Question

If $\frac{x^2+2y^2}{x^2-2y^2} = 17,$ find (i) $\frac{x}{y}$ (ii) $\frac{2x^3+3y^3}{2x^3-3y^3}$

Answer 1

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Answer 2

Solution:

$\frac{x^2+2y^2}{x^2-2y^2} = 17 \\ \\ {x}^{2} + 2 {y}^{2} = 17( {x}^{2} - 2 {y}^{2} ) \\ \\ {x}^{2} + 2 {y}^{2} = 17 {x}^{2} - 34 {y}^{2} \\ \\ {x}^{2} - 17 {x}^{2} = - 34 {y}^{2} - 2 {y}^{2} \\ \\ - 16 {x}^{2} = - 36 {y}^{2} \\ \\ squaring \: both \: sides \\ \\ 4x = 6y \\ \\ \frac{x}{y} = \frac{6}{4} \\ \\ \frac{x}{y} = \frac{3}{2}$
$\frac{2x^3+3y^3}{2x^3-3y^3} \\ \\ = \frac{2 ({ \frac{x}{y} })^{3} + 3}{2 { (\frac{x}{y} })^{3} - 3} \\ \\ = \frac{2 \times \frac{27}{8} + 3}{2 \times \frac{27}{8} - 3 } \\ \\ = \frac{ \frac{27}{4} + 3}{ \frac{27}{4} - 3 } \\ \\ = \frac{27 + 12}{27 - 12} \\ \\ = \frac{39}{15}\\\\ \frac{2x^3+3y^3}{2x^3-3y^3}=\frac{13}{5}$
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