Carcass wrote:
\(|a|<|b|\)
Quantity A |
Quantity B |
\(\frac{b}{a+b} - \frac{a}{a-b}\) |
\(\frac{a}{b+a} - \frac{b}{b-a }\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
I manipulated both sides to get to an answer:
\(\frac{b}{a+b} - \frac{a}{a-b} ? \frac{a}{b+a} - \frac{b}{b-a }\)
\(\frac{b-a}{a+b} - \frac{a}{a-b} ? - \frac{b}{b-a }\)
\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{a}{a-b}\)
\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{(-1)-a}{(-1)(b-a)}\)
\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{-a}{b-a}\)
\(\frac{b-a}{a+b} ? \frac{-b-a}{b-a}\)
\(\frac{b-a}{a+b} ? \frac{(-1)(a+b)}{b-a}\)
Here I crossed multiplied. You usually wouldn't want to do this because you run the risk of changing the sign (the '?' in our algebra above). However, this is a unique situation, and can be better seen in the result of the cross multiplication:
\((b-a)^2 ? (-1)(a+b)^2\)
We don't have to worry about the cases when \(b-a\) and \(a+b\) are positive or negative since they are being squared, so they are always positive.
Given that \((a+b)^2\) is being multiplied by \(-1\), it must be the case that:
\((b-a)^2 > (-1)(a+b)^2\)
So A is greater.The only case where this wouldn't be true is if they were equal, or Choice C. In other words:
\(b-a = a + b\)
But quick algebra rules this possibility out:
\(b-a = a + b\)
\(-a = a\)
\(0 = 2a\)
\(0 = a\)
Which would imply that \(b = 0\), but we know that both cannot be \(0\) because of the initial restriction given in the question: \(|a|<|b|\) (or by recognizing that the denominators in the fraction of both A and B cannot be 0).