The numerator of a fraction is four less than the denominator The sum of the fraction and its reciprocal is 5/2 Find the numerator and denominator if each is a positive integer
n = d - 4
[(d - 4) / d] + [d / (d - 4)] = 5/2
multiplying by [2 * d * (d - 4)]
[2 (d - 4)^2] + (2 d^2) = 5 d (d - 4)
2d^2 - 16d + 32 + 2d^2 = 5d^2 - 20d
0 = d^2 - 4d - 32
factoring __ 0 = (d - 8)(d + 4)
0 = d - 8 __ 8 = d
To find the numerator and denominator, let's assign variables to each unknown quantity.
Let's say the numerator is represented by 'x', and the denominator is represented by 'y'.
According to the given information, the numerator is four less than the denominator, so we can write the equation:
x = y - 4 (Equation 1)
The sum of a fraction and its reciprocal can be written as:
Fraction + Reciprocal = Sum
As per the question, the sum of the fraction and its reciprocal is 5/2. Therefore, we have:
x/y + y/x = 5/2 (Equation 2)
We can simplify equation 2 by finding a common denominator:
(2x + 2y) / (xy) = 5/2
Now, let's cross-multiply:
4x + 4y = 5xy/2
Multiplying both sides by 2 to eliminate the fraction:
8x + 8y = 5xy
Rearranging the equation:
5xy - 8x - 8y = 0
Now, we have a system of equations:
x = y - 4 (Equation 1)
5xy - 8x - 8y = 0 (Equation 3)
To solve this system of equations, we can substitute Equation 1 into Equation 3:
5(y - 4)y - 8(y - 4) - 8y = 0
5y^2 - 20y - 8y + 32 - 8y = 0
5y^2 - 36y + 32 = 0
Next, we can try factoring this quadratic equation:
(5y - 4)(y - 8) = 0
Setting each factor equal to zero:
5y - 4 = 0 or y = 8
If 5y - 4 = 0, then 5y = 4, and solving for y:
y = 4/5
However, since y needs to be a positive integer, we can't use 4/5 as a valid solution.
Therefore, the only valid solution is y = 8.
Plugging this value back into Equation 1:
x = 8 - 4 = 4
Hence, the numerator is 4 and the denominator is 8.