We are given the equation:

$\frac{3x-2}{6x-6} - \frac{5x-1}{3x+3} = \frac{7(25 - 10x + x^2)}{6 - 6x^2}$

Step 1: Simplify the denominators

First, let's factor the denominators where possible:

1. $6x - 6 = 6(x - 1)$
2. $3x + 3 = 3(x + 1)$
3. $6 - 6x^2 = 6(1 - x^2) = 6(1 - x)(1 + x)$

Now, we rewrite the equation using the factored forms:

$\frac{3x - 2}{6(x - 1)} - \frac{5x - 1}{3(x + 1)} = \frac{7(25 - 10x + x^2)}{6(1 - x)(1 + x)}$

Step 2: Simplify the right-hand side

The expression $25 - 10x + x^2$ is a perfect square trinomial:

$25 - 10x + x^2 = (5 - x)^2$

So, the equation becomes:

$\frac{3x - 2}{6(x - 1)} - \frac{5x - 1}{3(x + 1)} = \frac{7(5 - x)^2}{6(1 - x)(1 + x)}$

Step 3: Consider common denominators

At this point, we can multiply both sides by the least common denominator (LCD) of the fractions. The LCD is $6(x - 1)(x + 1)$, so we multiply each term by this expression.

On the left-hand side:

$\frac{3x - 2}{6(x - 1)} \times 6(x - 1)(x + 1) = (3x - 2)(x + 1)$ $\frac{5x - 1}{3(x + 1)} \times 6(x - 1)(x + 1) = 2(5x - 1)(x - 1)$

On the right-hand side:

$\frac{7(5 - x)^2}{6(1 - x)(1 + x)} \times 6(x - 1)(x + 1) = 7(5 - x)^2$

Step 4: Expand and simplify

Now, we expand both sides.

Left-hand side:

$(3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2$ $2(5x - 1)(x - 1) = 2(5x^2 - 5x - x + 1) = 10x^2 - 12x + 2$

So the left-hand side becomes:

$(3x^2 + x - 2) - (10x^2 - 12x + 2) = 3x^2 + x - 2 - 10x^2 + 12x - 2 = -7x^2 + 13x - 4$

Right-hand side:

$7(5 - x)^2 = 7(25 - 10x + x^2) = 175 - 70x + 7x^2$

Step 5: Set the equation equal and solve

Now, we equate both sides:

$-7x^2 + 13x - 4 = 7x^2 - 70x + 175$

Move all terms to one side:

$-7x^2 + 13x - 4 - 7x^2 + 70x - 175 = 0$ $-14x^2 + 83x - 179 = 0$

Now, solve the quadratic equation $-14x^2 + 83x - 179 = 0$.

Step 6: Solve the quadratic equation

Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -14$, $b = 83$, and $c = -179$. Substituting these values:

$x = \frac{-83 \pm \sqrt{83^2 - 4(-14)(-179)}}{2(-14)}$ $x = \frac{-83 \pm \sqrt{6889 - 10012}}{-28}$ $x = \frac{-83 \pm \sqrt{-3123}}{-28}$

Since the discriminant is negative ($-3123$), there are no real solutions for $x$.

Conclusion:

The equation has no real solutions because the discriminant is negative, indicating that the solutions are complex.

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Would you like more details on any of the steps or a deeper explanation on how to handle complex solutions?

Here are some related questions:

1. How do you solve quadratic equations with a negative discriminant?
2. Can you explain the significance of complex numbers in this context?
3. What are the different methods for solving rational equations?
4. How do you check for extraneous solutions when solving rational equations?
5. What is the general approach to factoring quadratic trinomials?

Tip: Always check the discriminant when solving quadratics to quickly determine if the solutions are real or complex!