A Possible Problem...?

Since $y = 0$ , we can eliminate the $y$ variable:

$xz + 13x + 13z = 0$

$xz + 13x + 13z = 0$

$xz + 13x - 13z = 0$

$xz - 13x - 13z = 0$

$xz - 13x - 13z = 0$

$xz - 13x + 13z = 0$

Now, since all $6$ equations are the same, we can eliminate $3$ of them to create $3$ simultaneous equations:

$xz - 13x - 13z = 0$

$xz - 13x + 13z = 0$

$xz + 13x - 13z = 0$

Now, since $x = -z$ , this must mean that $z = -x$ . We will now take two routes:

Route $1$ - $x = -z$ :

Substitute $x = -z$ into the simultaneous equations:

$-z(z) - 13(-z) - 13z = 0$

$-z(z) - 13(-z) + 13z = 0$

$-z(z) + 13(-z) - 13z = 0$

Simplify:

$-z + 13z - 13z = 0$

$-z + 13z + 13z = 0$

$-z - 13z - 13z = 0$

Simplify even further:

$-z = 0$

$-z + 26z = 0$

$-z - 26z = 0$

Simplify the second equation even further:

$25z = 0$ , $z = 0$

Simplify the third equation even further:

$-27z = 0$ , $z = 0$

Since $x = -z$ , $x = 0$

Route $2$ - $z = -x$ :

Substitute $z = -x$ into the simultaneous equations:

$x(-x) - 13x - 13(-x) = 0$

$x(-x) - 13x + 13(-x) = 0$

$x(-x) + 13x - 13(-x) = 0$

Simplify:

$-x - 13x + 13x = 0$

$-x - 13x - 13x = 0$

$-x + 13x + 13x = 0$

Simplify even further:

$-x + 26x = 0$

$-x - 26x = 0$

$-x = 0$

Simplify the second equation even further:

$25x = 0$ , $x = 0$

Simplify the third equation even further:

$-27x = 0$ , $x = 0$

Since $z = -x$ , $z = 0$

In both routes $x = z = 0$ . Now since $y = 0$ , $\fbox{x = y = z = 0}$ . We have a solution!

Therefore, the answer is $\fbox 1$ .

Only from the last three equations: $x=y=z=0$

x =y=z=0 and a suggestion brilliant gives you three chances and there are only two possibility :-P lol

As $y = 0$ , we can eliminate some expressions from the list of equations

$13x + 13z$

$13x - 13z$

$13z - 13x$

$-13z - 13x$

Replacing $x$ with $-z$

$13x - 13x$

$13x + 13x$

$-13x - 13x$

$13x - 13x$

The double of a number will never be equal to zero unless the number itself is zero

As $x$ is zero, we have also proved that $z = 0$ , as $x = -z$