Question

The sum of two number is 9 .the sum of their reciprocal is 1/2 find the number

Answer 1

Let one number be x
and other number be 9-x

a/q

1/x + 1/9-x = 1/2

9-x+x/x(9-x) = 1/2
9/x(9-x) = 1/2
x(9-x)=18
x^2-9x+18=0
x^2-6x-3x+18=0
x(x-6)-3(x-6)=0
(x-6)(x-3)=0
x=6,3
when one=6, other=3

Answer 2

Answer:

$\large{\underline{\boxed{\sf 6 \: and \: 3}}}$

Step-by-step explanation:

Given that -

The sum of two numbers is 9 and the sum of their reciprocal is 1/2.

Let the numbers be x and y respectively.

Sum of numbers is 9.

⇒ x + y = 9.... (i)

⇒ y = 9 - x .... (ii)

Sum of reciprocals is 1/2.

⇒ $\dfrac{1}{\text{x}} + \dfrac{1}{\text{y}}= \dfrac{1}{2}$.... (iii)

Now, on solving (iii),

$\dfrac{1}{\text{x}} + \dfrac{1}{\text{y}}= \dfrac{1}{2} \\ \\ \frac{ \text {x + y }}{ \text{xy}} = \frac{1}{2} \\ \\ \bf On \: cross \: multiplying : \\ \\2( \text{x + y) = 1xy}$

Putting the value of (i) and (ii) here, we get -

$2* 9= 1 \text{xy} \\ \\ \implies 18 = \text{x(9 - x)} \\ \\ \implies 18 = 9\text{x - x}{}^{2} \\ \\ \implies \text{x} {}^{2} - 9\text x + 18 \\ \\ \implies \text{x} {}^{2} - 6\text{x - 3x} + 18 = 0 \\ \\ \implies \text{x(x - 6) - 3(x - 6) = 0 } \\ \\ \implies \text{(x - 6)(x - 3) = 0} \\ \\ \boxed{\therefore \bf x = 6 \: or \: x = 3}$

Hence, the required numbers are 6 and 3.