Math, asked by HarshTharani, 10 months ago

8 root x- 15 root y= -root xy, 10/root x+ 8/root y=4 . solve for x and y

Answers

Answered by mysticd
0

 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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