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Difference between revisions of "1981 AHSME Problems/Problem 6"

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==Solution==
 
==Solution==
  
−
We can cross multiply both sides of the equation and simplify.
+
Cross-multiply and simplify:
 
<cmath>x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)</cmath>
 
<cmath>x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)</cmath>
 
<cmath>xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1</cmath>
 
<cmath>xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1</cmath>
 
<cmath>-x = -y^2 - 2y + 1</cmath>
 
<cmath>-x = -y^2 - 2y + 1</cmath>
−
<cmath>x = y^2 + 2y - 1</cmath>
+
<cmath>x = \boxed{(\textbf{A})\ y^2 + 2y - 1}</cmath>
−
 
 
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The answer is <math>\text{A.}</math>
 
  
 
-edited by coolmath34
 
-edited by coolmath34
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==See also==
 
==See also==
  
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{{AHSME box|year=1981|before=5|num-a=7}}
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{{AHSME box|year=1981|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:02, 18 August 2025

Problem

If $\frac{x}{x-1} = \frac{y^2 + 2y - 1}{y^2 + 2y - 2},$ then $x$ equals

$\text{(A)} \quad y^2 + 2y - 1$

$\text{(B)} \quad y^2 + 2y - 2$

$\text{(C)} \quad y^2 + 2y + 2$

$\text{(D)} \quad y^2 + 2y + 1$

$\text{(E)} \quad -y^2 - 2y + 1$

Solution

Cross-multiply and simplify: \[x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)\] \[xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1\] \[-x = -y^2 - 2y + 1\] \[x = \boxed{(\textbf{A})\ y^2 + 2y - 1}\]

-edited by coolmath34

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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