1986 AIME Problems/Problem 4

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 $\begin{align*} x_2-x_1&=6\\ x_3-x_17&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*}$ 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, $$ (Error compiling LaTeX. Unknown error_msg)3x_4+2x_5=3(x_1+42)=2(x_1+90)=\boxed{

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
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1986 AIME Problems/Problem 4

Contents

Problem

Solution

Solution 2

See also