Two rails are algebraically represented by the equations $3$ x - $5$ y - $20$ = $0$ and $6$ x - $10$ y + $40$ = $0$

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Solution

$\begin{matrix} A c c o r d i n g t o q u e s t i o n : G i v e n c o n d i t i o n i s p a r a l l e l l i n e e q a t i o n : (w e n e e d t o t w o s o l u t i o n x & y .) 3 x - 5 y - 20 = 0 \begin{matrix} A & B x & 0 & 5 y & - 4 & - 1 \end{matrix} \Rightarrow 3 x - 5 y = 20 - - - - - (i) s u p p o s e : \Rightarrow 3 \times 0 - 5 y = 20, i f x = 0 ∴ y = - 4 3 x - 5 y = 20, i f y = - 1 \Rightarrow x = \frac{20 - 5}{3} = \frac{15}{3} = 5 ∴ x = 5 A g a i n, 6 x - 10 y + 40 = 0 \begin{matrix} C & D x & 0 & - 5 y & 4 & 1 \end{matrix} \Rightarrow 6 x - 10 y = - 40 - - - - - (i i) \Rightarrow 6 x - 10 y = - 40, i f x = 0 ∴ y = 4 \Rightarrow 6 x - 10 y = - 40, i f y = 1 ∴ x = - 5 \end{matrix}$

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