If 5x+6y5x−6y=5u+6v5u−6v\dfrac{5x + 6y}{5x - 6y} = \dfrac{5u + 6v}{5u - 6v}5x−6y5x+6y=5u−6v5u+6v, show that xy=uv\dfrac{x}{y} = \dfrac{u}{v}yx=vu.
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Apply Componendo & Dividendo :
⇒(5x+6y)+(5x−6y)(5x+6y)−(5x−6y)=(5u+6v)+(5u−6v)(5u+6v)−(5u−6v)⇒5x+6y+5x−6y5x+6y−5x+6y=5u+6v+5u−6v5u+6v−5u+6v⇒10x12y=10u12v⇒xy=uv.\Rightarrow \dfrac{(5x + 6y) + (5x - 6y)}{(5x + 6y) - (5x - 6y)} = \dfrac{(5u + 6v) + (5u - 6v)}{(5u + 6v) - (5u - 6v)} \\[1em] \Rightarrow \dfrac{5x + 6y + 5x - 6y}{5x + 6y - 5x + 6y} = \dfrac{5u + 6v + 5u - 6v}{5u + 6v - 5u + 6v} \\[1em] \Rightarrow \dfrac{10x}{12y} = \dfrac{10u}{12v} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{u}{v}.⇒(5x+6y)−(5x−6y)(5x+6y)+(5x−6y)=(5u+6v)−(5u−6v)(5u+6v)+(5u−6v)⇒5x+6y−5x+6y5x+6y+5x−6y=5u+6v−5u+6v5u+6v+5u−6v⇒12y10x=12v10u⇒yx=vu.
Hence, proved that xy=uv\dfrac{x}{y} = \dfrac{u}{v}yx=vu.
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