(7x-5y):(3x+4y)=7:11 হলে, দেখাই যে (3x-2y):(3x+4y)=137:473


\(\cfrac{(7x-5y)}{(3x+4y)}=\cfrac{7}{11} \)
বা, \(77x-55y=21x+28y\)
বা, \(77x-21x=28y+55y\)
বা, \(56x=83y \)
বা, \(\cfrac{x}{y}=\cfrac{83}{56}\)
বা, \(x:y=83:56\)

ধরি, \(x=83p\) এবং \(y=56p\) [যেখানে, \(p\) একটি বাস্তব সংখ্যা এবং \(p≠0\)]

\(∴(3x-2y):(3x+4y)\)
\(=\cfrac{(3x-2y)}{(3x+4y)}\)
\(=\cfrac{3×83p-2×56p}{3×83p+4×56p}\)
\(=\cfrac{249p-112p}{249p+224p}\)
\(=\cfrac{137p}{473p}\)
\(= \cfrac{137}{473}\)

\(∴(3x-2y):(3x+4y)=137:473\) (প্রমাণিত)

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