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 number and the original number is 154. Find the original number.
The original number is 68 and the reversed number is 86.
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To find the original number, let's represent the tens digit as 'x' and the units digit as '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
When the digits are reversed, the new number is formed by placing the units digit in the tens place and the tens digit in the units place. This gives us a new number represented by '10y + x'.
The problem states that when the original number and the reversed number are added together, the sum is 154. Mathematically, we can express this as:
(10y + x) + (10x + y) = 154
Simplifying this equation, we can combine like terms:
11y + 11x = 154
Dividing both sides of the equation by 11, we get:
y + x = 14
Now we have a system of equations:
x = y - 2
y + x = 14
We can solve this system by substituting the value of 'x' from the first equation into the second equation:
y + (y - 2) = 14
Simplifying this equation:
2y - 2 = 14
Adding 2 to both sides:
2y = 16
Dividing both sides by 2:
y = 8
Now we can substitute the value of 'y' back into the first equation to find 'x':
x = 8 - 2 = 6
Therefore, the original number is 68.