# 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 explain the answer.

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### 1 Answer

**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

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-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)) )`