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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.

 

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

(Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple (a, b, c), then determine the possible solutions (x,y, z).)

 May 27, 2020
 #1
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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 \(c=-6z\).

 

Given that (x, y, z) is a solution to the original system of equations, determine all distinct possible values of x+y.
(Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple (a, b, c), then determine the possible solutions (x,y, z).)

 

Using the substitutions in part (a):

\(\begin{array}{|rcll|} \hline && \begin{array}{c} a=3x,\ b=5y,\ \text{and} \ c=-6z \\ \end{array} \\ \hline 3x+5y-6z &=&2 \\ \mathbf{a+b+c} &=& \mathbf{2} \\\\ \hline abc &=& 3x5y(-6z) \\ abc &=& -90xyz \quad | \quad xyz = 6 \\ abc &=& -90*6 \\ \mathbf{abc} &=& \mathbf{-540} \\ \hline \end{array} \begin{array}{|rcl|} \hline \boxed{ ab=15xy,\ bc=-30yz,\ \text{and} \ ac=-18xz \\ xy = \frac{ab}{15},\ yz = -\frac{bc}{30},\ xz = -\frac{ac}{18} } \\ \hline 5xy-10yz-6xz &=& -41 \\ 5*\frac{ab}{15}+10*\frac{bc}{30}+6*\frac{ac}{18} &=& -41 \\ \frac{ab}{3}+\frac{bc}{3}+\frac{ac}{3} &=& -41 \\ ab+bc+ac &=& -3*41 \\ \mathbf{ab+bc+ac} &=& \mathbf{-123} \\ \hline \end{array} \)

 

\(\begin{array}{|rcl|} \hline \begin{array}{r@{~}c@{~}l l} \mathbf{a+b+c} &=& \mathbf{2} \\ \mathbf{ab+bc+ac} &=& \mathbf{-123} \\ \mathbf{abc} &=& \mathbf{-540} \\ \end{array} \\ \hline \text{If $a$, $b$, $c$ are the roots of $(x-a)(x-b)(x-c)=0 \\$expandig we get $x^3-(a+b+c)x^2+(ab+bc+ac)x^2-abc = 0$} \\ \mathbf{x^3-2x^2-123x^2+540} &=& \mathbf{0} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline x^3-2x^2-123x^2+540 &=& 0 \quad | \quad \text{re-write as} \\ x^3 - 2x^2 - 63x - 60x + 540 &=& 0 \\ x(x^2 - 2x - 63) - 60x + 540 &=& 0 \\ x(x -9)(x + 7) - 60 (x - 9) &=& 0 \\ (x - 9) [ x(x + 7) - 60 ] &=& 0 \\ (x - 9) [ x^2 + 7x - 60] &=& 0 \\ (x - 9) ( x + 12)(x - 5) &=& 0 \\\\ \text{The roots: $5$, $9$, and $-12$ } \\ \hline \end{array}\)

re-write see by CPhill: https://web2.0calc.com/questions/in-this-multi-part-problem-we-will-consider-this

 

The solutions for a, b, c:

\(\begin{array}{|rcll|} \hline \text{possible values of $a = 5$, $9$, and $-12$ } \\ \text{possible values of $b = 5$, $9$, and $-12$ } \\ \text{possible values of $c = 5$, $9$, and $-12$ } \\ \hline \end{array}\)

 

There are 6 permutations for a,b,c:

\(\begin{array}{|r|r|r|r|r|r|r|} \hline & & & x & y & z \\ a & b & c & =\frac{a}{3} & =\frac{b}{5} & = -\frac{c}{6} & \mathbf{x+y} \\ \hline 5 & 9 & -12 & \frac{5}{3} & \frac{9}{5} & \frac{12}{6}=2 & \frac{52}{15} \\ \hline 9 & -12 & 5 & \frac{9}{3}=3 & -\frac{12}{5} & -\frac{5}{6} & \frac{3}{5} \\ \hline -12 & 5 & 9 & -\frac{12}{3}=-4 & \frac{5}{5}=1 & -\frac{9}{6}=-\frac{3}{2} & -3 \\ \hline 9 & 5 & -12 & \frac{9}{3}=3 & \frac{5}{5}=1 & \frac{12}{6}=2 & 4 \\ \hline 5 & -12 & 9 & \frac{5}{3} & -\frac{12}{5} & -\frac{9}{6}-\frac{3}{2} & -\frac{11}{15} \\ \hline -12 & 9 & 5 & -\frac{12}{3}=-4 & \frac{9}{5} & -\frac{5}{6} & -\frac{11}{5} \\ \hline \end{array}\)

 

laugh

 May 28, 2020

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