Question

in a two digit no sum of the digit is 9 if the digits are reversed the new no is increased by 9 find the original number
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Answer 1

Let the two digit number is represented by xy.

Given x + y = 9 when the digits are reversed we get yx. Now yx can also be written as 10y + x since y corresponds to 10 th place and x corresponds to unit place.

xy = 10x + y

given 10y + x -10x -y = 9 => - 9x +9y = 9

= > -x + y = 1. We also know x + y = 9

solving the above equations we get x = 4 and y = 5

=> the number is 45 which when reversed become 54 that is increased by 9.

Hence the digit is 45.

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Answer 2

Answer:

The number is 45.

Let the digits of the number be a, and b. The conditions given are:

a+b=9

The number is 10a+b.

10b+a=10a+b+9

Next, take the last condition and simplify it.

10b-b+a=10a+b-b+9

9b+a-a=10a-a+9

9b÷9=(9a+9)÷9

b=a+1

Replace b with its derived value in the first condition, and solve for a.

a+(a+1)=9

2a+1–1=9–1

2a÷2=8÷2

a=4, and since b=a+1, then b=5

So the number is 45. which when reversed become 54 that is increased by 9.

Now to test it.

4+5=9

40+5=45

50+4=40+5+9