The units digit of a two-digit positive integer is 4. If the tens digit and units digit of the number are switched, the new number is 45 less than the original number. What is the sum of the original and the new number?
let the tens digit of the original be x
so original number was 10x+4
number reversed = 40+x
10x+4 = 40+x + 45
9x = 81
x = 9
the old was 94, the new one is 49
Their sum = 143
To solve this problem, we need to set up a system of equations.
Let's call the tens digit of the original number 'x' and the units digit 'y'. Since the units digit is 4, we have y = 4.
The original number can be written as 10x + y, and the new number (with the digits switched) can be written as 10y + x.
According to the given information, the new number is 45 less than the original number. This can be expressed as an equation:
10x + y = (10y + x) - 45
Now, substitute y = 4 into the equation:
10x + 4 = (10 * 4 + x) - 45
Simplify the equation:
10x + 4 = 40 + x - 45
Combine like terms:
10x + 4 = x - 5
Subtract x from both sides:
9x + 4 = -5
Subtract 4 from both sides:
9x = -9
Divide both sides by 9:
x = -1
Now that we have the value of x, we can substitute it back into y = 4 to find the value of y:
y = 4
Therefore, the original number is 10(-1) + 4 = -6 and the new number is 10(4) + (-1) = 39.
The sum of the original and new number is -6 + 39 = 33.
Therefore, the sum of the original and the new number is 33.