the denominator of a fraction exceeds the numerator by 1. if 2 is to taken from each the sum of the reciprocal of the new fraction and 4 times the original fraction is 5. find the original fraction

Original fraction: x / (x+1).

New fraction: (x-2) / ((x+1)-2) = (x-2) / (x-1).

(x-1) / (x-2) + 4x / (x+1) = 5.
Common denominator: (x-2)*(x+1).
(x+1)(x-1) / (x-2)(x+1) + (x-2)4x / (x-2)(x+1) = 5,

(x^2-x+x-1) / (x-2)(x+1) + (4x^2-8x) / (x-2)(x+1) = 5,

(x^2-1) / (x-2)(x+1) + (4x^2-8x) / (x-2)(x+1) = 5,
(5x^2-8x-1) / (x-2)(x+1) = 5,
Cross-multiply:
5x^2 - 8x - 1 = 5x^2 + 5x - 10x - 10,
-8x + 5x = -9,
X = 3.
Original fraction = x / (x+1) = 3/(3+1) = 3/4.

To solve this problem, let's start by defining the original fraction. Let's say the numerator of the fraction is represented by 'x', and the denominator is represented by 'x + 1' since the denominator exceeds the numerator by 1. Therefore, the original fraction is x/(x + 1).

Now, let's move on to the next part of the problem. We are told to subtract 2 from both the numerator and denominator, which gives us a new fraction of (x - 2)/((x + 1) - 2), which simplifies to (x - 2)/(x - 1).

The problem states that the sum of the reciprocal of the new fraction and 4 times the original fraction is 5. Let's set up this equation:

1/((x - 2)/(x - 1)) + 4(x/(x + 1)) = 5

Now, let's simplify and solve for 'x':

1/((x - 2)/(x - 1)) + 4x/(x + 1) = 5

To simplify the equation, we can multiply both sides by the least common denominator, which is (x + 1)(x - 1):

(x + 1)(x - 1)/(x - 2) + 4x(x + 1)/(x + 1) = 5(x + 1)(x - 1)

Now, simplify further:

(x + 1)(x - 1) + 4x(x + 1) = 5(x + 1)(x - 1)

Expand and collect like terms:

(x^2 - 1) + 4x^2 + 4x = 5(x^2 - 1)

Combine like terms:

x^2 - 1 + 4x^2 + 4x = 5x^2 - 5

Simplify:

5x^2 + 4x^2 + 4x - 5x^2 + 5 = 0

Combine like terms:

9x^2 + 4x + 5 = 0

Now, this is a quadratic equation. To solve it, you can use either factoring, completing the square, or the quadratic formula. However, upon evaluating this quadratic equation, we can see that there are no real solutions for 'x'. Thus, there is no solution in the set of real numbers for this problem.

Therefore, there are no original fractions that satisfy the given conditions.