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Question
If `(a - 2b - 3c + 4d)/(a + 2b - 3c - 4d) = (a - 2b + 3c - 4d)/(a + 2b + 3c + 4d)`, show that: 2ad = 3bc.
Solution
`(a - 2b - 3c + 4d)/(a + 2b - 3c - 4d) = (a - 2b + 3c - 4d)/(a + 2b + 3c + 4d)`
Applying componendo and dividendo,
`\implies (a - 2b - 3c + 4d + a + 2b - 3c - 4d)/(a - 2b - 3c + 4d - a - 2b + 3c + 4d)= (a - 2b + 3c - 4d + a + 2b + 3c + 4d)/(a - 2b + 3c - 4d - a - 2b - 3c - 4d)`
`\implies (2a-6c)/(-4b+8d)=(2a+6c)/(-4b-8d)`
`\implies (2(a - 3c))/(-4(b - 2d)) = (2(a + 3c))/(4(b + 2d))`
`\implies (a - 3c)/(a + 3c) = (b - 2d)/(b + 2d)`
Applying componendo and dividendo,
`\implies (a - 3c + a + 3c)/(a - 3c - a - 3c) = (b - 2d + b + 2d)/(b - 2d - b - 2d)`
`\implies (2a)/(-6c) = (2b)/(-4d)`
`\implies` – 4da = – 6cb
`\implies` 2ad = 3bc
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