Question

pls solve it fast. it's urgent

Answer 1

Hope u like my process
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$= > \: ax + by = 1 \\ \\ or. \: \: ax = 1 - by \\ \\ or. \: \: x = \frac{1 - by}{a} ......(1) \\ \\ \\ now..given \: \: \: \\ \\ = > bx + ay = \frac{ {(a + b)}^{2} }{ {a}^{2} + {b}^{2} } - 1 \\ \\ or. \: \: b( \frac{1 - by}{a} ) + ay = \frac{ {(a + b)}^{2} - ( {a}^{2} + {b}^{2}) }{ {a}^{2} + {b}^{2} } \\ \\ or . \: \: \frac{(b - {b}^{2}y) + {a}^{2} y}{a} = \frac{ {a}^{2} + {b}^{2} + 2ab - {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \\ \\ or. \: \: b + {a}^{2} y - {b}^{2} y = \frac{a \times 2ab}{ { a }^{2} + {b}^{2} } \\ \\ or. \: \: y( {a}^{2} - {b}^{2} ) = \frac{2 {a}^{2}b }{ {a}^{2} + {b}^{2} } - b \\ \\ or. \: \: y( {a}^{2} - {b}^{2} ) = \frac{2 {a}^{2}b - {a}^{2}b - {b}^{3} }{( {a}^{2} + {b}^{2}) } \\ \\ or. \: \: y = \frac{ {a}^{2}b - {b}^{3} }{( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2}) } \\ \\ or. \: \: y = \frac{b( {a}^{2} - {b}^{2} ) }{( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2}) } \\ \\ or. \: \: y = \frac{b}{ {a}^{2} + {b}^{2} } \\ \\ \\ putting \: \: the \: value \: \: of \: \: y \: \: in \: equation \: \: (1) \\ ............................................................. \\ \\ = > x = \frac{1 - b( \frac{b}{ {a}^{2} + {b}^{2} } )}{a} \\ \\ or. \: \: x = \frac{( {a}^{2} + {b}^{2}) - {b}^{2} }{a( {a}^{2} + {b}^{2} ) } \\ \\ or. \: \: x = \frac{ {a}^{2} + {b}^{2} - {b}^{2} }{a( { {a}^{2} + {b}^{2}) }} = \frac{ {a}^{2} }{a( {a}^{2} + {b}^{2} )} \\ \\ or. \: \: x = \frac{a}{ {a}^{2} + {b}^{2} }$
Thus the value of x and y are...

$= > \: x = \frac{a}{ {a}^{2} + {b}^{2} } \\ \\ = > \: y = \frac{b}{ {a}^{2} + {b}^{2} }$
Hope this is ur required answer

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