Difference between revisions of "1967 AHSME Problems/Problem 14"

(Solution)
(Solution)
Line 13: Line 13:
  
 
<math>y=\frac{1}{1-x}</math>
 
<math>y=\frac{1}{1-x}</math>
 +
 
<math>\Rightarrow y(1-x)=1</math>
 
<math>\Rightarrow y(1-x)=1</math>
 +
 
<math>\Rightarrow y-yx=1</math>
 
<math>\Rightarrow y-yx=1</math>
 +
 
<math>\Rightarrow yx=y-1</math>
 
<math>\Rightarrow yx=y-1</math>
 +
 
<math>\Rightarrow x=\frac{y-1}{y}</math>
 
<math>\Rightarrow x=\frac{y-1}{y}</math>
  

Revision as of 13:13, 10 August 2020

Problem

Let $f(t)=\frac{t}{1-t}$, $t \not= 1$. If $y=f(x)$, then $x$ can be expressed as

$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$

Solution

Since we know that $y=f(x)$, we can solve for $y$ in terms of $x$. This gives us

$y=\frac{1}{1-x}$

$\Rightarrow y(1-x)=1$

$\Rightarrow y-yx=1$

$\Rightarrow yx=y-1$

$\Rightarrow x=\frac{y-1}{y}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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. AMC logo.png