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In this 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.

 

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

 Mar 5, 2022
 #2
avatar+117104 
0

I have forgotten how to do this.  crying

 

I would like to be reminded of the method though.  Any takers?

 Mar 6, 2022
 #3
avatar+32957 
+2

Here's one way (messy though!):

 Mar 6, 2022
 #4
avatar+67 
0

What would the other options be exactly?

shananigans  Mar 6, 2022
 #5
avatar+117104 
+2

Thanks very much Alan,

I finally managed it with a different method.

 

\(\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array} \)

 

Let 

\(a=3x \qquad\rightarrow \qquad x=\frac{a}{3}\\ b=5y \qquad\rightarrow \qquad y=\frac{b}{5}\\ c=-6z \qquad\rightarrow \qquad z=\frac{c}{-6}\\\)

 

The above equations become 

\(a+b+c=2\\ ab+bc+ac=-123\\ abc=-540\)

 

This gave me a hint:

If a,b and c are the roos of a monic cubic equations then we have.

 

Note: I used g because x has a different meaning in this question.

 

\(g^3-(a+b+c)g^2+(ab+bc+ac)g-(abcd)=0\\ g^3-(2)g^2+(-123)g-(-540)=0\\ g^3-2g^2-123g+540=0\\ \)

I used the wolfram|alpha calculator to factoize this and got

 

\((g-9)(g-5)(g+12)=0\)

 

so        a,b,c = -12,5,9     (any order)

These can be arranged in 3! = 6 ways so there will be 6 answers.

 

To show all the possible answers I made up a Excel table:

 

 

Alan thinks that there are 9 solutions but i don't know what the others could be ....

Notice that my last solution is the one that Alan found.  

 Mar 7, 2022
 #6
avatar+32957 
+2

Nicely done Melody.  Here are the answers found by Mathcad (I should have said 6, not 9):

Alan  Mar 7, 2022
 #7
avatar+117104 
0

Thanks Alan  laugh

Melody  Mar 7, 2022

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