Difference between revisions of "1958 AHSME Problems/Problem 4"

Latest revision as of 05:10, 3 October 2014

Problem

In the expression $\frac{x + 1}{x - 1}$ each $x$ is replaced by $\frac{x + 1}{x - 1}$ . The resulting expression, evaluated for $x = \frac{1}{2}$ , equals:

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ \text{none of these}$

Solution

When $x=\frac{1}{2}$ , $\frac{x+1}{x-1}=-3$ , substituting $-3$ for $x$ in the original equation we get:

$\frac{-3+1}{-3-1}=\frac{-2}{-4}=\frac{1}{2}\implies \boxed{\mathbf{(E)}\text{ None of these}}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 ==See also==
-{{AHSME box|year=1958|num-b=3|num-a=5}}
+{{AHSME 50p box|year=1958|num-b=3|num-a=5}}
+{{MAA Notice}}

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Difference between revisions of "1958 AHSME Problems/Problem 4"

Latest revision as of 05:10, 3 October 2014

Problem

Solution

See also