if the difference of two positive integer is 4,and the difference of their square is 80,what is the sum of ?
I'm
I'll give you a clue. Each of the numbers is between 5 and 15.
12 and 8
To solve this problem, let's assign variables to the two positive integers. Let's say the first integer is x, and the second integer is y.
We are given two pieces of information:
1. The difference between the two positive integers is 4, which can be expressed as: y - x = 4
2. The difference between the squares of the two positive integers is 80, which can be expressed as: y^2 - x^2 = 80
Now, let's use these equations to solve for x and y:
From equation 1 (y - x = 4), we can solve for y in terms of x by adding x to both sides: y = x + 4.
Substituting this value for y in equation 2, we get: (x + 4)^2 - x^2 = 80.
Expanding the square, we have: x^2 + 8x + 16 - x^2 = 80.
Simplifying the equation: 8x + 16 = 80.
Subtracting 16 from both sides: 8x = 64.
Dividing both sides by 8: x = 8.
Now that we know the value of x, we can substitute it back into equation 1 to find the value of y: y - 8 = 4.
Solving for y, we get: y = 12.
Therefore, the two positive integers are 8 and 12.
Finally, to find the sum of the two integers, we simply add them: 8 + 12 = 20.
So, the sum of the two integers is 20.