How to solve the following system of equations?:
$$\begin{align} 4a+7/2b+3c&=0 \\ a+5b+2c&=0\end{align}$$ I used the Gaussiam elimination and by subtracting the second row by the first one multiplied by $1/4$ I get the two equations $4a+7/2b+3c=0$ and $47/8b+11/4c=0$ which are still not solvable.
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In this case, $a,b,c$ can be expressed in terms of a parameter (say $k \in \Bbb R$), as we have $3$ variables but only $2$ equations.
$$4a+\frac72b+3c=0$$ $$a+5b+2c=0$$
By cross-multiplication,
$$\frac{a}{7-15}=\frac{b}{3-8}= \frac{c}{20-\frac72} = k \Rightarrow (a,b,c) = \left(-8k,-5k,\frac{33}2k\right)$$
You have fewer equations than unknowns, and so the system does not have a unique solution. It has a one-parameter family of solutions that in this case can be computed by the cross product: $$(4,7/2,3)×(1,5,2)=(-8,-5,33/2)$$ $$(a,b,c)\in\{(-8t,-5t,33t/2):t\in\mathbb R\}$$