Add 'em up!

You only need to sum up the three equations, which produces $2(x+y+z) = 12$ , thus $x+y+z = 6$ . More precisely, $(x, y, z) = (1, 2, 3)$

x=3-y => z+3-y=4 => z=4-3+y = 1+y

y + 1 + y = 5 so y=2 x=1 z=3

add them up to get 6 :)

A: x+y=3

B: y+z=5

C: z+x=4

Def A for x: x=3-y

A in C: z+(3-y)=4

z=1+y

Def B for y: y=5-z

To equations with to unknown:

1: z=1+y

2: y=5-z

2 in 1: z=1+(5-z)

        z=3

z=3 in 2: y=5-3

       y=2

y=2 in A to find x: x=3-2

      x=1

Answer: x+y+z=

            1+2+3=**6**

So simple. x+y=3, y+z=5, z+x=4, So, x+y+y+z+z+x=3+5+4.
2(x+y+z)=12.
x+y+z=12/2=6

By adding the equations, we get
2(x+y+z)=12 Therefore, x+y+z=6

Simply..... x+y+y+z+z+x=3+4+5 => 2(x+y+z) = 12 .'. x+y+z = 6 .

Consider the last equation, z + x = 4. If we assume here that neither z or x are zero, then either z = 3 and x = 1 or vice versa. But if z = 1, then y must be 4 since z + y = 5. This is impossible, because then x + y would be 7, but it says here that x + y = 3. So now we only have one value for z and x: z = 3 and x = 1. If x = 1, then y = 2 since x + y = 3.

x = 1, y = 2, and z = 3.

1 + 2 + 3 = 6.

solving the system of equations we get,x=1,y=2,z=3 so the sum is 6