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

Revision as of 12:13, 4 June 2011

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

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

1958 AHSME (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
All AHSME Problems and Solutions

@@ Line 11: / Line 11: @@
 ==Solution==
-{{solution}}
+When <math> x=\dfrac{1}{2}</math>, <math> \dfrac{x+1}{x-1}=-3</math>, substituting <math> -3</math> for <math> x</math> in the original equation we get:
+<math> \dfrac{-3+1}{-3-1}=\dfrac{-2}{-4}=\dfrac{1}{2}\implies \boxed{\mathbf{(E)}\text{ None of these}}</math>.
 ==See also==
 {{AHSME box|year=1958|num-b=3|num-a=5}}

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

Revision as of 12:13, 4 June 2011

Problem

Solution

See also