# What are x,y and z? x+y=-3 x+z=-2 xy+yz+xz=2

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We have to solve the equations:

x + y = -3 ...(1)

x + z = -2 ...(2)

xy + yz + xz = 2 ...(3)

From (1)

x + y = -3

=> y = -3 - x

From (2)

x + z = -2

=> z = -2 - x

Substitute these in (3)

xy + yz + xz = 2

=> x(-3 - x) + (-3 - x)(-2 - x) + x(-2 - x) = 2

=> -3x- x^2 + 6 + 3x + 2x + x^2 - 2x - x^2 = 2

=> - x^2 + 6 = 2

=> x^2 = 4

=> x = 2 and x = -2

y = -3 - x = -5 and -1

z = -2 - x = -4 and 0

**The values of x, y and z are (2, -5, -4) and (-2, -1, 0).**

First, we'll multiply the 1st and the 2nd equations:

(x+y)(x+z) = (-3)*(-2)

We'll remove the brackets:

x^2 + xz + xy + yz = 6

We notice that the sum xz + xy + yz can be replaced by the 3rd equation:

x^2 + 2= 6

x^2 = 6-2

x^2 = 4

x1 = 2 and x2 = -2

We'll substitute x1 in the 1st equation:

2 + y = -3

y = -2-3

y = -5

We'll substitute x1 in the 2nd equation:

2 +z = -2

z = -4

We'll substitute x2 in the 1st equation:

-2 + y = -3

y = -1

We'll substitute x2 in the 2nd equation:

-2 + z = -2

z = 0

**The pairs of solutions are: (2 ; -5 ; -4) and (-2 ;-1 ; 0).**