1960 AHSME Problems/Problem 28
Problem
The equation has:
Solution
Both terms have a term, so add to both sides. This results in .
However, note that if is plugged back into the original equation, it results in
Since dividing by zero is undefined, is an extraneous solution. That means there are no solutions, so the answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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