1966 AHSME Problems/Problem 33

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Problem

If $ab \ne 0$ and $|a| \ne |b|$, the number of distinct values of $x$ satisfying the equation

\[\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},\]

is:

$\text{(A) zero}  \quad \text{(B) one}  \quad \text{(C) two}  \quad \text{(D) three}  \quad \text{(E) four}$

Solution

$\fbox{D}$

Solution 2

Let $m=\frac{x-a}{b}$ and $n=\frac{x-b}{a}$ then we have \[m+n=\frac{1}{m}+\frac{1}{n}\] \[m+n=\frac{m+n}{mn}\] Notice that the equation is possible iff $m+n=0$ or $mn=1$.

If $m+n=0$ then \[\frac{x-a}{b}+\frac{x-b}{a}=0\] \[\frac{x-a}{b}=\frac{b-x}{a}\] \[x=\frac{a^2+b^2}{a+b}\] Which yields $1$ solution for $x$.

If $mn=0$ then \[(\frac{x-a}{b})(\frac{x-b}{a})=1\] \[x^2-(a+b)x=0\] Solving the quadratic gets another $2$ solutions for $x$.

Thus there are $\boxed{\text{(D) three}}$ solutions in total.

~ Nafer

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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