Question

a number consists of 2-digits whose sum is 7.if 9 is added to the number,the digits change their places, then the value is​

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

We have been given that a number consists of 2-digits whose sum is 7.

To Find:

We need to find their value after inter changing their places.

Solution:

Let the one's digit of the number be x.

As it is given that the sum of numbers is 7, so let the tens digit number be 7-x.

Therefore the number formed will be 10(x) + (7-x)

= 10x + 7-x

Now according to the question, we have

10x + (7-x) + 9 = 10(7-x) + x

=> 10x + 7 - x + 9 = 70 - 10x + x

=> 10x -x + 7 + 9 = 70 - 9x

=> 9x + 16 = 70 - 9x

=> 9x + 9x = 70 - 16

=> 18x = 54

=> x= 54/18

=> x = 3

Therefore, the ones digit is 3 and the tens digit is (7 - x) = (7 - 3) = 4.

Hence, the number formed will be

(10 × 3) + 4

= 30 + 4 = 34.

Hence, the required number is 34.

Answer 2

Answer:

let the unit place be x.

ten's digit=7-x

number=x+10(7-x)

=x+70-10x=70-9x

when the digits are interchanged

unit place=7-x

ten's digit=x

new number= 7-x+10x =7+9x

ATQ

7+9x+9=70-9x

16+9x=70-9x

9x+9x=70-16

18x=54

x=3

the number is 34 or 43 both are right.

hope this helps you

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Step-by-step explanation: