If $({3 x}^{2} - {3 y}^{2}) : ({2 x}^{2} + {5 y}^{2}) = 12 : 19$ , then $(x : y)$ is:

\frac{39}{11}

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\frac{9}{11}

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\frac{8}{13}

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\frac{7}{11}

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Solution

The correct option is A $\frac{39}{11}$
$\frac{3 x^{2} - 3 y^{2}}{2 x^{2} + 5 y^{2}} = \frac{12}{19}$
$⟹ \frac{x^{2} - y^{2}}{2 x^{2} + 5 y^{2}} = \frac{4}{19}$
$⟹ 19 x^{2} - 19 y^{2} = 8 x^{2} + 20 y^{2}$
$⟹ 11 x^{2} = 39 y^{2}$
$⟹ \frac{x}{y} = \frac{39}{11}$

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Similar questions

(\begin{matrix} 4 x^{2} - 3 y^{2} \end{matrix}) : (\begin{matrix} 2 x^{2} + 5 y^{2} \end{matrix}) = 12 : 19

x : y

Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

- 1 \frac{1}{7} + t a n^{-} 1 \frac{1}{13} = t a n^{-} 1 \frac{2}{9}

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