Difference between revisions of "1950 AHSME Problems/Problem 4"
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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== | ||
Revision as of 22:30, 4 October 2022
Problem
Reduced to lowest terms, 
is equal to:
Solution
We start off by simplifying the second term.
![\[-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.\]](http://latex.artofproblemsolving.com/b/d/a/bda210f1dcc7a20cc65ea176cc7eee74d817bf06.png)
Now create a common denominator and simplify.
![\[\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}\]](http://latex.artofproblemsolving.com/3/0/d/30de8e5e5a71aca066b9c9ffbbbabcae654da1c2.png)
obs: Assume that 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
