Find $x + y + z$

let $a = \frac{1}{x}$ ; $b = \frac{1}{y}$ ; $c = \frac{1}{z}$

$\frac{4}{x} + \frac{1}{y} + \frac{1}{z} = 5$ becomes $4a + b + c = 5$ (equation 1)

$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 4$ becomes $a + b - c = 4$ (equation 2)

$\frac{6}{x} - \frac{1}{y} - \frac{1}{z} = 9$ becomes $6a - b - c = 9$ (equation 3)

Subtract equation 2 from equation 1 to eliminate $b$ . We obtain

$3a + 2c = 1$ (equation 4)

Add equation 2 and equation 3 to eliminate b. We obtain

$7a - 2c = 13$ (equation 5)

Add equations 4 and 5 to eliminate c. We obtain

$10a = 14$

$a = \frac{14}{10} = \frac{7}{5}$

In equation 4, substitute $a = \frac{7}{5}$ to solve for $c$ . We obtain

$3a + 2c = 1$

$3(\frac{7}{5}) + 2c = 1$

$c = -\frac{8}{5}$

In equation 1, substitute $a = \frac{7}{5}$ and $c = -\frac{8}{5}$ to solve for $b$ .

$4a + b + c = 5$

$4(\frac{7}{5} + b - \frac{8}{5} = 5$

$b = 1$

From here, we can now find the values of $x, y$ and $z$ .

$x = \frac{5}{7}$ ; $y = 1$ and $z = -\frac{5}{8}$

$x + y + z = \frac{5}{7} + 1 - \frac{5}{8} =$ $\boxed{\frac{61}{56}}$