Q:

Eight times the sum of the digits of a certain two-digit number exceeds the number by 19. The tens digit is two less than the units digit. Find the number.

Accepted Solution

A:
Answer:number is 13Step-by-step explanation:Let the digit at units place is bthe digit at tens place is aThe number is: abNow According to condition:  The tens digit is two less than the units digita = b -2  ---------- eq1From the condition: Eight times the sum of the digits of a certain two-digit number exceeds the number by 19. 8(a+b)=10a+b+19 By simplifyingPutting a= b-28a+8b=10(b-2)+b+19 8(b-2)+8b=10b-20+b+19 8b-16+8b=10b-20+b+19 By adding like terms we get:5b=15 Dividing both sides by 5 b = 3Now putting value of b in eq1a = b - 2a = 3  -2 a =1hence the number is : 13i hope it will help you!u=3 t=u-2=1