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, this has no real solutions. $\boxed{(A)}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1980 AHSME Problems/Problem 8

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