Let's solve the equation:

\[
2 \frac{3}{x+1} + 1 \frac{4}{x} = 5 \frac{3}{x}
\]

### Step 1: Convert mixed numbers to improper fractions.

We have mixed numbers in the equation, so let's convert them into improper fractions first.

1. \(2 \frac{3}{x+1}\) becomes \(\frac{2(x+1) + 3}{x+1} = \frac{2x + 2 + 3}{x+1} = \frac{2x + 5}{x+1}\).

2. \(1 \frac{4}{x}\) becomes \(\frac{1 \cdot x + 4}{x} = \frac{x + 4}{x}\).

3. \(5 \frac{3}{x}\) becomes \(\frac{5 \cdot x + 3}{x} = \frac{5x + 3}{x}\).

### Step 2: Substitute improper fractions back into the equation.

The equation now looks like:

\[
\frac{2x + 5}{x+1} + \frac{x + 4}{x} = \frac{5x + 3}{x}
\]

### Step 3: Find a common denominator.

To simplify, we need a common denominator on the left-hand side.

The common denominator is \(x(x+1)\).

Rewriting each term with this denominator:

\[
\frac{2x + 5}{x+1} = \frac{(2x + 5) \cdot x}{x(x+1)} = \frac{x(2x + 5)}{x(x+1)} = \frac{2x^2 + 5x}{x(x+1)}
\]

\[
\frac{x + 4}{x} = \frac{(x + 4) \cdot (x+1)}{x(x+1)} = \frac{(x + 4)(x+1)}{x(x+1)} = \frac{x^2 + x + 4x + 4}{x(x+1)} = \frac{x^2 + 5x + 4}{x(x+1)}
\]

Now the left-hand side becomes:

\[
\frac{2x^2 + 5x}{x(x+1)} + \frac{x^2 + 5x + 4}{x(x+1)} = \frac{3x^2 + 10x + 4}{x(x+1)}
\]

### Step 4: Set the equation equal.

The equation is now:

\[
\frac{3x^2 + 10x + 4}{x(x+1)} = \frac{5x + 3}{x}
\]

### Step 5: Cross-multiply to eliminate fractions.

Cross-multiply to get rid of the denominators:

\[
(3x^2 + 10x + 4) \cdot x = (5x + 3) \cdot (x(x+1))
\]

Expanding both sides:

Left-hand side: \((3x^2 + 10x + 4) \cdot x = 3x^3 + 10x^2 + 4x\)

Right-hand side: \((5x + 3) \cdot (x(x+1)) = (5x + 3)(x^2 + x) = 5x^3 + 5x^2 + 3x^2 + 3x = 5x^3 + 8x^2 + 3x\)

Now, the equation is:

\[
3x^3 + 10x^2 + 4x = 5x^3 + 8x^2 + 3x
\]

### Step 6: Move all terms to one side.

\[
3x^3 + 10x^2 + 4x - 5x^3 - 8x^2 - 3x = 0
\]

Simplify:

\[
-2x^3 + 2x^2 + x = 0
\]

### Step 7: Factor the equation.

Factor out \(x\):

\[
x(-2x^2 + 2x + 1) = 0
\]

This gives two cases:

1. \(x = 0\)
2. Solve \(-2x^2 + 2x + 1 = 0\).

### Step 8: Solve the quadratic equation.

Use the quadratic formula to solve \(-2x^2 + 2x + 1 = 0\):

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

Where \(a = -2\), \(b = 2\), and \(c = 1\):

\[
x = \frac{-2 \pm \sqrt{2^2 - 4(-2)(1)}}{2(-2)} = \frac{-2 \pm \sqrt{4 + 8}}{-4} = \frac{-2 \pm \sqrt{12}}{-4}
\]

Simplify:

\[
x = \frac{-2 \pm 2\sqrt{3}}{-4} = \frac{1 \pm \sqrt{3}}{2}
\]

So, the two solutions from this are:

\[
x = \frac{1 + \sqrt{3}}{2}, \quad x = \frac{1 - \sqrt{3}}{2}
\]

### Final Answer:

The solutions to the equation are:

\[
x = 0, \quad x = \frac{1 + \sqrt{3}}{2}, \quad x = \frac{1 - \sqrt{3}}{2}
\]

Would you like more details on any step, or have any questions?

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Here are 5 related questions:
1. How do you solve other types of rational equations?
2. What happens if a solution leads to division by zero?
3. Can mixed numbers be solved without converting to improper fractions?
4. How can you simplify equations involving fractions faster?
5. How does the quadratic formula help solve second-degree equations?

**Tip:** Always check your solutions to ensure they don't result in undefined expressions, such as division by zero.

This problem explores the step-by-step solution of a rational equation with mixed numbers and improper fractions. It uses key algebraic techniques like converting mixed numbers to improper fractions, cross-multiplying to eliminate denominators, and solving a quadratic equation using the quadratic formula. Ideal for students in Grades 9-11.

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