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[flamathdummy]

I have worked this....but I am not sure as how to write my final solution

3/x+2 > 2/x-1

3(x-1)/(x+2)(x-1) -2(x+2)/x-1(x+2) > 0

3x-3/(x+2)(x-1) - 2x+4/(x + 1) (x + 2) > 0

x-7/(x + 2)(x -1) > 0

So is my final solution x-7/(x+2)(x-1) ???

<--(-5)--(-4)--(-3)--((-2)--(-1)--(0))--(1)--(2)--(3)--(4)--(5)--(6)--((7)-->)

 

[tkhunny]

3/x+2 > 2/x-1

You seem to be missing some of the idea.

First, worry about the Domain.

x = -2 is OUT of the Domain.
x = 1 is OUT of the Domain.

You need to know why this is so. Once these are removed, x+2 and x-1 cannot be zero and they are a little safer to work with.

Solve the EQUALITY

3/x+2 = 2/x-1
3(x-1)/(x+2) = 2
3*(x-1) = 2*(x+2)
3x - 3 = 2x + 4
x - 3 = 4
x = 7

NOW, let's look at the number line, marking all the known points of interest.


<------(-2)-------(1)----------(7)------------->

Now, what works? Try zero (0)

3/(0+2) > 2/(0-1) ???
3/2 > 2/(-1) ???
3/2 > -2 ??? Yes, indeed

Similarly, we can determine that values less than -2 do NOT work.
Similarly, we can determine that values greater than 1 and less than 7 do NOT work.
Similarly, we can determine that values greater than 7 DO work.

Notice how the regions alternate, working and not working. There are conditions to ensure this. Do you know them?

Solution? (-2,1)U(7,big)

 

[flamathdummy]

Notice how the regions alternate, working and not working. There are conditions to ensure this. Do you know them?

No, I do not know them.

 

[tkhunny]

Well, something to think about. It is not mandatory, but it is interesting.

Think on the "degree" of the term that causes the break in the number line. Would 1/(x+1) > 5 give a different type of result than 1/[(x+1)²] > 25