Question

the sum of two numbers is 18 . The sum of their reciprocals is 1/4.Find the numbers.​

Answer 1

$\bf\blue{\underline{\boxed{Question}}}$

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The sum of two numbers is 18 . The sum of their reciprocals is 1/4.Find the numbers.

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$\bf\blue{\underline{\boxed{Solution}}}$

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$\small\tt{Let\:the\:numbers\:be\:x\:and\:y}$

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$\small\tt{We\:are\: given\:that\:the\: sum\:}\\{\small\tt{of\: two\: numbers\: is\: 18}}$

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$\small\tt{So,}$

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$\small\tt{x+y\:=\:18\:\:\:---(1)}$

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$\small\tt{We\:are\: given\:that\:the\: sum\:}\\{\small\tt{of\: their\: reciprocal\: is\: {\frac{1}{4}}}}$

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$\small\tt{So,}$

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$\small\tt{\frac{1}{x}}+{\frac{1}{1-x}}\:=\:{\frac{1}{4}}$

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$\small\tt{Substitute\:the\:value\: of\:y\:from\:1}$

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$\small\tt{\frac{18-x+x}{x(18-x)}}\:=\:{\frac{1}{4}}$

$\small\tt{\frac{18}{x(18-x)}}\:=\:{\frac{1}{4}}$

$\small\tt{\frac{18 \times 4}{x(18-x)}}\:=\:1$

$\small\tt{\frac{72}{18x-x^{2}}}\:=\:1$

$\small\tt{72:=\:18x-x^{2}}$

$\small\tt{x^{2}-18x +72\:=\:0}$

$\small\tt{x^{2}-12x-6x +72\:=\:0}$

$\small\tt{x(x-12)-6(x -12)\:=\:0}$

$\small\tt{(x-6)(x-12)\:=\:0}$

$\small\tt{x\:=\:6,12}$

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$\small\tt{\underline{\underline{when\:x\: is\: 6}}}$

$\small\tt{y\:=\:18-x}$

$\small\tt{y\:=\:18-6}$

$\small\tt\red{\underline{\boxed{y\:=\:12}}}$

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$\small\tt{\underline{\underline{when\:x\: is\: 12}}}$

$\small\tt{y\:=\:18-x}$

$\small\tt{y\:=\:18-12}$

$\small\tt\red{\underline{\boxed{y\:=\:6}}}$

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$\bf\blue{\underline{\boxed{Hope\:It\: Help\:You}}}$

Answer 2

Step-by-step explanation:

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