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Question

Solve:
(i) 16x220x+98x2+12x+21=4x52x+3
(ii) 5y2+40y125y+10y24=y+81+2y

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Solution

(i) 16x220x+98x2+12x+21=4x52x+3

Let, k=16x220x+98x2+12x+21=4x52x+3

Multiplying numerator and denominator of 4x52x+3, we get,

16x220x+98x2+12x+21=4x(4x5)4x(2x+3)

16x220x+98x2+12x+21=16x220x8x2+12x

Using Theorems on Equal ratios ie. ab=cd=acbd, we get,

16x220x+98x2+12x+21=16x220x8x2+12x=16x220x+916x2+20x8x2+12x+218x212x

k=921=37

Hence, 16x220x+98x2+12x+21=4x52x+3=37

(ii) 5y2+40y125y+10y24=y+81+2y

Let, z=5y2+40y125y+10y24=y+81+2y

Multiplying numerator and denominator of y+81+2y by 5y, we get,

5y2+40y125y+10y24=5y(y+8)5y(1+2y)

5y2+40y125y+10y24=5y2+40y5y+10y2

Using Theorems on Equal ratios, we get,

5y2+40y125y+10y24=5y2+40y5y+10y2=5y2+40y125y240y5y+10y245y10y2

z=124=3

Hence, 5y2+40y125y+10y24=y+81+2y=3


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