Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1954 AHSME Problems/Problem 2"

(Created page with "== Problem 2== The equation <math>\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0</math> can be transformed by eliminating fractions to the equation <math>x^2-5x+4=0</ma...")
 
 
Line 12: Line 12:
  
 
Thus the only solution to the equation is <math>x=4</math>, <math>\fbox{C}</math>
 
Thus the only solution to the equation is <math>x=4</math>, <math>\fbox{C}</math>
 +
 +
==See Also==
 +
 +
{{AHSME 50p box|year=1954|num-b=1|num-a=3}}
 +
 +
{{MAA Notice}}

Latest revision as of 19:34, 17 February 2020

Problem 2

The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$. The roots of the latter equation are $4$ and $1$. Then the roots of the first equation are:

$\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root}$

Solution

We plug $x=4, x=1$ into the original equation $x=4\implies\frac{8}{-3}+\frac{15}{3}+\frac{-20}{-3}\implies\frac{15+20-8}{4}+1=0$ $x=1\implies\frac{2}{0}-\frac{9}{3}+\frac{-2}{0}+1=0$, which is undefined because $\frac{k}{0}$ is undefined.

Thus the only solution to the equation is $x=4$, $\fbox{C}$

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_2&oldid=117885"