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System of equations
Entrepreneur   2
N Yesterday at 8:57 AM by lbh_qys
Solve for $x,y\;\&\;z:$ $$\color{blue}{\frac xa+\frac{y+z}{b+c}=\frac yb+\frac{z+x}{c+a}=\frac zc+\frac{x+y}{a+b}=2.}$$
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Entrepreneur
Nov 10, 2024
lbh_qys
Yesterday at 8:57 AM
System of equations
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Entrepreneur
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Solve for $x,y\;\&\;z:$ $$\color{blue}{\frac xa+\frac{y+z}{b+c}=\frac yb+\frac{z+x}{c+a}=\frac zc+\frac{x+y}{a+b}=2.}$$
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Entrepreneur
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Bump....
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lbh_qys
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Rewrite the system of equations as:

\[
\frac{b+c}{a} x + y + z = 2(b+c)
\]\[
x + \frac{c+a}{b} y + z = 2(c+a)
\]\[
x + y + \frac{a+b}{c} z = 2(a+b)
\]
The determinant of the coefficient matrix is:

\[
\left| \begin{matrix} 
\frac{b+c}{a} & 1 & 1 \\
1 & \frac{c+a}{b} & 1 \\
1 & 1 & \frac{a+b}{c}
\end{matrix} \right| = 
\prod \frac{b+c}{a} + 2 - \sum \frac{b+c}{a} = \frac{\prod(b+c) + 2abc - \sum (b+c)bc}{abc} = \frac{4abc}{abc} = 4 \neq 0
\]
This shows that the system of equations has a unique solution. Clearly, the solution is \(x = a\), \(y = b\), and \(z = c\).
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