Question

the sum of two digit number and the number obtained by reversing the digits is 66 if the digit of a number differ by 2 find that number how many such numbers of there?
BY USING THE ELIMINATION METHOD
please help me. ​

Answer 1

Heya

Any two digit number say ( ab ) can be written like this

10a + b

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EXP:-

Let the two digit number be xy

ACCORDING TO THE QUESTION

10x + y + 10y + x = 66

=>

11x + 11y = 66

=>

x + y = 6 .... Equation ( i )

And

x - y = 2 ..... Equation ( ii )

Now, Add both the equations we get

2x = 8

x = 4 And y = 2

So, the two digit number is 42

one one two digit number can be formed from this given information.

Answer 2

$hello \: mate$

Let the number at the ones place be x and the number at the tens place be y.

➡So, the number will be 10y+x

➡After reversing:10x+y

Now, according to the question,

➡10y+x+10x+y=66

➡11x+11y=66

➡x+y=6------(1)

Also, it is given that the digits differ by 2, so

➡x-y=2-----(2)

Adding (1) and (2),

↪2x=8

↪x=4

Now substituting x=4 in (1),

4+y=6

↪y=2

Hence, the numbers are

↪10(2)+4=24

↪10(4)+2=42(on reversing)