Difference between revisions of "2002 AMC 10B Problems/Problem 13"
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<math>8xy = 12y</math> | <math>8xy = 12y</math> | ||
| − | There is only one solution, namely <math>x = \boxed{\dfrac{3}{2}}</math> or <math>\text{D}</math> | + | There is only one solution, namely <math>x = \boxed{\dfrac{3}{2}}</math> or <math>\text{(D)}</math> |
== See Also == | == See Also == | ||
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{{AMC10 box|year=2002|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2002|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category: Introductory Algebra Problems]] | ||
Latest revision as of 16:51, 18 October 2025
Contents
Problem
Find the value(s) of 
such that 
is true for all values of 
.
Solution
We have 
.
As 
must be true for all , we must have 
, hence 
.
Solution 2
Since we want only the -variable to be present, we move the terms only with the -variable to one side, thus constructing to 
. For there to be infinite solutions for and there is no , we simply find a value of such that the equation is symmetrical. Therefore,
There is only one solution, namely 
or 
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
