If th equestion meant: (x+y+z)^2
The expansion is:
(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
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xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
Yes.
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)
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
What are we solving here for, Z, X, or Y?If solve for Z then the answer would be z = x + yIf solve for X then the answer would be z = z - yIf solve for Y than the answer would y = z - xHope this help.