We are given a system of three linear equations with three unknowns $x$, $y$, and $z$. The system is: $x - y + z = 0$ $(a+b)x - (a+c)y + (b+c)z = 0$ $abx - acy + bcz = 1$ We need to solve for $x$, $y$, and $z$.

1. Problem Description

We are given a system of three linear equations with three unknowns $x$, $y$, and $z$. The system is:

$x - y + z = 0$

$(a+b)x - (a+c)y + (b+c)z = 0$

$abx - acy + bcz = 1$

We need to solve for $x$, $y$, and $z$.

2. Solution Steps

From the first equation, we have $z = y - x$. Substituting this into the second and third equations gives:

$(a+b)x - (a+c)y + (b+c)(y-x) = 0$

$abx - acy + bc(y-x) = 1$

Simplifying the second equation:

$(a+b)x - (a+c)y + (b+c)y - (b+c)x = 0$

$(a+b-b-c)x + (-a-c+b+c)y = 0$

$(a-c)x + (b-a)y = 0$

$(a-c)x = (a-b)y$

$y = \frac{a-c}{a-b} x$

Now, simplifying the third equation:

$abx - acy + bcy - bcx = 1$

$(ab-bc)x + (bc-ac)y = 1$

$b(a-c)x + c(b-a)y = 1$

Substitute $y = \frac{a-c}{a-b} x$ into the simplified third equation:

$b(a-c)x + c(b-a)\frac{a-c}{a-b} x = 1$

$[b(a-c) + c(b-a)\frac{a-c}{a-b}]x = 1$

$[b(a-c)(a-b) + c(b-a)(a-c)]x = (a-b)$

$(a-c)[b(a-b) + c(b-a)]x = (a-b)$

$(a-c)[ab - b^2 + cb - ca]x = (a-b)$

$(a-c)[ab - b^2 + cb - ca]x = (a-b)$

$x = \frac{a-b}{(a-c)(ab-b^2+bc-ac)}$

Now, we find y:

$y = \frac{a-c}{a-b} x = \frac{a-c}{a-b} \frac{a-b}{(a-c)(ab-b^2+bc-ac)} = \frac{1}{ab-b^2+bc-ac}$

Finally, we find z:

$z = y - x = \frac{1}{ab-b^2+bc-ac} - \frac{a-b}{(a-c)(ab-b^2+bc-ac)}$

$z = \frac{(a-c)-(a-b)}{(a-c)(ab-b^2+bc-ac)} = \frac{a-c-a+b}{(a-c)(ab-b^2+bc-ac)} = \frac{b-c}{(a-c)(ab-b^2+bc-ac)}$

We can factor $(ab-b^2+bc-ac)$ as:

$ab-b^2+bc-ac = b(a-b) - c(a-b) = (a-b)(b-c)$

So $x = \frac{a-b}{(a-c)(a-b)(b-c)} = \frac{1}{(a-c)(b-c)}$

$y = \frac{1}{(a-b)(b-c)}$

$z = \frac{b-c}{(a-c)(a-b)(b-c)} = \frac{1}{(a-c)(a-b)}$

3. Final Answer

$x = \frac{1}{(a-c)(b-c)}$

$y = \frac{1}{(a-b)(b-c)}$

$z = \frac{1}{(a-c)(a-b)}$