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Difference between revisions of "1950 AHSME Problems/Problem 4"

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== Solution ==
 
== Solution ==
  
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We start off by simplifying the second term.
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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>
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obs: Assume that <math>a \not = 0, b\not = 0, a\not = b</math>.
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:21, 10 January 2023

Problem

Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}}$ is equal to:

$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$

Solution

We start off by factoring the second fraction.

\[-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.\]

Now create a common denominator and simplify.

\[\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}\]

obs: Assume that $a \not = 0, b\not = 0, a\not = b$.

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

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