Question

if 4a+7b/4c+7d=4a-7b/4c-7d then prove that a:b::c:d​

Answer 1

proved a:b::c:d if 4a+7b/4c+7d=4a-7b/4c-7d

Step-by-step explanation:

4a+7b/4c+7d=4a-7b/4c-7d

(4a+7b)(4c-7d) =(4a-7b)(4c + 7d)

16ac -28ad + 28bc -49bd = 16ac + 28ad - 28bc - 49bd

56bc = 56ad

bc = ad

c/d = a/b

a/b = c/d

QED Proved

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Answer 2

Hence proved

$\dfrac{4a+7b}{4c+7d}=\dfrac{4a-7b}{4c-7d}\Rightarrow$ a:b :: c:d

Step-by-step explanation:

Given: $\dfrac{4a+7b}{4c+7d}=\dfrac{4a-7b}{4c-7d}$

To find: a:b::c:d

$\dfrac{4a+7b}{4c+7d}=\dfrac{4a-7b}{4c-7d}$

Move a and b same side/ c and d same side

$\dfrac{4a+7b}{4a-7b}=\dfrac{4c+7d}{4c-7d}$

Apply componendo & dividendo

$\dfrac{4a+7b+4a-7b}{4a+7b-4a+7b}=\dfrac{4c+7d+4c-7d}{4c+7d-4c+7d}$

Simplify

$\dfrac{8a}{14b}=\dfrac{8c}{14d}$

Divide both side by 8 and multiply 14

$\dfrac{8a}{14b}\times \dfrac{14}{8}=\dfrac{8c}{14d}\times \dfrac{14}{8}$

$\dfrac{a}{b}=\dfrac{c}{d}$

Therefore, a:b :: c:d

Hence proved

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