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).
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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).
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