There are 8 different subsets.
The null set.
{x}
{y}
{z}
{x y}
{x z}
{y z} {x y z}
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xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
7
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
x + y + z = 12 y = 1 x - y - z = 0 Substitute y = 1 in the other two equations: x + 1 + z = 12 so that x + z = 11 and x - 1 - z = 0 so that x - z = 1 Add these two equations: 2x + 0z = 12 which implies that x = 6 and then x + z = 11 gives z = 5 So the solution is (x, y, z) = (6, 1, 5)
Given x=k1y and x=k2/z x=125 ,y=5 then k1=25 x=125 , z=4 then k2=125(4)=500 If y=4 ,z=5 then x=25y = 100