You should rewrite the third equation such that:

`1/x + 1/y + 1/z = (xy+yz+xz)/(xyz)`

Since the problem provides the value of xyz, you need to substitute 1 for xyz such that:

`(xy+yz+xz)/(xyz)= (xy+yz+xz)`

Since `1/x + 1/y + 1/z = 7/2 => xy+yz+xz = 7/2`

Notice that you may solve this system of equations using Vieta's relations, considering x,y,z as the roots of an equation `t^3 + (x+y+z)t^2 + (xy+yz+xz)t + xyz = 0` such that:

`t^3 + (7/2)t^2 + (7/2)t +1 = 0 `

`2t^3 + 7t^2 + 7t +2 = 0`

Notice that t=-1 is a root of this equation such that:

`(t+1)(at^2+bt+c) = 0`

You need to find the polynomial `at^2+bt+c` such that:

`2t^3 + 7t^2 + 7t + 2 = (t+1)(at^2+bt+c)`

`2t^3 + 7t^2 + 7t + 2 = at^3 + bt^2 + ct + at^2 + bt + c`

`2t^3 + 7t^2 + 7t + 2 = at^3 + t^2(a+b) + t(c+b) + c`

Equating coefficients of like powers yields:

`a = 2`

`a+b = 7 => b = 7-2 => b = 5`

`c = 2`

Hence, evaluating the coefficients a,b,c yields:

`at^2+bt+c = 2t^2+5t+2`

You need to use quadratic formula to find the zeroes of `2t^2+5t+2 = 0` such that:

`t_(1,2) = (-5+-sqrt(25 - 16))/4 => t_(1,2) = (-5+-sqrt9)/4`

`t_(1,2) = (-5+-3)/4 => t_1 = -1/2 ; t_2 = -2`

Notice that x, y, z represents the solutions to equation `2t^3 + 7t^2 + 7t + 2 = 0 .`

**Hence, evaluating the solutions to the given system of equations yields `x = -1 , y = -1/2 , z = -2` .**