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I would like an explanation as to why the solutions of the simultaneous equations $$ax+by+c = 0$$ and $$dx+ey+f = 0$$ have rational expressions in {$a,b,c,d,e,f$} whenever $ae\neq bd$.

I'm reading Saul Stahl's Introductory to Modern Algebra textbook and came across this problem. He used an example from his textbook to describe what a rational expression is. Here is his explanation:

"Let $z$ be a complex number. If $c \neq 1$, then $\frac {(2-ab)}{(1+c)}$ has a rational expression in {a,b,c} with n = $4$ where $$z_1 = ab, z_2 = 2 -z_1, z_3 = 1+c$$ and $$z=z_4=\frac{z_2}{z_1}"$$

I understand what a rational expression is but I'm unable to apply it in order to solve the initial problem.

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    When $ae \ne bd$ the system has a unique solution, which can be easily calculated. As for the second part of the question, it's hard to guess how that relates to the first, or what $z, n, z_k$ etc are. – 2017-02-07
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    The second part was just something I quoted from the book to mostly remind myself of what a rational expression is. – 2017-02-07
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    Hint (for the first part): eliminate $y$ between the equations, then solve the remaining linear equation for $x\,$. For example, multiply the first equation by $e$, the second one by $b$ and subtract the two. – 2017-02-07
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    Okay I'll give this a shot. – 2017-02-07
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    I got $x = \frac{bf-ec}{ae-bd}$ – 2017-02-07
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    Ahhh so that's why $ae\neq bd$; the denominator will be $0$ and the solution will be undefined. – 2017-02-07
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    That's a very rational expression. If $ae=bd$ then the system will either have no solutions, or infinitely many. – 2017-02-07
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    That's actually the next problem in the textbook, when $ae=bd$ but your explanation really clarified everything. – 2017-02-07

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