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

Adding the two fractions on the left side yields $\frac{a+b}{ab}=\frac{1}{a+b}$ .

Cross-multiplying yields $a^2+2ab+b^2=ab$ .

Subtracting $ab$ from both sides yields $a^2+ab+b^2=0$ .

Suppose $b$ is constant and $a$ is variable. Then the discriminant of this quadratic equation is $b^2 - 4 \cdot 1 \cdot b^2 = -3b^2$ , which is negative if $b \neq 0$ . Therefore, for each $b \neq 0$ , this equation has no real solutions, and the answer is $\boxed{\text{(\textbf{A})\ none}}$ .

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

Problem

Solution

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