Systems of Equations

Given x+y=z; x-y=z; Therefore ; x+y=x-y; Due to which 2y=0; So we can conclude that x=0(as x-y=0)and z=0(as xy=z)

$x + y = z \quad\quad\quad\quad\quad$ Eq. (1)

$x - y = z \quad\quad\quad\quad\quad$ Eq. (2)

$xy = z \quad\quad\quad\quad\quad\quad\;$ Eq. (3)

Eq. (1) $-$ Eq. (2):

$(x+y) - (x-y) = z - z\\ 2y=0 \implies y=0$

Substitute this into Eq. (3):

$0x = z \implies z=0$

Finally, substitute both values to find $x$ :

$x + 0 = 0 \implies x=0$

$x=y=z=0$ , therefore:

$3x + 2y + z = 3(0) + 2(0) + 0 = \boxed{0}$

Adding first two equations we get $2x=2z$ or $x=z$ . Thus we get $y=0$ .

From the third equation we get $xy=z$ or $0=z$ .

Thus $x=y=z=0$ .

Hence the value of $3x+2y+z=\boxed{0}$ .

Squaring the first two equations we get:

$z^2=(x+y)^2=x^2+2xy+y^2, z^2=(x-y)^2=x^2-2xy+y^2$

Taking these from each other: $4xy=0, z=xy \Rightarrow 4z=0 \Rightarrow z=0$

Going back to the original equations:

$x+y=z, x-y=z \Rightarrow 2x=2z \Rightarrow x=0$

Also $x-y=z \Rightarrow y=0$

So $x=y=z=0 \Rightarrow 3x+2y+z=0$