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

Revision as of 17:45, 29 May 2016

(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.

Revision as of 23:28, 4 August 2014 (view source) BENJAMINDJON (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 17:45, 29 May 2016 (view source) Niu (talk \| contribs) (→‎See also) Newer edit →
Line 23:		Line 23:

	==See also==		==See also==
		+	lol