If $x - 2y + 4z = 6$, $3x + y + 5z = 4$ and $2x - y + 3z = 4$, then what is the value of $(2x – 3y + 5z)$?

Given the system of equations:

(1) $x - 2y + 4z = 6$

(2) $3x + y + 5z = 4$

(3) $2x - y + 3z = 4$

Add (2) and (3):

$(3x+y+5z) + (2x-y+3z) = 4+4$

$5x + 8z = 8$ (Equation A)

Multiply (3) by 2: $4x - 2y + 6z = 8$ (Equation 3')

Subtract (1) from (3'):

$(4x - 2y + 6z) - (x - 2y + 4z) = 8 - 6$

$3x + 2z = 2$ (Equation B)

From (B), $2z = 2 - 3x$. Substitute this into (A):

$5x + 4(2z) = 8$

$5x + 4(2 - 3x) = 8$

$5x + 8 - 12x = 8$

$-7x = 0 \Rightarrow x = 0$

Substitute $x=0$ into (B): $3(0) + 2z = 2 \Rightarrow 2z = 2 \Rightarrow z = 1$.

Substitute $x=0$ and $z=1$ into (2):

$3(0) + y + 5(1) = 4$

$y + 5 = 4 \Rightarrow y = -1$.

Now, calculate $2x - 3y + 5z$:

$2(0) - 3(-1) + 5(1) = 0 + 3 + 5 = 8$.