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>. | ||
| − | <math>\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \ | + | <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>. | ||
| − | 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>. |
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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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:12, 30 November 2019
Contents
Problem
Find the value(s) of 
such that 
is true for all values of 
.
Solution
We have 
.
As 
must be true for all , we must have 
, hence 
.
Solution 2
Since we want only the -variable to be present, we move the terms only with the -variable to one side, thus constructing to 
. For there to be infinite solutions for and there is no , we simply find a value of such that the equation is symmetrical. Therefore,
There is only one solution, namely 
or 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
