2008 iTest Problems/Problem 8

Problem

(story eliminated)

Given the system of equations

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

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

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

find $x+y+z$ .

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:

10,203.