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 "1960 AHSME Problems/Problem 28"

(Solution to Problem 28)
 
m (→‎See Also)
Line 15: Line 15:
  
 
==See Also==
 
==See Also==
−
{{AHSME box|year=1960|num-b=27|num-a=29}}
+
{{AHSME 40p box|year=1960|num-b=27|num-a=29}}

Revision as of 19:15, 10 May 2018

Problem

The equation $x-\frac{7}{x-3}=3-\frac{7}{x-3}$ has:

$\textbf{(A)}\ \text{infinitely many integral roots}\qquad\textbf{(B)}\ \text{no root}\qquad\textbf{(C)}\ \text{one integral root}\qquad$ $\textbf{(D)}\ \text{two equal integral roots}\qquad\textbf{(E)}\ \text{two equal non-integral roots}$

Solution

Both terms have a $-\frac{7}{x-3}$ term, so add $\frac{7}{x-3}$ to both sides. This results in $x = 3$.

However, note that if $3$ is plugged back into the original equation, it results in \[3-\frac{7}{0}=3-\frac{7}{0}\]

Since dividing by zero is undefined, $3$ is an extraneous solution. That means there are no solutions, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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
All AHSME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_28&oldid=94401"