Difference between revisions of "1958 AHSME Problems/Problem 4"
(→Solution) |
m (→See also) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
| − | When <math> x=\ | + | When <math> x=\frac{1}{2}</math>, <math> \frac{x+1}{x-1}=-3</math>, substituting <math> -3</math> for <math> x</math> in the original equation we get: |
| − | <math> \ | + | <math> \frac{-3+1}{-3-1}=\frac{-2}{-4}=\frac{1}{2}\implies \boxed{\mathbf{(E)}\text{ None of these}}</math>. |
==See also== | ==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}} | ||
Latest revision as of 06:10, 3 October 2014
Problem
In the expression 
each 
is replaced by . The resulting expression, evaluated for 
, equals:
Solution
When 
, 
, substituting 
for in the original equation we get:
.
See also
| 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. 
