Difference between revisions of "1966 AHSME Problems/Problem 33"

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
 
 
== Solution 2 ==
 
  
 
Let <math>m=\frac{x-a}{b}</math> and <math>n=\frac{x-b}{a}</math> then we have
 
Let <math>m=\frac{x-a}{b}</math> and <math>n=\frac{x-b}{a}</math> then we have
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If <math>m+n=0</math> then
 
If <math>m+n=0</math> then
<math></math>\frac{
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<cmath>\frac{x-a}{b}+\frac{x-b}{a}=0</cmath>
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<cmath>\frac{x-a}{b}=\frac{b-x}{a}</cmath>
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<cmath>x=\frac{a^2+b^2}{a+b}</cmath>
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Which yields <math>1</math> solution for <math>x</math>.
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If <math>mn=0</math> then
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<cmath>(\frac{x-a}{b})(\frac{x-b}{a})=1</cmath>
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<cmath>x^2-(a+b)x=0</cmath>
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Solving the quadratic gets another <math>2</math> solutions for <math>x</math>.
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Thus there are <math>\boxed{\text{(D) three}}</math> solutions in total.
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~ Nafer
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1966|num-b=32|num-a=34}}   
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{{AHSME 40p box|year=1966|num-b=32|num-a=34}}   
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:26, 29 July 2024

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

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 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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