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

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We have <math>8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)</math>.  
 
We have <math>8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)</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>.
+
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\ \mathrm{ (D)}}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:54, 16 July 2014

Problem

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

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

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 $\boxed{x = \frac 32\ \mathrm{ (D)}}$.

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

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