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9x+8y+7z=0, 6x+5y+4z=0, 3x+2y+z=0

Have you learned about matixes & determinates?  If not:

You can multiply the third equation by three and subtract it from the first equation.  The result will be an equation that has variables y and z.  (x is eliminated.)

Similarly, multiply the third equation by two and subtract from the second equation.  This second result also will be devoid of the variable x.

You now have two equations with two unknowns (y and z).  Solve them for one of the two variables, say for z.  Substitute z back in to find y.  Substitute y and z into one of the original three equations to find x.

Oh man i hate these. You have to do simulatenous for all of them.
9x+8y+7z=0 (1)
6x+5y+4z=0 (2)
3x+2y+z=0 (3)

Make z the subject of (3)
Therefore z=3y+2y (4)

Sub (4) into (1)

9x+8y+7(3x+2y)=0
9x+8y+21x+14y=0
30x+22y=0 (5)
30x=-22y
x=-22y/30
 =-11/15

Sub x=-11/15 into (5)
20x+22y=0
30(-11/15)+22y=0
22y-22=0
22y=22
y=1

Sub y=1, z=-11/15 into (3)
z = 3(-11/15) + 2(1)
z = -1/5

Therefore x=-11/15, y=1 and z=-1/5

:D
i better get best answer ><"

(1) 9x+8y+7z=0
(2) 6x+5y+4z=0
(3) 3x+2y+z=0

From (3) isolate z
z = -3x -2y
Plug z in (1) and (2)

(1.1) 9x + 8y + 7(-3x -2y) = 0
9x +8y -21x -14y = 0
-12x -6y = 0 # divide by -6
(1.2) 2x + y = 0 ===> y = -2x

(2.1) 6x + 5y + 4(-3x -2y) = 0
6x +5y -12x -8y = 0
-6x -3y = 0 # divide by -3
(2.2) 2x + y = 0

Since 2x + y = 0 ===> y = -2x

Now (2.1) = (2.2)
Thus there's not a single solution.
The linear system ist true where
y = -2x
and
z = -3x -2(-2x) = -3x +4x = x

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