Find 2 positive numbers who differ by 8 and whose reciprocals add to be 1/3
eh whatever, I figured it out.
To solve this problem, let's call the two positive numbers x and y.
According to the given information, the numbers differ by 8. This can be expressed mathematically as:
x - y = 8
The second piece of information states that the reciprocals of the numbers add up to 1/3. The reciprocal of a number is found by flipping it upside-down. So mathematically, we have:
1/x + 1/y = 1/3
To solve this system of equations, we can use substitution or elimination method. Let's use substitution:
From equation 1, we can express x in terms of y by adding y to both sides:
x = y + 8
Now we substitute this value of x into equation 2:
1/(y + 8) + 1/y = 1/3
To get rid of fractions, we can multiply every term by 3y(y + 8):
3y + 3(y + 8) = (y + 8)y
Simplifying the equation:
3y + 3y + 24 = y^2 + 8y
Combining like terms:
6y + 24 = y^2 + 8y
Rearranging terms:
y^2 + 2y - 24 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Fortunately, this equation can be factored as:
(y - 4)(y + 6) = 0
Setting each factor to zero:
y - 4 = 0 or y + 6 = 0
Solving for y gives two possible values:
y = 4 or y = -6
However, we are looking for positive numbers, so we discard the negative solution. Therefore, y = 4.
Now we substitute this value back into equation 1 to find x:
x = y + 8
x = 4 + 8
x = 12
So the two positive numbers that satisfy the given conditions are 12 and 4.