Question

a, b, c, d प्रमाणात असतील, तर सिद्ध करा: $\frac{11a^2+9ac}{11b^2+9bd} = \frac{a^2+3ac}{b^2+3bd}$

Answer 1

Solution :

$\frac{11a^2+9ac}{11b^2+9bd} = \frac{a^2+3ac}{b^2+3bd}$

Divide both sides by ( b/a) ,

we get

$\implies\frac{11a+9c}{11b+9d} = \frac{a+3c}{b+3d}$

Do the cross multiplication,

we get

=> (11a+9c)(b+3d)=(a+3c)(11b+9d)

=> 11ab+33ad+9bc+27cd

= 11ab+9ad+33bc+27cd

=> 33ad+9bc = 9ad + 33bc

=> 33ad - 9ad = 33bc - 9bc

=> 24ad = 24bc

=> ad = bc

Or

a/b = c/d

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