To solve the equation:

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

### Step 1: Eliminate the denominators
Multiply through by the least common denominator (LCD), which is \(x(x+3)\), to get rid of the fractions:

\[
x(x+3) \cdot \left( \frac{4}{x} - \frac{5}{x+3} \right) = x(x+3) \cdot 1
\]

Simplifying each term:

\[
4(x+3) - 5x = x(x+3)
\]

### Step 2: Expand both sides
Expand both sides of the equation:

\[
4x + 12 - 5x = x^2 + 3x
\]

### Step 3: Simplify the equation
Combine like terms on the left side:

\[
-x + 12 = x^2 + 3x
\]

### Step 4: Move all terms to one side
Rearrange the equation to form a quadratic equation:

\[
x^2 + 3x + x - 12 = 0
\]

This simplifies to:

\[
x^2 + 4x - 12 = 0
\]

### Step 5: Solve the quadratic equation
Use the quadratic formula to solve for \(x\), where \(a = 1\), \(b = 4\), and \(c = -12\):

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

Substitute the values of \(a\), \(b\), and \(c\):

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

\[
x = \frac{-4 \pm \sqrt{16 + 48}}{2}
\]

\[
x = \frac{-4 \pm \sqrt{64}}{2}
\]

\[
x = \frac{-4 \pm 8}{2}
\]

This gives two solutions:

\[
x = \frac{-4 + 8}{2} = 2 \quad \text{or} \quad x = \frac{-4 - 8}{2} = -6
\]

### Step 6: Check for extraneous solutions
Substitute both values of \(x\) back into the original equation to ensure they are valid solutions.

- For \(x = 2\):  
  \[
  \frac{4}{2} - \frac{5}{2+3} = 1
  \]
  \[
  2 - 1 = 1 \quad \text{(Valid solution)}
  \]

- For \(x = -6\):  
  \[
  \frac{4}{-6} - \frac{5}{-6+3} = 1
  \]
  This results in division by zero, so \(x = -6\) is not a valid solution.

### Final Answer:
The solution is:

\[
x = 2
\]

Do you want more details or have any questions?

Here are 5 related questions:
1. How do you solve quadratic equations using factorization?
2. What is the quadratic formula and how is it derived?
3. How can you check if a quadratic equation has real or complex solutions?
4. What is an extraneous solution and why does it occur?
5. How do you solve rational equations without the quadratic formula?

**Tip:** Always check your solutions in the original equation, especially for rational equations, to avoid extraneous solutions.

Explore how to solve the rational equation 4/x - 5/(x+3) = 1 with step-by-step guidance. Learn key algebraic concepts like eliminating denominators, using the quadratic formula, and identifying extraneous solutions. Ideal for students in Grades 9-10.

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