MHB Simplify Complex Rational Expression

AI Thread Summary
The discussion revolves around simplifying the rational expression (3/x-2) - (4/x+2) / (7/x^2-4). One participant simplified it to -x+14/7, while the book provided x-14/7 as the answer. It was clarified that the correct interpretation of the expression was the second option, confirming that the book's answer is incorrect. Additionally, a domain restriction was noted, stating that |x| cannot equal 2. The conversation emphasizes the importance of correctly interpreting complex rational expressions.
PaperStSoap
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(3/x-2) - (4/x+2) / (7/x2-4)

I got it down to...

-x+14/7

but the book is showing

x-14/7
 
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Welcome to MHB, Paper! :D

Which of these expressions did you mean:

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

\dfrac{\dfrac{3}{x-2} - \dfrac{4}{x+2}}{\dfrac{7}{x^2-4}}

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

But the book is showing: .$\dfrac{x-14}{7}$ . The book is wrong!
 
Fantini said:
Welcome to MHB, Paper! :D

Which of these expressions did you mean:

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

My apologies, the problem was the second one.
 
PaperStSoap said:
My apologies, the problem was the second one.

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

(working in spoiler)
$\dfrac{\frac{3}{x-2} - \frac{4}{x+2}}{\frac{7}{x^2-4}}$

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

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

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

$\frac{-x+14}{7}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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