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How can I determine all solutions of the following system of equations?

$$(6\alpha -2x-3y)x = 0$$ and $$(20-5y-4x)y = 0$$

I can determine the solutions $(x_1,y_1)=(0,0), (x_2,y_2)=(0,4)$ and $(x_3,y_3)=(3\alpha,0)$ quite straight forwardly. However I am not sure how to determine the final solution $(x_4,y_4)=(30-15\alpha,12\alpha-20)$.

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If $x,y\neq0$ then we can divide the equations by $x,y$ respectively so we get $$(6\alpha-2x-3y)x=0\iff 6\alpha-2x-3y=0\\(20-5y-4x)y=0\iff 20-5y-4x=0$$ Since the last solution set corresponds to $x\neq 0,y\neq 0$ we have that $$6\alpha-2x-3y=0\\20-5y-4x=0\\2x+3y=6\alpha\\4x+5y=20$$ Now if we multiply the first equation by $2$ get $$4x+6y=12\alpha\tag{1}$$$$4x+5y=20\tag{2}$$ Now if we subtract first $(1)$ from the second $(2)$ we get $$y=12\alpha-20$$ From this plugging in into any of the two equations we get $x=30-15\alpha$

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  • $\begingroup$ Thats the one, thank you. I was trying to follow a similar method but without removing the x and y outside the brackets and it got complicated very quickly. Why is it the case you can just remove them both by the way? So I can get a better understanding. $\endgroup$ – Aesler May 20 '17 at 14:23
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    $\begingroup$ @Aesler you divide both sides by $x$, because $\frac{0}{x}=0$, while $x\neq 0$, the same goes for $y$ $\endgroup$ – Mr. Xcoder May 20 '17 at 14:25
  • $\begingroup$ @Aesler What Mr.Xcoder said, since we already checked the cases when $x=0,y=0$ and $x=0,y\neq 0$ and $x\neq 0,y=0$ so the only case left is $x\neq 0,y\neq 0$ $\endgroup$ – kingW3 May 20 '17 at 14:29

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