If 25x4 + 61x2y2 + 4ay4 = 114 5x2 + 3xy + 7y2 = 19 Find \(\frac{5x}{y}\) + \(\frac{7y}{x}\). |
\(\frac{25}{13}\) \(\frac{75}{13}\) \(\frac{150}{13}\) None |
\(\frac{75}{13}\) |
5x2 + 3xy + 7y2 = 19 5x2 - 3xy + 7y2 = \(\frac{114}{19}\) = 6 - + - 6xy = 13 xy = \(\frac{13}{6}\) Now, 5x2 + 7y2 + 3 × \(\frac{13}{6}\) = 19 5x2 + 7y2 = 19 - \(\frac{13}{2}\) = \(\frac{25}{2}\) Now → \(\frac{5x}{y}\) + \(\frac{7y}{x}\) = \(\frac{5x^2 + 7y^2}{xy}\) ⇒ \(\frac{25}{2}\) × \(\frac{6}{13}\) = \(\frac{75}{13}\) |