Difference between revisions of "2008 iTest Problems/Problem 8"

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== Problem ==
 
== Problem ==
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Given the system of equations
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The math team at Jupiter Falls Middle School meets twice a month during the Summer, and the math team coach, Mr. Fischer, prepares some Olympics-themed problems for his students. One of the problems Joshua and Alexis work on boils down to a system of equations:
  
 
<center><math>2x + 3y + 3z = 8</math>,
 
<center><math>2x + 3y + 3z = 8</math>,
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==Solution==
 
==Solution==
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<cmath>2x + 3y + 3z = 8</cmath>
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<cmath>3x + 2y + 3z = 808</cmath>
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<cmath>3x + 3y + 2z = 80808</cmath>
  
==See also==
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The Solution can be found by summing the three equations to get:
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<cmath>8x + 8y + 8z = 81624</cmath>
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then solving for <math>x+y+z</math>, divide by 8 to get:
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<math>\boxed{10,203}</math>
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==See Also==
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{{2008 iTest box|num-b=7|num-a=9}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 23:59, 21 June 2018

Problem

The math team at Jupiter Falls Middle School meets twice a month during the Summer, and the math team coach, Mr. Fischer, prepares some Olympics-themed problems for his students. One of the problems Joshua and Alexis work on boils down to a system of equations:

$2x + 3y + 3z = 8$,

$3x + 2y + 3z = 808$,

$3x + 3y + 2z = 80808$,

find $x+y+z$.

Solution

\[2x + 3y + 3z = 8\] \[3x + 2y + 3z = 808\] \[3x + 3y + 2z = 80808\]

The Solution can be found by summing the three equations to get:

\[8x + 8y + 8z = 81624\]

then solving for $x+y+z$, divide by 8 to get:

$\boxed{10,203}$

See Also

2008 iTest (Problems)
Preceded by:
Problem 7
Followed by:
Problem 9
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