The problem asks us to solve the given systems of equations. We will solve system (a). The system is: $x+y=a$ $y+z=b$ $z+x=c$

1. Problem Description

The problem asks us to solve the given systems of equations. We will solve system (a). The system is:

$x+y=a$

$y+z=b$

$z+x=c$

2. Solution Steps

We have the following system of equations:

$x+y=a$ (1)

$y+z=b$ (2)

$z+x=c$ (3)

Adding the three equations, we get:

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

$2x + 2y + 2z = a+b+c$

$2(x+y+z) = a+b+c$

$x+y+z = \frac{a+b+c}{2}$ (4)

From equation (4) and (1), we can find $z$:

$x+y+z = \frac{a+b+c}{2}$

$a+z = \frac{a+b+c}{2}$

$z = \frac{a+b+c}{2} - a = \frac{a+b+c-2a}{2} = \frac{b+c-a}{2}$

From equation (4) and (2), we can find $x$:

$x+y+z = \frac{a+b+c}{2}$

$x+b = \frac{a+b+c}{2}$

$x = \frac{a+b+c}{2} - b = \frac{a+b+c-2b}{2} = \frac{a+c-b}{2}$

From equation (4) and (3), we can find $y$:

$x+y+z = \frac{a+b+c}{2}$

$c+y = \frac{a+b+c}{2}$

$y = \frac{a+b+c}{2} - c = \frac{a+b+c-2c}{2} = \frac{a+b-c}{2}$

3. Final Answer

$x = \frac{a+c-b}{2}$

$y = \frac{a+b-c}{2}$

$z = \frac{b+c-a}{2}$