Question

The sum of two numbers is 15 and the sum of their reciprocals is

3
/10

. Find the numbers.​

Answer 1

Step-by-step explanation:

Hey.... mate here is you answer.

Solution :-

Let the 1st natural number be x.

And the 2nd natural number is (15 - x).

Sum of their reciprocals = 3/10

According to the Question,

⇒ 1/x + 1/(15 - x) = 3/10

⇒ 15/(15x - x²) = 3/10

⇒ 150 = 45x - 3x²

⇒ 3x² - 45x + 150 = 0

⇒ x² - 15x + 50 = 0

⇒ x² - 5x - 10x + 50 = 0

⇒ (x - 10) (x - 5) = 0

⇒ x - 10 = 0 or x - 5 = 0

⇒ x = 10, 5

When x = 10

1st number = x = 10

2nd number = 15 - x = 15 - 10 = 5

When x = 5

1st number = x = 5

2nd number = 15 - x = 15 - 5 = 10

Hence, the numbers are 10 and 5 or 5 and 10.

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Answer 2

$\huge \sf \color{purple}{\underline{\underline{Answer}}} :$

Two numbers are 5 and 10

$\huge \sf \color{purple}{\underline{\underline{Solution}}}:$

✯ Let the number be x .

✯ Then the other number is 15 - x.

$\sf \color{brown}{Using \: the \: given \: information \: , we\: get }$

$\sf \frac{1}{x}+\frac{1}{15-x}=\frac{3}{10} \\ \\ \sf \frac{15-x+x}{x(15-x)}=\frac{3}{10} \\ \\ \sf \frac{15}{(15x-x²)}=\frac{3}{10} \\ \\ \sf 150=45x-3x² \\ \\ \sf 3(x²-15+50)=0 \\\\ \sf x²-15+50=0 \\ \\ \sf x²-5x-10x+50 =0 \\ \\ \sf (x-10)(x-5)=0 \\ \\ \sf x-10=0 \: \: \ \ \ OR \ \ \ \ \sf x-5 \\ \\ \sf \boxed{ \colorbox{lightgreen}{x=10}} \ \ \ \ or \ \ \ \ \boxed{\colorbox{lightgreen}{ x = 5 }}$

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When x = 10 , then the other number is 15-5 = 10

When x = 5, then the other number is 15-5 = 10

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$\sf \color{blue}{Hence, \ the \ two \ numbers \ are \ 5 \ and \ 10 }$

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