Difference between revisions of "2002 AMC 10B Problems/Problem 13"

(Solution)
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As <math>(4y + 1)(2x - 3) = 0</math> must be true for all <math>y</math>, we must have <math>2x - 3 = 0</math>, hence <math>\boxed{x = \frac 32}</math>.
 
As <math>(4y + 1)(2x - 3) = 0</math> must be true for all <math>y</math>, we must have <math>2x - 3 = 0</math>, hence <math>\boxed{x = \frac 32}</math>.
 
(Too bad there is no such option -- maybe a typo when transcribing the options? Or the question is not formulated correctly? Note that the other fraction in option <math>B</math> would be the answer to the complementary question "find the value <math>y</math> such that ... for all <math>x</math>".)
 
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2002|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2002|ab=B|num-b=12|num-a=14}}

Revision as of 17:05, 30 December 2009

Problem

Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$.

$\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac34 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14$

Solution

We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$.

As $(4y + 1)(2x - 3) = 0$ must be true for all $y$, we must have $2x - 3 = 0$, hence $\boxed{x = \frac 32}$.

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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
All AMC 10 Problems and Solutions