Difference between revisions of "1986 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
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Determine <math>\displaystyle 3x_4+2x_5</math> if <math>\displaystyle x_1</math>, <math>\displaystyle x_2</math>, <math>\displaystyle x_3</math>, <math>\displaystyle x_4</math>, and <math>\displaystyle x_5</math> satisfy the system of equations below.
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<center><math>\displaystyle 2x_1+x_2+x_3+x_4+x_5=6</math></center>
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<center><math>\displaystyle x_1+2x_2+x_3+x_4+x_5=12</math></center>
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<center><math>\displaystyle x_1+x_2+2x_3+x_4+x_5=24</math></center>
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<center><math>\displaystyle x_1+x_2+x_3+2x_4+x_5=48</math></center>
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<center><math>\displaystyle x_1+x_2+x_3+x_4+2x_5=96</math></center>
  
 
== Solution ==
 
== Solution ==
 
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{{solution}}
 
== See also ==
 
== See also ==
 
* [[1986 AIME Problems]]
 
* [[1986 AIME Problems]]
  
 
{{AIME box|year=1986|num-b=3|num-a=5}}
 
{{AIME box|year=1986|num-b=3|num-a=5}}

Revision as of 19:52, 10 February 2007

Problem

Determine $\displaystyle 3x_4+2x_5$ if $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$, $\displaystyle x_4$, and $\displaystyle x_5$ satisfy the system of equations below.

$\displaystyle 2x_1+x_2+x_3+x_4+x_5=6$
$\displaystyle x_1+2x_2+x_3+x_4+x_5=12$
$\displaystyle x_1+x_2+2x_3+x_4+x_5=24$
$\displaystyle x_1+x_2+x_3+2x_4+x_5=48$
$\displaystyle x_1+x_2+x_3+x_4+2x_5=96$

Solution

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See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions