Question

solve:
8 root x-15 root y=root xy
10/root x+8/root y=4

Answer 1

/* There is a mistake in the question. It may be like this */

$Given \: 8\sqrt{x} - 15\sqrt{y} = - \sqrt{xy}$

$Dividing \: each \: term \: by \: \sqrt{xy} \: we \\get$

$\implies \frac{8}{\sqrt{y}} - \frac{15}{\sqrt{x}} = - 1 \ : --(1)$

$and \: \frac{10}{\sqrt{x}} + \frac{8}{\sqrt{y}} = 4 \ : --(2)$

$Let \: \frac{1}{\sqrt{x}} = a \: and \: \frac{1}{\sqrt{y}} = b \: now \: it \: becomes$

$8b - 15a = -1 \: --(3)$

$and \: 10a + 8b = 4 \: --(4)$

/* Subtract equation (3) from equation (4), we get */

$\implies 25a = 5$

$\implies a = \frac{5}{25}$

$\implies \pink {a = \frac{1}{5}} \: --(5)$

$put \: value \: of \: a = \frac{1}{5} \: in \\ equation\: (3) \: we \:get$

$\implies 8b - 15 \times \frac{1}{5} = -1$

$\implies 8b - 3 = -1$

$\implies 8b = -1 + 3$

$\implies 8b = 2$

$\implies b = \frac{2}{8}$

$\implies b = \frac{1}{4}\: --(6)$

$Now, a = \frac{1}{\sqrt{x}} = \frac{1}{5}$

/* On squaring both sides, we get */

$\implies \frac{1}{x} = \frac{1}{25}$

$and \: b = \frac{1}{\sqrt{y}} = \frac{1}{4}$

/* On squaring both sides, we get */

$\implies \frac{1}{y} = \frac{1}{16}$

Therefore.,

$\red { Value \:of \: x } \green { = 25 }$

$\red { and \:Value \:of \: y} \green { = 16 }$

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