Question

The sum of two numbers is 13 and the sum of their reciprocal is 13/42. Find the numbers​

Answer 1

Step-by-step explanation:

see the attached document

Answer 2

Answer:

6 and 7

Step-by-step explanation:

Given:-

The sum of two numbers is 13 and the sum of their reciprocal is 13/42.

To Find:-

The numbers

Solution:-

Let first number be x

Other number = 13-x

Reciprocal of 1st number = 1/x

Reciprocal of 2nd number = 1/(13-x)

The sum of their reciprocals is 13/42. So,

ACQ

$\therefore \: \rm \: \frac{1}{x} + \frac{1}{13 - x} = \frac{13}{42} \\ \rm \: LCM \: of \:( x )\: and \: (13 - x) =x(13 - x) \\ \rightharpoonup \rm \: \frac{ \cancel{(x)}(13 - x)} {\cancel{x} }+ \frac{(x) \cancel{(13 - x)}} { \cancel{13 - x}} = \frac{13}{42} (x)(13 - x) \\ \rightharpoonup \rm \: 13 - x + x = \frac{13}{42} (13x - {x}^{2} ) \\ \rm \: By \: cross - multiplying \: we \: get, \\ \rightharpoonup \rm \:42(13) = 13(13x - {x}^{2} ) \\ \rightharpoonup \rm \:546 = 169x - 13 {x}^{2} \\ \rightharpoonup \rm \:13 {x}^{2} - 169x - 546 = 0 \\ \rm \: Dividing \: 13 \: to \: all \\ \rightharpoonup \rm \: {x}^{2} - 13x + 42 = 0 \\ \rightharpoonup \rm \: {x}^{2} - 6x - 7x + 42 = 0 \\ \rightharpoonup \rm \:x(x - 6) - 7(x - 6) = 0 \\ \rightharpoonup \rm \:(x - 6)(x - 7) = 0$

If x-6=0

x=6

If x-7=0

x=7

So two number are 6 and 7