Difference between revisions of "1967 AHSME Problems/Problem 14"
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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 | 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</math> | + | <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> |
== See also == | == See also == |
Revision as of 14:12, 10 August 2020
Problem
Let ,
. If
, then
can be expressed as
Solution
Since we know that , we can solve for
in terms of
. This gives us
See also
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 |
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