MHB What is the solution (rational function/interval table)?

[eleventhxhour]

What is the solution of |x/(x-2)| < 5 ?

So, I did this the usual way of moving over the 5 to the left side and then cross multiplying and simplifying etc. However, I keep getting the wrong answer. I got x < 5/3 and x > 2, while the answer in the book says that it's x< 5/3 and x > 5/2.

What did I do wrong? I'm assuming it has something to do with the absolute value sign, but I'm not sure how to figure it out...

 

[Siron]

What do yo mean with cross multiplying? I'm not familiar with this.

I would solve it this way: to get rid of the absolut value signs you can square both sides, that is

$$\left | \frac{x}{x-2}\right | < 5 \Rightarrow \frac{x^2}{(x-2)^2} < 25$$
Solving this inequality gives the desired result.

 

[topsquark]

Another way to get rid of the absolute value:
[math]\left | \frac{x}{x - 2} \right | < 5[/math]

splits into the two inequalities:
[math]\frac{x}{x - 2} < 5[/math]

and
[math]-\frac{x}{x - 2} < 5[/math]

and solve them separately.

The trouble with cross multiplying when using inequalities is that we need to know the sign of what we are multiplying by. For example -4/x < 1 when x is positive, but -4/x > 1 when x is negative. The same thing will occur for the x - 2 term.

-Dan

 

[eleventhxhour]

Another way to get rid of the absolute value:
[math]\left | \frac{x}{x - 2} \right | < 5[/math]

splits into the two inequalities:
[math]\frac{x}{x - 2} < 5[/math]

and
[math]-\frac{x}{x - 2} < 5[/math]

and solve them separately.

The trouble with cross multiplying when using inequalities is that we need to know the sign of what we are multiplying by. For example -4/x < 1 when x is positive, but -4/x > 1 when x is negative. The same thing will occur for the x - 2 term.

-Dan

So I solved each of those and for the first one I got that it's negative at x<2 and x>5/2. For the second I got that it's negative at x<5/3 and x>2. Is this correct? The book had a different answer (x<5/3 and x>5/2).

 

[topsquark]

So I solved each of those and for the first one I got that it's negative at x<2 and x>5/2. For the second I got that it's negative at x<5/3 and x>2. Is this correct? The book had a different answer (x<5/3 and x>5/2).

All of this has to come together. Take a look first at the lower limits of x. We have x < 2 and x < 5/3. In order for both of these to be true then we require that x < 5/3, because 5/3 is smaller than 2...both conditions are satisfied by this. See if you can do the upper limits of x based on a similar argument.

-Dan