: $(x- 2)(x)> (x+ 3)(x+ 1)$. The same calculations as before go through with the changed inequality sign: $x< -2$. That tells us that the orignal inequality is true for $-3< x< -2$. 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: $(x- 2)(x)< (x+ 3)(x+ 1)$ and again we get $x> -1$. 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.