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

(Solution 2)
(Solution 2)
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<math>8xy = 12y</math>
 
<math>8xy = 12y</math>
  
There is only one solution, namely <math>x = \boxed{\dfrac{3}{2}} ( \text{D} )</math>
+
There is only one solution, namely <math>x = \boxed{\dfrac{3}{2}}     ( \text{D} )</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 13:12, 30 November 2019

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

Solution 2

Since we want only the $y$-variable to be present, we move the terms only with the $y$-variable to one side, thus constructing $8xy - 12y + 2x - 3 = 0$ to $8xy + 2x = 12y + 3$. For there to be infinite solutions for $y$ and there is no $x$, we simply find a value of $x$ such that the equation is symmetrical. Therefore,

$2x = 3$

$8xy = 12y$

There is only one solution, namely $x = \boxed{\dfrac{3}{2}}     ( \text{D} )$

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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