Let's solve the equation step-by-step:

The equation is given as:

$\frac{4x}{x + 3} - \frac{1}{x} = \frac{1}{x}$

Step 1: Combine terms with $\frac{1}{x}$

First, let's move all the $\frac{1}{x}$ terms to one side of the equation. To do that, add $\frac{1}{x}$ to both sides:

$\frac{4x}{x + 3} = \frac{1}{x} + \frac{1}{x}$

Simplifying the right-hand side:

$\frac{4x}{x + 3} = \frac{2}{x}$

Step 2: Eliminate the fractions by cross-multiplying

Now, to eliminate the fractions, multiply both sides of the equation by $x(x + 3)$ (the denominators of both fractions):

$x(x + 3) \cdot \frac{4x}{x + 3} = x(x + 3) \cdot \frac{2}{x}$

Simplifying both sides:

$4x^2 = 2(x + 3)$

Step 3: Expand the right side

Distribute the 2 on the right-hand side:

$4x^2 = 2x + 6$

Step 4: Move all terms to one side

Subtract $2x + 6$ from both sides to set the equation equal to zero:

$4x^2 - 2x - 6 = 0$

Step 5: Solve the quadratic equation

Now, let's solve the quadratic equation. We can divide through by 2 to simplify it:

$2x^2 - x - 3 = 0$

Next, we solve this using factoring, the quadratic formula, or completing the square. Let's try factoring.

We need to find two numbers that multiply to $2 \times -3 = -6$ and add to $-1$. These numbers are $-3$ and $2$.

So, we can rewrite the equation as:

$2x^2 - 3x + 2x - 3 = 0$

Now, factor by grouping:

$x(2x - 3) + 1(2x - 3) = 0$

Factor out the common binomial:

$(2x - 3)(x + 1) = 0$

Step 6: Solve for $x$

Now, set each factor equal to zero:

1. $2x - 3 = 0$ $\Rightarrow$ $x = \frac{3}{2}$
2. $x + 1 = 0$ $\Rightarrow$ $x = -1$

Step 7: Check for extraneous solutions

We need to check if either of these solutions makes any denominators in the original equation zero.

• For $x = \frac{3}{2}$, neither denominator becomes zero, so this is a valid solution.
• For $x = -1$, the denominators $x + 3$ and $x$ would both become zero, making this an extraneous solution.

Thus, the only valid solution is:

$x = \frac{3}{2}$

Final Answer:

$x = \frac{3}{2}$

Would you like to explore further details or have any additional questions?

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Here are 5 related questions to consider:

1. How do you solve quadratic equations by factoring?
2. What is the quadratic formula, and how is it derived?
3. How do you identify extraneous solutions in rational equations?
4. What are the steps to simplify rational expressions?
5. How do you verify solutions in equations involving fractions?

Tip: When solving rational equations, always check for potential restrictions on the variables (like values that make denominators zero).