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

Revision as of 13:12, 10 August 2020

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

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

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

$y=\frac{1}{1-x}$ $\Rightarrow y(1-x)=1$ $\Rightarrow y-yx=1$ $\Rightarrow yx=y-1$ $\Rightarrow x=\frac{y-1}{y}$

1967 AHSME (Problems • Answer Key • Resources)
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

@@ Line 12: / Line 12: @@
 Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us
-<math>y=\frac{1}{1-x}\Rightarrow y(1-x)=1\Rightarrow y-yx=1\Rightarrow yx=y-1\Rightarrow x=\frac{y-1}{y}</math>
+<math>y=\frac{1}{1-x}</math>
+<math>\Rightarrow y(1-x)=1</math>
+<math>\Rightarrow y-yx=1</math>
+<math>\Rightarrow yx=y-1</math>
+<math>\Rightarrow x=\frac{y-1}{y}</math>
 == See also ==