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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.
+
Determine <math>3x_4+2x_5</math> if <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, <math>x_4</math>, and <math>x_5</math> satisfy the system of equations below.
−
<center><math>\displaystyle 2x_1+x_2+x_3+x_4+x_5=6</math></center>  
+
<center><math>2x_1+x_2+x_3+x_4+x_5=6</math></center>  
−
<center><math>\displaystyle x_1+2x_2+x_3+x_4+x_5=12</math></center>
+
<center><math>x_1+2x_2+x_3+x_4+x_5=12</math></center>
−
<center><math>\displaystyle x_1+x_2+2x_3+x_4+x_5=24</math></center>
+
<center><math>x_1+x_2+2x_3+x_4+x_5=24</math></center>
−
<center><math>\displaystyle x_1+x_2+x_3+2x_4+x_5=48</math></center>
+
<center><math>x_1+x_2+x_3+2x_4+x_5=48</math></center>
−
<center><math>\displaystyle x_1+x_2+x_3+x_4+2x_5=96</math></center>
+
<center><math>x_1+x_2+x_3+x_4+2x_5=96</math></center>
  
 
== Solution ==
 
== Solution ==
−
Adding all five [[equation]]s gives us <math>6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)</math> so <math>x_1 + x_2 + x_3 + x_4 + x_5 = 31</math>.  Subtracting this from the fourth given equation gives <math>x_4 = 17</math> and subtracting it from the fifth given equation gives <math>x_5 = 65</math>, so our answer is <math>3\cdot17 + 2\cdot65 = 181</math>.
+
Adding all five [[equation]]s gives us <math>6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)</math> so <math>x_1 + x_2 + x_3 + x_4 + x_5 = 31</math>.  Subtracting this from the fourth given equation gives <math>x_4 = 17</math> and subtracting it from the fifth given equation gives <math>x_5 = 65</math>, so our answer is <math>3\cdot17 + 2\cdot65 = \boxed{181}</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Subtracting the first equation from every one of the other equations yields
 +
<cmath>\begin{align*}
 +
x_2-x_1&=6\\
 +
x_3-x_1&=18\\
 +
x_4-x_1&=42\\
 +
x_5-x_1&=90
 +
\end{align*}</cmath>
 +
Thus
 +
<cmath>\begin{align*}
 +
2x_1+x_2+x_3+x_4+x_5&=6\\
 +
2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\
 +
6x_1+156&=6\\
 +
x_1&=-25
 +
\end{align*}</cmath>
 +
Using the previous equations,
 +
<cmath>3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}</cmath>
 +
 
 +
~ Nafer
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=3|num-a=5}}
 
{{AIME box|year=1986|num-b=3|num-a=5}}
 +
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:25, 17 November 2019

Problem

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

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

Solution

Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$.

Solution 2

Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, \[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}\]

~ Nafer

See also

1986 AIME (Problems • Answer Key • Resources)
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

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