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Difference between revisions of "2002 AMC 10B Problems/Problem 13"
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Find the value(s) of <math>x</math> such that <math>8xy - 12y + 2x - 3 = 0</math> is true for all values of <math>y</math>.
 
Find the value(s) of <math>x</math> such that <math>8xy - 12y + 2x - 3 = 0</math> is true for all values of <math>y</math>.
  
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<math>\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</math>
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<math>\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</math>
  
 
== Solution ==
 
== Solution ==
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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>.  
  
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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>.
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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\ \mathrm{ (D)}}</math>.
  
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(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>".)
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== Solution 2 ==
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Since we want only the <math>y</math>-variable to be present, we move the terms only with the <math>y</math>-variable to one side, thus constructing <math>8xy - 12y + 2x - 3 = 0</math> to <math>8xy + 2x = 12y + 3</math>. For there to be infinite solutions for <math>y</math> and there is no <math>x</math>, we simply find a value of <math>x</math> such that the equation is symmetrical. Therefore,
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<math>2x = 3</math>
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<math>8xy = 12y</math>
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There is only one solution, namely <math>x = \boxed{\dfrac{3}{2}}</math> or <math>\text{(D)}</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}}
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{{MAA Notice}}

Latest revision as of 12:12, 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 $x = \boxed{\dfrac{3}{2}}$ or $\text{(D)}$

See Also

2002 AMC 10B (Problems • Answer Key • Resources)
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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