Grade 8 - Algebra #3

Divide the equations with $ab,~bc,~ca$ respectively and you get:

$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{2}~\cdots~A$

$\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{3}~\cdots~B$

$\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{1}{4}~\cdots~C$

Add up all of them and divide the equation by $2$ , then you get $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{13}{24}$ .

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Subtract $A,~B,~C$ respectively from the equation we've got.

$\dfrac{1}{c}=\dfrac{1}{24}~~~\therefore 3c=72$

$\dfrac{1}{a}=\dfrac{5}{24}~~~\therefore 10a=48$

$\dfrac{1}{b}=\dfrac{7}{24}~~~\therefore 7c=24$

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$\therefore~10a+7b+3c=48+24+72=\boxed{144}$ .

We find a, and c in the first and second eqs. in term of b, a=2b/(b-2), and c=3b/(b-3).Put these in the third eq. then we have 6b^2/[(b-2)(b-3)] = 4(2b/(b-2) +3b/(b-3), the we get b=24/7, then a=48/10, and c=72/3. So 10a +7b+3c=144