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$$\begin{cases}7a+b+2c=0 \\ 3a+5b-6c=0 \end{cases}$$ Solving given equations, we obtain: $$\frac{a}{1}=\frac{b}{-3}=\frac{c}{-2}=k$$

I am confused how the two equations give the third one. It would be very helpful if someone could explain me the process.

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    do you mean $$7a+b+2c=0$$ and $$3a+5b-6c=0$$? – 2017-01-25
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    Might be best to type the equations into the problem, rather than have the big awkward picture there. – 2017-01-25
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    @Dr. Sonnhard Graubner: yes that is what I meant, how do I solve these to give the third equation? The answer is also there but what I am confused on is the process. – 2017-01-25
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    see my calculations in my post – 2017-01-25
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    Can you provide me the link to the post? – 2017-01-25

3 Answers 3

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Hint -

Put a = k then find value of b and c in terms of k.

You got c = $\frac{-32k}{16} = -2k$

Then we have from first equation,

7k + b + 2(-2k) = 0

b = -3k

Then,

k = a

k = $\frac{b}{-3}$

k = $\frac{c}{-2}$

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    Its work. You can try. – 2017-01-25
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When you have two equations in three unknowns you expect a single parameter family of solutions, which you have here. They are giving the parameter the name $k$, but it is the same as $a$. We can then solve the two equations assuming $a$ is a constant. Adding three times the first to the second gives $$24a+8b=0\\b=-3a\\a=\frac b{-3}$$ Then substituting this in gives $$-12a-6c=0\\c=-2a\\a=\frac c{-2}$$ Which is the result you quote.

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multiplying the first eqaution by $3$ and adding to the second we get $$24a+8b=0$$ therefore $$a=-\frac{1}{3}b$$ plugging this in the equation above we get $$-b-\frac{5}{3}b=6c$$ and we get $$c=-\frac{4}{9}b$$