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.
PLEASE ANSWER QUICKLEY!!!
let the tens digit be x, and the unit digit be y
and your number is 10x + y
reversed number is 10y + x
10x + y + 10y + x = 154
11x + 11y = 154
x + y = 14
but x = y-2
y-2 + y = 14
2y = 16
y = 8 , then x = 6
now just the original number using the definitions
To solve this problem, let's start by assuming that the tens digit is represented by 'x' and the units digit is represented by 'y'. According to the problem, we know that the tens digit is two less than the units digit, so we can write the equation:
x = y - 2
We are also given that when the digits are reversed, the sum of the reversed number and the original number is 154. In other words, if we reverse the digits 'x' and 'y', the resulting number will be (10 * y) + x. Therefore, we can write the second equation:
(10 * y) + x + (10 * x) + y = 154
Now we have a system of two equations. We can substitute the value of x from the first equation into the second equation to solve for y:
(10 * y) + (y - 2) + (10 * (y - 2)) + y = 154
Simplifying the equation:
10y + y - 2 + 10y - 20 + y = 154
Combining like terms:
22y - 22 = 154
Adding 22 to both sides:
22y = 176
Dividing both sides by 22:
y = 8
Now that we have the value of y, we can substitute it back into the first equation to find the value of x:
x = 8 - 2 = 6
Therefore, the original number is 68.