Let's solve the equation step by step:

The given equation is:

\[ \frac{5x}{2x+2} + 1 = \frac{6}{x} + \frac{15x}{2x+2} + 1 = \frac{6}{x} + 1 \]

However, it looks like there might be some confusion because the equation seems to be written in a duplicated manner and is missing proper separation or parentheses. I'll address it assuming you're working with this part:

\[ \frac{5x}{2x+2} + 1 = \frac{6}{x} + \frac{15x}{2x+2} + 1 \]

Let's separate it first as needed and simplify the equation down:

Given equation:

\[ \frac{5x}{2x+2} + 1 = \frac{6}{x} + \frac{15x}{2x+2} + 1 \]

First, notice both sides have \( +1 \), which we can subtract from both sides, simplifying the equation to:

\[ \frac{5x}{2x+2} = \frac{6}{x} + \frac{15x}{2x+2} \]

To combine the fractional terms on the right-hand side, notice that a common denominator for the fractions on the right would need to include \( x(2x+2) \).

The equation reorganizes to:

Both sides get the common denominator \( x(2x+2) \):

\[ \frac{5x \cdot x}{x(2x+2)} = \frac{6(2x+2) + 15x \cdot x}{x(2x+2)} \]

Simplifying the terms:

The left side becomes:

\[ \frac{5x^2}{x(2x+2)} \]

and the right side becomes:

\[ \frac{12x + 12 + 15x^2}{x(2x+2)} \]

Equate the numerators, given the denominators are the same:

\[ 5x^2 = 12x + 12 + 15x^2 \]

Rearrange and simplify:

\[ 0 = 10x^2 + 12x + 12 \]

\[ 10x^2 + 12x + 12 = 0 \]

Divide each term by 2 to make it simpler:

\[ 5x^2 + 6x + 6 = 0 \]

Attempt to solve this quadratic equation:

This doesn't factor easily, so apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5, b = 6, \) and \( c = 6 \).

Calculate the discriminant:

\[ b^2 - 4ac = 6^2 - 4(5)(6) = 36 - 120 = -84 \]

Since the discriminant is negative, the equation has no real solutions. Thus, in the real number system, there are no solutions for \( x \) that satisfy this equation given the structure and conditions.