The sum of the digits of a two-digit number is equal to 15. If these digits switch places then you will get a number that is 27 less then the original number. Find these numbers.
a+b=15
10b+a = 10a+b - 27
69 = 96-27
To solve this problem, let's assume the original two-digit number is represented as "10a + b," where 'a' is the tens digit and 'b' is the ones digit.
According to the problem, the sum of the digits of the two-digit number is 15, so we can write the equation as:
a + b = 15 (Equation 1)
If the digits switch places, the new number will be represented as "10b + a." It is given that this new number is 27 less than the original number, so we can write the equation as:
10b + a = (10a + b) - 27 (Equation 2)
Now, we can solve this system of equations to find the values of 'a' and 'b'.
First, let's simplify Equation 2:
10b + a = 10a + b - 27
9b - 9a = -27
b - a = -3 (Dividing both sides by 9)
Next, we can rewrite Equation 1 as:
a = 15 - b
Substituting this value of 'a' into Equation 2:
b - (15 - b) = -3
2b - 15 = -3
2b = 12
b = 6
Substituting the value of 'b' back into Equation 1:
a + 6 = 15
a = 9
Therefore, the tens digit 'a' is 9 and the ones digit 'b' is 6. So, the original two-digit number is 96.