A number consists of two digits. The sum of the digits is 9. If 45 is subtracted from the number, its digits are interchanged. What is the number?

The correct answer is 72.

Let's solve the problem step-by-step.

Let the tens digit of the two-digit number be represented by $x$ and the units digit by $y$.
A two-digit number can be written algebraically in terms of its digits as:
$10x+y$

From the first condition, the sum of the digits is 9. We can write this as:
$x+y=9$      --- (Equation 1)

From the second condition, if 45 is subtracted from the number, the digits are interchanged. The number with interchanged digits is represented as:
$10y+x$

According to this condition, we can form the equation:
$(10x+y)-45=10y+x$

Let's simplify this equation by moving all the variables containing $x$ and $y$ to one side:
$10x-x+y-10y=45$
$9x-9y=45$

Dividing the entire equation by 9, we get:
$x-y=5$      --- (Equation 2)

Now, we have a system of two linear equations:
1) $x+y=9$
2) $x-y=5$

Adding Equation 1 and Equation 2:
$(x+y)+(x-y)=9+5$
$2x=14$
$x=7$

Substitute the value of $x=7$ back into Equation 1:
$7+y=9$
$y=9-7$
$y=2$

Since the tens digit $x=7$ and the units digit $y=2$, the number is:
$10x+y=10\left(7\right)+2=72$