Let the digits at unit’s place be a and ten’s place be v, respectively. Thus, the number we need to find is $10b+a$.
As per the given question, the sum of the two digit number is $15$. Thus, we have;
$a+b=15$ ……………(i)
Upon interchanging the digit’s place, the new number will be $10a+b$.
It is also given that, the new number obtained exceeds from the original number by $9$. Therefore, we can write this as;
$10a+b=10b+a+9$
⇒ $10a+b–10b–a=9$
⇒ $9a–9b=9$
⇒ $9(a–b)=9$
⇒ $a–b=9/9$
⇒ $a–b=1$………………….. (ii)
On solving equation (i) and (ii), we can find value of a and b.
Now, adding the equations (i) and (ii), we get;
$(a+b)+(a–b)=15+1$
⇒ $a+b+a–b=16$
⇒ $2a=16$
⇒ $a=16/2$
⇒ $a=8$
Putting the value of a in the equation (i), we get value of b
$8+b=5$
⇒ $b=15–8$
⇒ $b=7$
Hence, the required number is, $10\times 7+8=78$