1960 AHSME Problems/Problem 28

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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