Question

Solve this algebra Question below

Answer 1

$\frac{ {a}^{2} }{x} - \frac{ {b}^{2} }{y} = 0...........(1) \\ \frac{ {a}^{2} b}{x} + \frac{ {b}^{2} a}{y} = a + b......(2) \\ from \: (1) \: we \: get \\ \frac{ {a}^{2} }{x} = \frac{ {b}^{2} }{y} \\ \frac{1}{x} = \frac{ {b}^{2} }{y {a}^{2} } \\ x = \frac{y {a}^{2} }{ {b}^{2} } .....................(3) \\ subsitute \: (3) \: in \: (2) \\ \frac{ {a}^{2} b}{x} + \frac{ {b}^{2}a }{y} = a + b \\ \frac{ {a}^{2} b}{ \frac{y {a}^{2} }{ {b}^{2} } } + \frac{ {b}^{2} a}{y} = a + b \\ {a}^{2} b( \frac{ {b}^{2} }{y {a}^{2} } ) + \frac{ {b}^{2} a}{y} = a + b \\ \frac{ {b}^{3} }{y} + \frac{ {b}^{2}a }{y} = a + b \\ \frac{ {b}^{2} (b + a)}{y} = a + b \\ \frac{ {b}^{2} }{y} = \frac{a + b}{a + b} \\ \frac{ {b}^{2} }{y} = 1 \\ y = {b}^{2} \\ so \: x \: = \: \frac{y {a}^{2} }{ {b}^{2} } = \frac{ {b}^{2} . {a}^{2} }{ {b}^{2} } = {a}^{2}$
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