Difference between revisions of "1950 AHSME Problems/Problem 4"
(Created page with "== Problem == Reduced to lowest terms, <math> \frac{a^{2}-b^{2}}{ab} </math> - <math> \frac{ab-b^{2}}{ab-a^{2}} </math> is equal to: <math> \textbf{(A)}\ \frac{a}{b}\qquad\text...") |
(→Solution) |
||
Line 8: | Line 8: | ||
We can use difference of two squares to expand <math>a^2-b^2=(a-b)(a+b)</math> and factor to get <math>ab-b^{2}=b(a-b)</math> and <math>a^2-ab=a(a-b)</math> | We can use difference of two squares to expand <math>a^2-b^2=(a-b)(a+b)</math> and factor to get <math>ab-b^{2}=b(a-b)</math> and <math>a^2-ab=a(a-b)</math> | ||
<math> \dfrac{a^{2}-b^{2}}{ab} - \dfrac{ab-b^{2}}{ab-a^{2}} =\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> | <math> \dfrac{a^{2}-b^{2}}{ab} - \dfrac{ab-b^{2}}{ab-a^{2}} =\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> | ||
+ | |||
+ | We can further factor to get <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> If we assume b is not equal to a, this is equal to <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b}{a}=\dfrac{(a-b)(a+b)}{ab}-\dfrac{b^2}{ab}=\dfrac{(a-b)(a+b)-b^2}{ab}=\dfrac{a^2-2b^2}{ab}</math> | ||
+ | |||
+ | The answer is | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1950|num-b=3|num-a=5}} | {{AHSME box|year=1950|num-b=3|num-a=5}} |
Revision as of 13:47, 28 October 2011
Problem
Reduced to lowest terms, - is equal to:
Solution
We can use difference of two squares to expand and factor to get and
We can further factor to get If we assume b is not equal to a, this is equal to
The answer is
See Also
1950 AHSME (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 | ||
All AHSME Problems and Solutions |