Question

Solve the 18th one. I couldn't understand it

Answer 1

rationalize both x and y,then take square
$x = \frac{(5 - \sqrt{3}) \times (5 - \sqrt{3} ) }{(5 + \sqrt{3} )(5 - \sqrt{3}) } \\ = \frac{ {(5 - \sqrt{3}) }^{2} }{ {(5)}^{2} - ( { \sqrt{3}) }^{2} } \\ = \frac{25 + 3 - 10 \sqrt{3} )}{25 - 3} \\ = \frac{(28 - 10 \sqrt{3}) }{22} \\ x \: = \frac{14 - 5 \sqrt{3} }{11} \\ y = \frac{(5 + \sqrt{3}) \times (5 + \sqrt{3} ) }{(5 - \sqrt{3}) \times (5 + \sqrt{3} ) } \\ = \frac{ {(5 + \sqrt{3}) }^{2} }{ {(5)}^{2} - ( { \sqrt{3}) }^{2} } \\ = \frac{25 + 3 + 10 \sqrt{3} )}{25 - 3} \\ = \frac{(28 + 10 \sqrt{3}) }{22} \\ y \: = \frac{14 + 5 \sqrt{3} }{11} \\ {x}^{2} = \frac{( {14 - 5 \sqrt{3}) }^{2} }{121} \\ = \frac{(196 + 75 - 140 \sqrt{3}) }{121} \\ = \frac{271 - 140 \sqrt{3} }{121} \\ \: by \: the \: same \: way \\ {y}^{2} = \frac{271 + 140 \sqrt{3} }{121} \\ {x}^{2} - {y}^{2} = \frac{271 - 140 \sqrt{3} }{121} - \frac{271 + 140 \sqrt{3} }{121} \\ = \frac{271 - 140 \sqrt{3} - 271 - 140 \sqrt{3} }{121} \\ = \frac{ - 280 \sqrt{3} }{121}$=(28/11)(-10√3/11)
is the correct answer,there is mistake in this answer.