1966 AHSME Problems/Problem 5

Revision as of 12:37, 13 January 2008 by Azjps (talk | contribs) (s)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

The number of values of $x$ satisfying the equation \[\frac {2x^2 - 10x}{x^2 - 5x} = x - 3\] is:

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$

Solution

Since $x^2 - 5x$ is in the denominator, $x \neq 0,5$. Simplifying,

\[2\left(\frac{x^2 - 5}{x^2 - 5}\right) = 2 = x-3\]

Thus $x = 5$, which isn't in the domain of the equation. Thus there are no values of $x \Rightarrow \mathrm{(A)}$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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
All AHSME Problems and Solutions