Question

The units digit of a two digit number is two more than 4 times the tens digit. When the digits are reversed, the new number is 13 more than 3 times the original number. What is the original number? Any help on how to show this would be greatly appreciated. I know what the answer is but can't figure out how to show it.

Answer 1

let the original number is AB. noticee that the AB can be represented to
10A + B. If B = 4A + 2, then that number equals
10A + B = 10A + 4A + 2 = 14A + 2

after reversed :
BA ---> 10B + A
according information above, When the digits are reversed, the new number is 13 more than 3 times the original number. it means
10B + A = 13 + 3(14A + 2)

(remember that B = 4A + 2), so we get

10(4A + 2) + A = 13 + 3(14A + 2)

simplify by combine the similar terms, then solve for A
40A + 20 + A = 13 + 42A + 6
41A + 20 = 42A + 19
1 = A

because B = 4A + 2, so
B = 4A+2 = 4(1) + 2 = 6

therefore that number is AB = 16

Answer 2

hoped that make sense ^_^

Answer 3

yes it did thank you.