+24 Hi I need help solving:
(a) In this multi-part problem, we will consider this system of simultaneous equations:
\(\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array}\)
Let \(a=3x\),\(b=5y\), and \(c=-6z\)
Determine the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c. Make sure to enter 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 \(x+y\).
Thank you very much!!
+310 Part (a):
Let's rewrite the system of equations in terms of a, b, and c:
a + b + c = 2
ab + bc + ac = -41
abc = -108
We know that for a cubic polynomial with roots a, b, and c, it can be written in the form: p(t) = (t-a)(t-b)(t-c)
Expanding this, we get: p(t) = t^3 - (a+b+c)t^2 + (ab+bc+ac)t - abc
Substituting the values from our system of equations: p(t) = t^3 - 2t^2 - 41t + 108
Therefore, the monic cubic polynomial is: p(t) = t^3 - 2t^2 - 41t + 108
+310 Part (b):
From the previous part, we know that the cubic polynomial with roots a, b, and c is: p(t) = t^3 - 2t^2 - 41t + 108
To find the possible values of x+y, we need to find the possible pairs of roots that add up to (a+b).
We can use Vieta's formulas to find the sum of the roots and the product of the roots:
Sum of roots: a + b + c = 2
Product of roots: abc = -108
From the product of roots, we can see that one of the roots must be negative.
Let's consider the possible scenarios:
One root is -1:
If one root is -1, the other two roots must multiply to 108.
The possible pairs are (9, 12) and (-9, -12).
One root is -2:
If one root is -2, the other two roots must multiply to 54.
The possible pairs are (6, 9) and (-6, -9).
One root is -3:
If one root is -3, the other two roots must multiply to 36.
The possible pairs are (4, 9) and (-4, -9).
One root is -4:
If one root is -4, the other two roots must multiply to 27.
The possible pairs are (3, 9) and (-3, -9).
One root is -6:
If one root is -6, the other two roots must multiply to 18.
The possible pairs are (3, 6) and (-3, -6).
One root is -9:
If one root is -9, the other two roots must multiply to 12.
The possible pairs are (3, 4) and (-3, -4).
From these pairs, we can calculate the possible values of a+b:
9 + 12 = 21
-9 - 12 = -21
6 + 9 = 15
-6 - 9 = -15
4 + 9 = 13
-4 - 9 = -13
3 + 9 = 12
-3 - 9 = -12
3 + 6 = 9
-3 - 6 = -9
3 + 4 = 7
-3 - 4 = -7
Therefore, the distinct possible values of x+y (which is equivalent to (a+b)/5) are: 21/5, -21/5, 3, -3, 13/5, -13/5, 12/5, -12/5, 9/5, -9/5, 7/5, -7/5
