Let's solve this problem with algebra. Let the two numbers be x and y, where x is greater than y.
Given:
1. The difference between the numbers is 4, so you can write an equation as x - y = 4.
2. The sum of their reciprocals is 2/3, so you can write an equation as 1/x + 1/y = 2/3.
Now, you have a system of two equations with two variables:
1. x - y = 4
2. 1/x + 1/y = 2/3
You can solve this system of equations simultaneously. First, solve the first equation for x: x = y + 4.
Now, substitute this expression for x in the second equation:
1/(y + 4) + 1/y = 2/3
To clear the fractions, you can find a common denominator:
(3y(y + 4) + 3(y + 4))/[(y + 4)y] = 2
Now, simplify the equation:
3y(y + 4) + 3(y + 4) = 2(y + 4)y
Expand and simplify:
3y^2 + 12y + 3y + 12 = 2y^2 + 8y
Combine like terms:
3y^2 + 15y + 12 = 2y^2 + 8y
Subtract 2y^2 and 8y from both sides:
y^2 + 7y + 12 = 0
Now, you have a quadratic equation. Factor it:
(y + 3)(y + 4) = 0
Set each factor equal to zero:
1. y + 3 = 0, which gives y = -3
2. y + 4 = 0, which gives y = -4
So, there are two pairs of numbers (x, y) that satisfy the given conditions:
1. (x, y) = (1, -3)
2. (x, y) = (0, -4)
These are the pairs of numbers whose difference is 4, and the sum of their reciprocals is 2/3.
Hope this helps <3