The digits of a two-digit number sum to 8. When the digits are reversed, the resulting number is 18 less than the original number. What is the original number?
If the digits are a and b, then we have
a+b = 8
10b+a = 10a+b - 18
To solve this problem, we need to set up an equation based on the given information and then solve for the original number.
Let's say the original two-digit number is represented as "10x + y," where x and y are the tens and units digits, respectively.
According to the problem, the sum of the digits is 8, so we can write the equation: x + y = 8 ---(Equation 1)
When the digits are reversed, the resulting number is 18 less than the original number. This means that (10y + x) = (10x + y) - 18.
Simplifying this equation, we get: 9y - 9x = -18
Dividing by 9, we have: y - x = -2 ---(Equation 2)
Now, we have a system of equations with Equation 1 and Equation 2:
Equation 1: x + y = 8
Equation 2: y - x = -2
We can solve this system of equations using the method of substitution or elimination.
If we substitute Equation 2 to Equation 1, we get:
x + (x - 2) = 8
2x - 2 = 8
2x = 10
x = 5
Substituting the value of x into Equation 1:
5 + y = 8
y = 8 - 5
y = 3
Therefore, the original number is 53.