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Difference between revisions of "1959 AHSME Problems/Problem 16"

(Created page with "== Problem 16== The expression<math> \frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},</math> when simplified is: <math>\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qq...")
 
m (formatted answer)
 
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== Solution ==
 
== Solution ==
 
Factoring each of the binomials in the expression <math> \frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},</math> will yield the result of <cmath> \frac{(x-2)(x-1)}{(x-3)(x-2)}\div \frac{(x-4)(x-1)}{(x-3)(x-4)},</cmath>
 
Factoring each of the binomials in the expression <math> \frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},</math> will yield the result of <cmath> \frac{(x-2)(x-1)}{(x-3)(x-2)}\div \frac{(x-4)(x-1)}{(x-3)(x-4)},</cmath>
−
We can eliminate like terms to get <math>\frac {x-1}{x-3}\div \frac{x-1}{x-3}</math>, which, according to identity property, is equivalent to the answer (D) 1.
+
We can eliminate like terms to get <math>\frac {x-1}{x-3}\div \frac{x-1}{x-3}</math>, which, according to identity property, is equivalent to the answer <math>\boxed{\textbf{(D) }1}</math>.
 +
 
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=15|num-a=17}}
 
{{AHSME 50p box|year=1959|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category: AHSME]][[Category:AHSME Problems]]
 
[[Category: AHSME]][[Category:AHSME Problems]]

Latest revision as of 11:28, 21 July 2024

Problem 16

The expression$\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified is: $\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution

Factoring each of the binomials in the expression $\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ will yield the result of \[\frac{(x-2)(x-1)}{(x-3)(x-2)}\div \frac{(x-4)(x-1)}{(x-3)(x-4)},\] We can eliminate like terms to get $\frac {x-1}{x-3}\div \frac{x-1}{x-3}$, which, according to identity property, is equivalent to the answer $\boxed{\textbf{(D) }1}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AHSME Problems and Solutions

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