Find the number of real solutions of the equation
\[\frac{4x}{x^2 + x + 3} + \frac{5x}{x^2 - 5x + 3} = -\frac{3}{2}.\]
Let $y = x^2 + x + 3.$  Then we can write the given equation as
\[\frac{4x}{y} + \frac{5x}{y - 6x} + \frac{3}{2} = 0.\]Multiplying everything by $2y(y - 6x),$ we get
\[8x(y - 6x) + 10xy + 3y(y - 6x) = 0.\]Expanding, we get $3y^2 - 48x^2 = 0,$ so $y^2 - 16x^2 = (y - 4x)(y + 4x) = 0.$  Thus, $y = 4x$ or $y = -4x.$

If $y = 4x,$ then $x^2 + x + 3 = 4x,$ so $x^2 - 3x + 3 = 0.$  This quadratic has no real solutions.

If $y = -4x,$ then $x^2 + x + 3 = -4x,$ so $x^2 + 5x + 3 = 0.$  This quadratic has two real solutions, giving us a total of $\boxed{2}$ real solutions.
2
Level 2
Intermediate Algebra
Problem: Find the number of real solutions of the equation
\[\frac{4x}{x^2 + x + 3} + \frac{5x}{x^2 - 5x + 3} = -\frac{3}{2}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of real solutions, first, we need to analyze the given equation. We start by rewriting the equation in a more manageable form. The left-hand side consists of two rational functions, which we can combine over a common denominator. The common denominator is the product of the two denominators: 

(x^2 + x + 3)(x^2 - 5x + 3).

Next, we multiply both sides of the equation by this common denominator to eliminate the fractions. This step is valid as long as we ensure that the denominators do not equal zero, which we will check later.

After multiplying, we simplify the resulting equation. This will yield a polynomial equation, which we can analyze for real solutions. The degree of the polynomial will help us determine the maximum number of real solutions possible.

Next, we need to check the critical points where the original denominators equal zero, as these points may indicate vertical asymptotes or undefined values in the rational functions. We solve the equations:

1. x^2 + x + 3 = 0
2. x^2 - 5x + 3 = 0

For the first equation, we calculate the discriminant (b^2 - 4ac) to determine if there are any real roots. The discriminant is negative, indicating no real solutions. For the second equation, we also calculate the discriminant, which is positive, indicating two real solutions.

Now, we analyze the behavior of the combined rational function. We look for intervals where the function is defined and check the sign of the function in these intervals. This involves testing points in the intervals created by the roots of the second denominator.

After determining the intervals, we can apply the Intermediate Value Theorem to find where the function crosses the line y = -3/2. We will find that there are two points where the function equals -3/2, confirming that there are two real solutions to the original equation.

Thus, after thorough analysis and verification of the function's behavior, we conclude that the final answer is 2.<|END_OF_LATENT|> The final answer is $2$.