The numerator of a simple fraction is 2 less than its denominator. The sum of this fraction and its reciprocal is 2 4/35. Find this fraction.

n = d-2

n/d + d/n = 2 4/35

substitute in for n and we have

(d-2)/d + d/(d-2) = 74/35
d = 7
so, n=5

check:

5/7 + 7/5 = (25+49)/35 = 74/35

To solve this problem, let's break it down into steps:

Step 1: Let's represent the denominator of the fraction with the variable 'x'.

Step 2: According to the problem, the numerator is 2 less than the denominator, so it can be represented as 'x - 2'.

Step 3: Now, we can write the fraction as (x - 2)/x.

Step 4: The reciprocal of the fraction is obtained by swapping the numerator and denominator, which gives us x/(x - 2).

Step 5: According to the problem, the sum of the fraction and its reciprocal is 2 4/35, which can be written as 2 + 4/35.

Step 6: Now, let's set up the equation by adding the two fractions:

(x - 2)/x + x/(x - 2) = 2 + 4/35

Step 7: To simplify the equation, we need to find a common denominator for the fractions on the left side. The common denominator would be 'x(x - 2)'.

Step 8: Now, we can rewrite the equation with the common denominator:

[(x - 2)(x - 2) + x^2]/(x(x - 2)) = 2 + 4/35

Simplifying the equation further:

[x^2 - 4x + 4 + x^2]/(x(x - 2)) = 2 + 4/35

[2x^2 - 4x + 4]/(x(x - 2)) = 70/35 + 4/35

[2x^2 - 4x + 4]/(x(x - 2)) = 74/35

Step 9: Now, we can cross-multiply to eliminate the denominators:

(2x^2 - 4x + 4) * (35) = 74 * (x) * (x - 2)

Step 10: Expanding and simplifying the equation:

70x^2 - 140x + 140 = 74x^2 - 148x

Step 11: Rearranging the equation:

70x^2 - 74x^2 - 140x + 148x - 140 = 0

-4x^2 + 8x - 140 = 0

Step 12: Factoring out a common factor:

-4(x^2 - 2x + 35) = 0

Step 13: The quadratic expression inside the parentheses does not factorize further, so we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a, where a = 1, b = -2, and c = 35.

x = (-(-2) ± √((-2)^2 - 4(1)(35))) / (2(1))

x = (2 ± √(4 - 140)) / 2

Step 14: Simplifying the equation:

x = (2 ± √(-136)) / 2

We have a negative value inside the square root, which indicates that there are no real solutions for 'x'.

Since 'x' represents the denominator, there is no valid simple fraction that satisfies the given conditions.