Solve it if you can

Add the first two equations to find that $(x + y)*(z + 1) = 15z$ . Note now that since $z \gt 0$ we can divide through by $(z + 1)$ .

Thus $x + y = \dfrac{15z}{z + 1}$ . Substitute this into the third equation to find that

$\dfrac{15z}{z + 1} + z = 12 \Longrightarrow 15z + z^{2} + z = 12z + 12 \Longrightarrow z^{2} + 4z - 12 = 0 \Longrightarrow (z + 6)(z - 2) = 0$ .

Now since $z \gt 0$ we can conclude that $z = 2$ . Substituting this value into the first and last equations gives us that

$(2x + y) - (x + y) = 14 - 10 = 4 \Longrightarrow x = 4$ , and so $y = 6$ , giving us

$xyz = 4*6*2 = \boxed{48}$ .

Hello,

xz + y = 7z(1st)

yz + x = 8z(2nd)

x + y + z = 12(3rd)

let's take x = 12 -y -z from (3rd),then substitute into (1st) & (2nd),

yz + 12 -y -z = 8z

zy + 12 -y - 9z = 0

zy = 9z + y -12(4th)

z(12-y-z) + y = 7z

12z - zy - z^2 + y = 7z

5z - zy - z^2 + y = 0(5th)

substitute (4th) into (5th),

5z - (9z + y -12) - z^2 + y = 0

5z - 9z - y + 12- z^2 + y = 0

-4z + 12 - z^2 = 0

z^2 + 4z -12 = 0

(z + 6)(z - 2) = 0

z = -6, z = 2

since z>0, take z=2(subsitute z=2 into (1st) & (2nd)),

2x + y = 14

y = 14 -2x(6th),

2y + x = 16(7th),

substitute (6th) into (7th),

2(14 - 2x) + x = 16

28 - 4x + x = 16

-3x = -12

x = 4

y = 14 -(2 x4) = 6, therefore x = 4, y = 6, z =2 , also check that x + y + z = 4 + 6 + 2 =12,

xyz=(4)(6)(2)=48

thanks...