+389 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 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.
a) Substitue each term with a variable and put it in the system of equations.
\(\begin{array}{r@{~}c@{~}l l} a+b+c &=& 2, & (\textrm{i}') \\ \frac{ab}3+\frac{bc}3+\frac{ac}3 &=& -41, & (\textrm{ii}') \\ -\frac{abc}{90}&=&6. & (\textrm{iii}') \end{array} \)
After we get rid of the denominators we get
\(\begin{array}{r@{~}c@{~}l l} a+b+c &=& 2, & (\textrm{i}{'}{'}) \\ ab+bc+ac &=& -123, & (\textrm{ii}{'}{'}) \\ abc&=&-540. & (\textrm{iii}{'}{'}) \end{array}\)
Then use we simply use Vieta's formulas to get:
t^3 - 2t^2 - 123t + 540 = 0
