Question

please solve this
$if \: \frac{ {x}^{3} + 4x {y}^{2} }{ {4x}^{2} y + {y}^{3} } = \frac{15}{16} \\ find \: x:y$

please solve with full explanation
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Answer 1

Answer:

x : y = ( 60x^2 + 15y^2 ) : ( 16x^2 + 64y^2 )

Step-by-step explanation:

$\frac{ {x}^{3} + 4x {y}^{2} }{4 {x}^{2}y + {y}^{3} } = \frac{15}{16} \\ by \: cross \: multiplication \\ 16 {x}^{3} + 64x {y}^{2} = 60 {x}^{2} y + 15 {y}^{3} \\ 16x( {x}^{2} + 4 {y}^{2} ) = 15y(4 {x}^{2} + {y}^{2} ) \\ \frac{16x}{15y} = \frac{4 {x}^{2} + {y}^{2} }{ {x}^{2} + 4 {y}^{2} } \\ \frac{x}{y} = \frac{60 {x}^{2} +15 {y}^{2} }{16 {x}^{2} + 64 {y}^{2} }$

which means x : y = ( 60x^2 + 15y^2 ) : ( 16x^2 + 64y^2 )