1960 AHSME Problems/Problem 28

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 $\boxed{\textbf{(B)}}$ .

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

Page

Toolbox

Search

1960 AHSME Problems/Problem 28

Problem

Solution

See Also