Carcass wrote:
If \(a < 0\) and \(b < 0\), what are the possible values for \(\frac{a+b}{ab }\) ?
Indicate all that apply.
A 2
B 1
C 0
D -1
E -2
Useful (and often-tested) fraction property: \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\)When we apply the above property, we get: \(\frac{a+b}{ab } = \frac{a}{ab } + \frac{b}{ab } = \frac{1}{b} + \frac{1}{a}\)
Since \(a\) and \(b\) are both NEGATIVE, we know that \(\frac{1}{a}\) and \(\frac{1}{b}\) are both NEGATIVE, which means \(\frac{1}{a} + \frac{1}{b}\) must be NEGATIVE.
So we can immediately and eliminate answer choices A, B, and C.
Now recognize that, if \(a = -2\) and \(b = -2\), then \(\frac{1}{a} + \frac{1}{b}=\frac{1}{-2} + \frac{1}{-2} = \frac{2}{-2} = -1\), which means D is a possible answer.
Also, if \(a = -1\) and \(b = -1\), then \(\frac{1}{a} + \frac{1}{b}=\frac{1}{-1} + \frac{1}{-1} = (-1) + (-1) = -2\), which means E is a possible answer.
Answer: D and E