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You have given two equations only and there are three variables. Hence one cannot solve for x as a numerical value. But x can be found in terms of y and z.

x + y + z = 15 --- 4

(x + z) y + z x = 72

(15 - y) y + z x = 72

x = ( y^2 - 15 y + 72 ) / z = 15 - y - z --- 3

then y and z satisfy : y^2 + z^2 - 15 (y+z) + y z +72 = 0 - -- 2

you can solve for y in terms of z.

Further,

(x+y+z)^2 - 2 (xy+yz+zx) = 15^2 - 2 * 72

x^2 + y^2 + z^2 = 81

81 - x^2 - 15 (y + z) + y z + 72 = 0

x^2 = 153 - 15 ( y + z ) + y z --- 1

substitue the value of y in terms of z, then x can be obtained in terms of z.

x + y + z = 15 --- 4

(x + z) y + z x = 72

(15 - y) y + z x = 72

x = ( y^2 - 15 y + 72 ) / z = 15 - y - z --- 3

then y and z satisfy : y^2 + z^2 - 15 (y+z) + y z +72 = 0 - -- 2

you can solve for y in terms of z.

Further,

(x+y+z)^2 - 2 (xy+yz+zx) = 15^2 - 2 * 72

x^2 + y^2 + z^2 = 81

81 - x^2 - 15 (y + z) + y z + 72 = 0

x^2 = 153 - 15 ( y + z ) + y z --- 1

substitue the value of y in terms of z, then x can be obtained in terms of z.