Difference between revisions of "1957 AHSME Problems/Problem 45"
(created solution page) |
(→Solution) |
||
| Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
| − | <math>\boxed{\textbf{(A)}}</math>. | + | From the initial equation <math>\tfrac x y = x-y</math>, we can solve for <math>y</math> in terms of <math>x</math> as follows: |
| + | \begin{align*}\ | ||
| + | \frac x y &= x-y \\ | ||
| + | x &= xy-y^2 \\ | ||
| + | y^2-xy+x &= 0 \\ | ||
| + | y = \frac{x \pm \sqrt{x^2-4x}}2 | ||
| + | \end{align*} | ||
| + | Because <math>y</math> is real, the [[discriminant]] of the above expression for <math>y</math> must be <math>\geq 0</math>, so <math>x^2-4x=x(x-4) \geq 0</math>. This is true when <math>x \geq 4</math> and <math>x \leq 0</math>, so we choose answer <math>\boxed{\textbf{(A)}}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 12:06, 27 July 2024
Problem
If two real numbers 
and 
satisfy the equation 
, then:
Solution
From the initial equation 
, we can solve for in terms of as follows:
\begin{align*}\
\frac x y &= x-y \\
x &= xy-y^2 \\
y^2-xy+x &= 0 \\
y = \frac{x \pm \sqrt{x^2-4x}}2
\end{align*}
Because is real, the discriminant of the above expression for must be 
, so 
. This is true when 
and 
, so we choose answer 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 44 |
Followed by Problem 46 | |
| 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. 
