Heres the problem:
The tens digit of a two-digit number is 1 more than 3 times the ones digit. Subtracting 45 from the number reverses it. Find the original number.
well,
If the 1's digit is 0, then 3*0+1 = 1, so the number is 10
If the 1's digit is 1, then 3*1+1 = 4, so the number is 41
If the 1's digit is 2, then 3*2+1 = 7, so the number is 72
If the 1's digit is 3, then 3*3+1 = 10, this is not a two digit number, so you are left with the first three cases.
Just check which one meets the criteria of "Subtracting 45 from the number reverses it"
Hello, sp0rtskid55!
Let $\displaystyle t$ = tens digit
Let $\displaystyle u$ = ones digit
The number is: .$\displaystyle 10t + u$
Its reversal is: .$\displaystyle 10u + t$
The tens digit of a two-digit number is 1 more than 3 times the ones digit.
$\displaystyle {\color{blue}t \:=\:3u+1}$
Subtracting 45 from the number reverses it.
$\displaystyle {\color{blue} 10t + u - 45 \:=\:10u + t}$
Find the original number.
The second equation is: .$\displaystyle 9t\:=\:9u + 45\quad\Rightarrow\quad t \:=\:u + 5$
Equate to the first equation: .$\displaystyle 3t + 1 \:=\:u + 5\quad\Rightarrow\quad u \:=\:2$
. . Then: .$\displaystyle t = 7$
Therefore, the original number is 72.