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Part a) In this multi-part problem, we will consider this system of simultaneous equations: 

3x+5y-6z=2,
5xy-10yz-6xz=-41
xyz=6.

Let a=3x, b=5y, and c=-6z.

Determine the monic cubic polynomial in terms of a variable t whose roots are a, b, and c. Make sure to enter your answer in terms of t and only t, in expanded form.

 

Part b) Given that (x,y,z) is a solution to the original system of equations, determine all distinct possible values of x+y.

 

I've solved part a and got the equations a+b+c=2, ab+bc+ac=-123, and abc=-540. Can someone help with part b?

 Aug 20, 2018
 #1
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b)

My CAS system gives the follwing values:
x = -4, y = 1, z = -3/2
x = -4, y = 9/5,  z = -5/6
x = 5/3, y = -12/5,  z = -3/2
x = 5/3,  y = 9/5,  z = 2
x = 3,  y = -12/5,  z = -5/6
x = 3,  y = 1,  z = 2

 Aug 20, 2018
edited by Guest  Aug 20, 2018
 #2
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Part a)
In this multi-part problem, we will consider this system of simultaneous equations:

 

\(\begin{array}{lcr} 3x + 5y - 6z &=& 2 \\ 5xy - 10yz - 6xz &=& -41 \\ xyz &=& 6 \\ \end{array}\)

 

\(\text{Let $a = 3x$, $b = 5y$, and $c = -6z.$}\)

 

Determine the monic cubic polynomial in terms of a variable \(t\) whose roots are \(a\), \(b\), and \(c\).
Make sure to enter your answer in terms of \(t\) and only \(t\), in expanded form.

 

\(\begin{array}{|rcll|} \hline (t-a)(t-b)(t-c) &=& 0 \quad | \quad a=3x, \quad b = 5y \quad c=-6z \\ \left(t^2-t(a+b)+ab\right)(t-c) &=& 0 \\ t^3-t^2c-t^2(a+b)+t(a+b)c+tab-abc &=& 0 \\ t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ \\ \hline a+b+c &=& 3x + 5y -6z \\ &=& 2 \\ \\ \hline ac+bc+ab &=& -18xz-30yz+15xy \\ &=& 3\cdot(5xy-10yz-6xz) \\ &=& 3\cdot(-41) \\ &=& -123 \\ \\ \hline abc &=& 3x5y(-6z) \\ &=& -90xyz \\ &=& -90\cdot 6 \\ &=& -540 \\ \\ \hline t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ t^3-t^2\cdot 2+t(123)-(-540) &=& 0 \\ \mathbf{t^3-2t^2 -123t +540} & \mathbf{=} & \mathbf{0} \\ \text{The roots are } t_1 = -12, t_2 = 5, t_3 = 9 \\ \hline \end{array}\)

 

 

Part b)
Given that \((x,y,z)\) is a solution to the original system of equations,

determine all distinct possible values of \(x+y\).

 

Roots are \(a\), \(b\), and \(c\)

 

\(\mathbf{x =\ ?}\)

\(\begin{array}{|rcll|} \hline a = 3x_1 &=& t_1 \\ 3x_1 &=& -12 \\ x_1 &=& -4 \\ \\ a = 3x_2 &=& t_2 \\ 3x_2 &=& 5 \\ x_2 &=& \frac53 \\\\ a = 3x_3 &=& t_3 \\ 3x_3 &=& 9 \\ x_3 &=& 3 \\\\ \hline x &=& \{-4,\frac53,3 \} \\ \hline \end{array} \)

 

\(\mathbf{y =\ ?}\)

\(\begin{array}{|rcll|} \hline b = 5y_1 &=& t_1 \\ 5y_1 &=& -12 \\ y_1 &=& -\frac{12}{5} \\ \\ b = 5y_2 &=& t_2 \\ 5y_2 &=& 5 \\ y_2 &=& 1 \\\\ b = 5y_3 &=& t_3 \\ 5y_3 &=& 9 \\ y_3 &=& \frac95 \\\\ \hline y &=& \{-\frac{12}{5},1,\frac95 \} \\ \hline \end{array}\)

 

\(\mathbf{z =\ ?}\)

\(\begin{array}{|rcll|} \hline c = -6z &=& t_1 \\ -6z &=& -12 \\ z_1 &=& 2 \\ \\ c = -6z &=& t_2 \\ -6z &=& 5 \\ z_2 &=& -\frac{5}{6} \\\\ c = -6z &=& t_3 \\ -6z &=& 9 \\ z_3 &=& -\frac32 \\\\ \hline z &=& \{2,-\frac{5}{6},-\frac32 \} \\ \hline \end{array}\)

 

Solutions:

\(\begin{array}{|r|r|r|r|} \hline x & y & z & \text{all distinct possible values of } x+y \\ \hline -4 & 1& -\frac32 & -3 \\ \hline -4 & \frac95& -\frac56 & -2.2 \\ \hline \frac53&-\frac{12}{5}& -\frac32 & -\frac{11}{15} \\ \hline \frac53& \frac95& 2 & \frac{52}{15} \\ \hline 3&-\frac{12}{5}& -\frac56 & \frac35 \\ \hline 3& 1& 2 & 4 \\ \hline \end{array}\)

 

laugh

 Aug 21, 2018

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