How Can Reversing Digits Triple the Tens Place?

[karush]

the sum of the digits of a two digit number is 6, If the digits are reversed, the new number is tree times the original tens number. find the original number.

well just playing with the numbers I got 51 as the original number since 15 is 3 times 5
but doing the problem with equations ?

I tried

$t + u = 6$
$3u = 3t$
but this not got it.

thanks ahead

 

[soroban]

Hello, karush!

The sum of the digits of a two-digit number is 6.
If the digits are reversed, the new number is three times the original ten's-digit.
Find the original number.

The original number is: $10t + u.$

We are told: .$t + u \,=\,6$ [1]

Also that: .$10u + t \:=\:3t \quad\Rightarrow\quad 10u \,=\,2t \quad\Rightarrow\quad t \,=\,5u$ [2]

Substitute [2] into [1]: .$5u + u \:=\:6 \quad\Rightarrow\quad 6u \,=\,6 \quad\Rightarrow\quad \boxed{u \,=\,1}$

Substiute into [2]: .$t \,=\,5(1) \quad\Rightarrow\quad \boxed{t \,=\,5}$Therefore, the original number is: .$10t + u \:=\:10(5) + 1 \:=\:51$