+389 So, sorry guys for reposting a question. But, it got deleted so I can't see it.
In this multi-part problem, we will consider this system of simultaneous equations:
3x+5y-6z=2, (i)
5xy-10yz-6xz=-41, (ii)
xyz=6. (iii)
Let a=3x, b=5y, and c=-6z.
(a) Determine the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c. Make sure your answer in terms of t and only t, in expanded form.
(b) Given that (x,y,z) is a solution to the original system of equations, determine all distinct possible values of x+y.
+33661 Replace x, y and z in the equations by a, b and c:
a + b + c = 2
a*b/3 + b*c/3 + a*c/3 = -41. So. a*b + b*c + a*c = -123
-a*b*c/(3*5*6) = 6. So. a*b*c = -540
(t - a)(t - b)(t - c) = 0. ⇒ t^3 - (a + b + c)t^2 + (a*b + b*c + a*c)t - a*b*c = 0
Hence: t^3 - 2t^2 - 123t + 540 = 0
Solving this gives: a = -12, b = 5, c = 9 or variants of these.
So: x = a/3 → -4, y = b/5 → 1, z = -c/6 → -3/2
x + y = -3 This is just one possibility. The others can be found by interchanging the values allocated to a, b and c.
+15000 Hello dabae!
3x+5y-6z=2
5xy-10yz-6xz=-41
xyz=6
x = 1,666666666667
y = 1,8
z = 2
Solved with computers for nonlinear equations.
Probably not the expected. Sorry!
Greeting asinus :- ) 
!
+33661 Replace x, y and z in the equations by a, b and c:
a + b + c = 2
a*b/3 + b*c/3 + a*c/3 = -41. So. a*b + b*c + a*c = -123
-a*b*c/(3*5*6) = 6. So. a*b*c = -540
(t - a)(t - b)(t - c) = 0. ⇒ t^3 - (a + b + c)t^2 + (a*b + b*c + a*c)t - a*b*c = 0
Hence: t^3 - 2t^2 - 123t + 540 = 0
Solving this gives: a = -12, b = 5, c = 9 or variants of these.
So: x = a/3 → -4, y = b/5 → 1, z = -c/6 → -3/2
x + y = -3 This is just one possibility. The others can be found by interchanging the values allocated to a, b and c.
