1985 AHSME Problems/Problem 8
Problem
Let and be real numbers with and nonzero. The solution to is less than the solution to if and only if
Solution
The solution to is and the solution to is . The first solution is less than the second if . We can get rid of the negative signs if we reverse the inequality sign, so we have . All our steps are reversible, so our answer is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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