A two digit number is three less than seven times the sum of its digits. if the digits are reversed, the new number is 18 less than the original number. what is the new number?

To solve this problem, let's assume that the two-digit number is represented as "10a + b", where "a" and "b" are the tens and units digits, respectively.

According to the given information, the two-digit number is three less than seven times the sum of its digits. So, we can write the equation as:

10a + b = 7(a + b) - 3

Now, we are also given that the reversed number is 18 less than the original number. The reversed number is represented as "10b + a". Thus, we can write the second equation as:

10a + b = (10b + a) - 18

Now, we have a system of two equations that we can solve simultaneously to find the values of "a" and "b" and ultimately determine the new number (10b + a).

Let's solve the system of equations:

Equation 1: 10a + b = 7(a + b) - 3
Simplifying, we get:
10a + b = 7a + 7b - 3
Rearranging terms, we get:
10a - 7a + b - 7b = -3
3a - 6b = -3

Equation 2: 10a + b = (10b + a) - 18
Simplifying, we get:
10a + b = 10b + a - 18
Rearranging terms, we get:
10a - a + b - 10b = -18
9a - 9b = -18
Dividing both sides by 9, we get:
a - b = -2

Now we have a system of equations:

Equation 1: 3a - 6b = -3
Equation 2: a - b = -2

We can now use various methods such as substitution or elimination to solve the system of equations and find the values of "a" and "b". Once we find the values of "a" and "b", we can substitute them into the expression (10b + a) to calculate the new number.