# 2000 AMC 12 Problems/Problem 11

The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.

## Problem

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?

$\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac { -1 }{2} \qquad \textbf{(C)} \ \frac {1}{3} \qquad \textbf{(D)} \ \frac {1}{2} \qquad \textbf{(E)} \ 2$

## Solution 1

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$.

Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

## Solution 2

This simplifies to $ab+b-a=0 \Rightarrow (a+1)(b-1) = -1$. The two integer solutions to this are $(-2,2)$ and $(0,0)$. The problem states than $a$ and $b$ are non-zero, so we consider the case of $(-2,2)$. So, we end up with $\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = 2 \Rightarrow \boxed{\text{E}}$

## Solution 3

Just realize that two such numbers are $a = 1$ and $b = \frac{1}{2}$. You can see this by plugging in $a = 1$ and then solving for b. With this, you can solve and get $2 \Rightarrow\boxed{\text{E}}$

## Solution 4

Set $a$ to some nonzero number. In this case, I'll set it to $4$.

Then solve for $b$. In this case, $b=0.8$.

Now just simply evaluate. In this case it's 2. So since 2 is a possible value of the original expression, select $\boxed{E}$.

~hastapasta

## Solution 5

Notice that $a=\frac{a}{b}-1$ and $b=1-\frac{b}{a}$. Then, $\frac{a}{b}=1+a$ and $\frac{b}{a}=1-b$.

$\frac{a}{b}+\frac{b}{a}-ab=(1+a)+(1-b)-(a-b)=2$. The answer is $\boxed{E}$.

~ pi_is_3.14

## Video Solution by Daily Dose of Math

~Thesmartgreekmathdude