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Find all numbers whose difference is 4 and the sum of whose reciprocals is 2/3

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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
User Ronin
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