Question

the tens digit of a certain number is 3 more than the one digit. The sum of the squares of the digits is 29. find the number.

Answer 1

Let x be in the tens digit of the certain number
so according to your question
one's digit will be=x-3
and sum of square is 29
ie x^2 +(x-3)^2= 29

x^2+(x^2+9-6x)=29
\[2x^2 +9 -6x=29\]

\[2x^2 -6x=29 -9\]

\[2x^2 -6x=20\]

@luhan hope you get it...
can you try to solve now

Answer 2

yeah, thanks :)