Question

anybody give the answer​

Answer 1

Answer:

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Step-by-step explanation:

$here \: given \: equations \: are \: \\ 5 \div \sqrt{x} - 3 \div \sqrt{ y} = 1 \div 6 \\ 3 \div \sqrt{x} + 4 \sqrt{y} = 3$

$so \: \: 5 \div \sqrt{x} - 3 \sqrt{y} =1 \div 6 \: can \: be \: written \: as \: 5 \times 1 \div \sqrt{x} - 3 \times 1 \div \sqrt{y} = 1 \div 3 \\ \\ similarly \: 3 \div \sqrt{x} + 4 \sqrt{y} = 3 \: can \: be \: written \: as \: 3 \times 1 \div \sqrt{x} + 4 \times 1 \div \sqrt{y} = 3$

$so \: let \: \: 1 \div \sqrt{x} = m \\ 1 \div \sqrt{y} = n$

$resubstituting \: these \: values \: we \: get \\ 5m - 3n = 1 \div 6 \\ ie \: 6(5m - 3n) = 1 \\ 30m - 18n = 1 \: \: \: \: \: \: \: \: \: (1) \\ \\ 3m + 4n = 3 \: \: \: \: \: \: \: \: \: (2)$

$multiplying \: (2) \: by \: 10 \\ we \: get \\ 30m + 40n = 30 \: \: \: \: \: \: \: \: (3) \\ \\ subtracting \: (1) \: from \: (3) \\ we \: get \\ 58n = 29 \\ ie \: n = 1 \div 2$

$substitute \: value \: of \: n \: in \: any \: of \: three \: equations \: \\ we \: get \\ m = 1 \div 3 \\ \\ resubstituting \: the \: values \: of \: m \: and \: n \: as \: 1 \div \sqrt{x} \: and \: \: 1 \div \sqrt{y} \: \: respectively \\ we \: get \\ 1 \div \sqrt{x} = m \\ 1 \div \sqrt{y} = n \\ so \: 1 \div \sqrt{x} = 1 \div 3 \\ 1 \div \sqrt{y} = 1 \div 2$

$so \: hence \: \\ \sqrt{x} = 3 \\ \sqrt{y} = 2 \\ squaring \: both \: sides \\ we \: get \\ x = 9 \\ y = 4$

$so \: thus \\ (x \: and \: y) = (9 \: and \: 4) \: respctively \: is \: solution \: of \: above \: linear \: equations$