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determine the value of \(a,b,c\) that satisfy the following system of equations: \[(a+b)(a+b+c)=7\] \[(b+c)(a+b+c)=8\] \[(a+c)(a+b+c)=9\]
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First, divide all equations by \(a+b+c\), that will help a bit.
Now, \[ (a+b) - (b+c) = a-c \]And \[ (a-c) + (a+c) = 2a \]
So \[ a = \frac{(a+b)-(b+c)+(a+c)}{2} = \frac{7-8+9}{2(a+b+c)} = \frac{4}{a+b+c} \]It works by symmetry for \(b\) and \(c\). \[ b = \frac{8-9+7}{2(a+b+c)} = \frac{3}{a+b+c} \\ c= \frac{9-7+8}{2(a+b+c)} = \frac{5}{a+b+c} \]
Now we add them all up: \[ a+b+c = \frac{4+3+5}{a+b+c} = \frac{12}{a+b+c}\implies(a+b+c)^2=12\\ a+b+c = \sqrt{12} \]
This leads us with: \[ a= \frac{4}{\sqrt{12}},\quad b=\frac{3}{\sqrt{12}},\quad c=\frac{5}{\sqrt{12}} \]
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