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1966 AHSME Problems/Problem 5

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Problem

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

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qqua...$

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 (Problems)
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1966_AHSME_Problems/Problem_5"

Category: Introductory Algebra Problems

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