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If 7a+8b7c+8d=7a−8b7c−8d\dfrac{7a + 8b}{7c + 8d} = \dfrac{7a - 8b}{7c - 8d}, prove that :

(i) a : b = c : d

(ii) 4a−5b4c−5d=3a+4b3c+4d\dfrac{4a - 5b}{4c - 5d} = \dfrac{3a + 4b}{3c + 4d}

Ratio Proportion

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Answer

(i) Given,

7a+8b7c+8d=7a−8b7c−8d\dfrac{7a + 8b}{7c + 8d} = \dfrac{7a - 8b}{7c - 8d}

Applying alternendo,

7a+8b7a−8b=7c+8d7c−8d\dfrac{7a + 8b}{7a - 8b} = \dfrac{7c + 8d}{7c - 8d}

Applying componendo and dividendo we get :

⇒7a+8b+7a−8b7a+8b−(7a−8b)=7c+8d+7c−8d7c+8d−(7c−8d)⇒14a7a−7a+8b+8b=14c7c−7c+8d+8d⇒14a16b=14c16d⇒ab=14c16d×1614⇒ab=cd.\Rightarrow \dfrac{7a + 8b + 7a - 8b}{7a + 8b - (7a - 8b)} = \dfrac{7c + 8d + 7c - 8d}{7c + 8d - (7c - 8d)} \\[1em] \Rightarrow \dfrac{14a}{7a - 7a + 8b + 8b} = \dfrac{14c}{7c - 7c + 8d + 8d} \\[1em] \Rightarrow \dfrac{14a}{16b} = \dfrac{14c}{16d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{14c}{16d} \times \dfrac{16}{14} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.

Hence, proved that a : b = c : d.

(ii) We know that :

ab=cd\dfrac{a}{b} = \dfrac{c}{d}

To prove :

4a−5b4c−5d=3a+4b3c+4d\dfrac{4a - 5b}{4c - 5d} = \dfrac{3a + 4b}{3c + 4d}

Solving L.H.S. of the above equation :

⇒4a−5b4c−5d⇒4a−5b5d(4c5d−1)⇒4a−5b5d(4a5b−1)⇒4a−5b5d(4a−5b5b)⇒5b(4a−5b)5d(4a−5b)⇒bd.\Rightarrow \dfrac{4a - 5b}{4c - 5d} \\[1em] \Rightarrow \dfrac{4a - 5b}{5d\Big(\dfrac{4c}{5d} - 1\Big)} \\[1em] \Rightarrow \dfrac{4a - 5b}{5d\Big(\dfrac{4a}{5b} - 1\Big)} \\[1em] \Rightarrow \dfrac{4a - 5b}{5d\Big(\dfrac{4a - 5b}{5b}\Big)} \\[1em] \Rightarrow \dfrac{5b(4a - 5b)}{5d(4a - 5b)} \\[1em] \Rightarrow \dfrac{b}{d}.

Solving R.H.S. of the above equation :

⇒3a+4b3c+4d⇒3a+4b4d(3c4d+1)⇒3a+4b4d(3a4b+1)⇒3a+4b4d(3a+4b4b)⇒4b(3a+4b)4d(3a+4b)⇒bd.\Rightarrow \dfrac{3a + 4b}{3c + 4d} \\[1em] \Rightarrow \dfrac{3a + 4b}{4d\Big(\dfrac{3c}{4d} + 1\Big)} \\[1em] \Rightarrow \dfrac{3a + 4b}{4d\Big(\dfrac{3a}{4b} + 1\Big)} \\[1em] \Rightarrow \dfrac{3a + 4b}{4d\Big(\dfrac{3a + 4b}{4b}\Big)}\\[1em] \Rightarrow \dfrac{4b(3a + 4b)}{4d(3a + 4b)} \\[1em] \Rightarrow \dfrac{b}{d}.

Since, L.H.S. = R.H.S. = bd\dfrac{b}{d}.

Hence, proved that 4a−5b4c−5d=3a+4b3c+4d\dfrac{4a - 5b}{4c - 5d} = \dfrac{3a + 4b}{3c + 4d}.

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