- Thread starter
- #1

#### PaperStSoap

##### New member

- Jun 26, 2012

- 9

(3/x-2) - (4/x+2) / (7/x

I got it down to...

-x+14/7

but the book is showing

x-14/7

^{2}-4)I got it down to...

-x+14/7

but the book is showing

x-14/7

- Thread starter PaperStSoap
- Start date

- Thread starter
- #1

- Jun 26, 2012

- 9

(3/x-2) - (4/x+2) / (7/x^{2}-4)

I got it down to...

-x+14/7

but the book is showing

x-14/7

I got it down to...

-x+14/7

but the book is showing

x-14/7

- Feb 29, 2012

- 342

Which of these expressions did you mean:

$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$

[tex]\dfrac{\dfrac{3}{x-2} - \dfrac{4}{x+2}}{\dfrac{7}{x^2-4}}[/tex]

I got it down to: .$\dfrac{-x+14}{7}$ . You are right!

But the book is showing: .$\dfrac{x-14}{7}$ . The book iswrong!

- Thread starter
- #4

- Jun 26, 2012

- 9

My apologies, the problem was the second one.

Which of these expressions did you mean:

$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$

- Mar 1, 2012

- 249

You are right and the book is wrong. It's worth mentioning though that there is a restriction on the domain: $|x| \neq 2$.My apologies, the problem was the second one.

(working in spoiler)

$\left(\frac{3}{x-2} - \frac{4}{x+2}\right) \cdot \frac{(x-2)(x+2)}{7}$

$\left(\frac{3(x+2)-4(x-2)}{(x-2)(x+2)}\right) \cdot \frac{(x-2)(x+2)}{7}$

$\frac{3x+6-4x+8}{7}$

$\frac{-x+14}{7}$