Solve for x, y, and z: x + y + z = 0 (b+c)x + (c+a)y + (a+b)z = 0 bcx + cay + abz = 1 (all equations under 1 bracket)Show complete solution and...

Solve the following system for x,y,and
z:
(i) 
x+y+z=0
(ii)
(b+c)x+(a+c)y+(a+b)z=0
(iii)
bcx+acy+abz=1

We proceed by
substitution:

(1) From (i) we get z=-x-y. Plugging this
into (ii) yields
(b+c)x+(a+c)y+(a+b)(-x-y)=0 which simplifies
to
(I) 
-(a-c)x-(b-c)y=0

Plugging into (iii)
yields
bcx+acy+ab(-x-y)=1 which simplifies to
(II) 
b(c-a)x-a(b-c)y=1

(2) We now solve (I) and
(II) simultaneously using linear
combinations:

-b(a-c)x-b(b-c)y=0   multiply (I) by
-b
-b(a-c)x-a(b-c)y=1
--------------------
-a(b-c)y+b(b-c)y=1 
subtract (II)-b(I) ; which simplifies
to

`y=-1/((a-b)(b-c))`

(3)
Plug this value for y into (I) to get
x:

`-(a-c)x+((b-c))/((a-b)(b-c))=-(a-c)x+1/(a-b)=0`

So
`x=1/((a-b)(a-c))`

(4) Take these values for x and y and
plug into (i) to get
z:

`z=-x-y=-1/((a-b)(a-c))+1/((a-b)(b-c))=(-b+c+a-c)/((a-b)(a-c)(b-c))=1/((a-c)(b-c))`

So
the values for (x,y,z) are `( 1/((a-b)(a-c)),(-1)/((a-b)(b-c)),1/((a-c)(b-c))
)`