Difference between revisions of "1950 AHSME Problems/Problem 4"
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== Solution == | == Solution == | ||
− | We start off by | + | We start off by factoring the second fraction. |
<cmath>-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.</cmath> | <cmath>-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.</cmath> | ||
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<cmath>\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}</cmath> | <cmath>\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}</cmath> | ||
+ | |||
+ | obs: Assume that <math>a \not = 0, b\not = 0, a\not = b</math>. | ||
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1950|num-b=3|num-a=5}} | + | {{AHSME 50p box|year=1950|num-b=3|num-a=5}} |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:21, 10 January 2023
Problem
Reduced to lowest terms, is equal to:
Solution
We start off by factoring the second fraction.
Now create a common denominator and simplify.
obs: Assume that .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AHSME Problems and Solutions |
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