In a two digit number the tens digit is two less than the units digit.If the digits are reversed, the sum of the reversed and the original number of 154.Find the original number.
t = u-2
10t+u + 10u+t = 154
154 = 11*14 so t+u=14
the number is 68
To solve this problem, we can use algebraic equations to represent the given information and find the original number.
Let's assume the tens digit is represented by variable 'x' and the units digit is represented by variable 'y'.
According to the given information, the tens digit is two less than the units digit. So, we can write the equation: x = y - 2.
If the digits are reversed, the new number can be represented as 10y + x, and the original number can be represented as 10x + y.
The sum of the reversed and the original numbers is given as 154. So, we can write the equation: (10y + x) + (10x + y) = 154.
Now, let's substitute the value of x from the first equation into the second equation:
(10y + (y - 2)) + (10(y - 2) + y) = 154.
Simplifying this equation, we get:
11y - 4 + 10y - 20 + y = 154.
Combining like terms, we get:
22y - 24 = 154.
Adding 24 to both sides, we get:
22y = 178.
Dividing both sides by 22, we get:
y = 8.
Now, substitute the value of y back into the first equation to find x:
x = 8 - 2 = 6.
Therefore, the original number is 68.