Difference between revisions of "1983 AHSME Problems/Problem 1"
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==Solution== | ==Solution== | ||
| + | From <math>\frac{x}{4} = 4y</math>, we get <math>x=16y</math>. Plugging in the other equation, <math>\frac{16y}{2} = y^2</math>, so <math>y^2-8y=0</math>. Factoring, we get <math>y(y-8)=0</math>, so the solutions are <math>0</math> and <math>8</math>. Since <math>x \neq 0</math>, we also have <math>y \neq 0</math>, so <math>y=8</math>. Hence <math>x=16\cdot 8 = \boxed{\textbf{(E)}\ 128}</math>. | ||
| + | ==See Also== | ||
| + | {{AHSME box|year=1983|before=First Question|num-a=2}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 23:38, 19 February 2019
Problem
If 
and 
, then 
equals
Solution
From , we get 
. Plugging in the other equation, 
, so 
. Factoring, we get 
, so the solutions are 
and 
. Since 
, we also have 
, so 
. Hence 
.
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
