# Difference between revisions of "1980 AHSME Problems/Problem 8"

## Problem

How many pairs $(a,b)$ of non-zero real numbers satisfy the equation $$\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}$$ $\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$ $\text{(E)} \ \text{two pairs for each} ~b \neq 0$

## Solution

We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$. $$a+b=\frac{ab}{a+b}$$ $$a^2+2ab+b^2=ab$$ $$a^2+ab+b^2=0.$$

By the quadratic formula and checking the discriminant (imagining one of the variables to be constant), we see that this has no real solutions. Thus the answer is $\boxed{\text{(A)none}}$.

## See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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