https://file.aiquickdraw.com/mathbot/file/1727402549614oigx4aeq.jpg

Let's focus on solving part **6(a)** from the provided image. The equation in 6(a) is:

\[
\frac{x^2 + 15x}{4x + 24} \div \frac{3x}{3x + 18}
\]

### Step 1: Factor the expressions
Factor both the numerator and the denominator for each term.

- For \(\frac{x^2 + 15x}{4x + 24}\):
    - The numerator \(x^2 + 15x = x(x + 15)\)
    - The denominator \(4x + 24 = 4(x + 6)\)

Thus, the first fraction becomes:
\[
\frac{x(x + 15)}{4(x + 6)}
\]

- For \(\frac{3x}{3x + 18}\):
    - The numerator is already \(3x\).
    - The denominator \(3x + 18 = 3(x + 6)\).

Thus, the second fraction becomes:
\[
\frac{3x}{3(x + 6)}
\]

### Step 2: Rewrite the division as multiplication
We now rewrite the division as multiplication by the reciprocal:

\[
\frac{x(x + 15)}{4(x + 6)} \times \frac{3(x + 6)}{3x}
\]

### Step 3: Simplify
Cancel out common factors:

- The \((x + 6)\) cancels from the numerator and denominator.
- The \(3x\) cancels with the \(3x\) in the other fraction.

We are left with:
\[
\frac{x + 15}{4}
\]

### Step 4: State the restrictions
The restrictions on the variable come from the values that make any denominator zero.

1. \(4(x + 6) \neq 0 \implies x \neq -6\)
2. \(3(x + 6) \neq 0 \implies x \neq -6\)
3. \(x \neq 0\) from the \(x\) in the second fraction.

Thus, the restrictions are:
\[
x \neq 0 \quad \text{and} \quad x \neq -6
\]

### Final Answer:
\[
\frac{x + 15}{4}, \quad \text{with restrictions} \ x \neq 0 \ \text{and} \ x \neq -6.
\]

Would you like further details on any steps or clarifications? Here are some related questions:

1. How do we factor quadratic expressions?
2. Why is division by zero undefined?
3. How does multiplication by the reciprocal work in fractions?
4. Can we always cancel terms directly across fractions?
5. What would happen if no restrictions were applied?

**Tip:** Always factor both numerators and denominators fully before simplifying in algebraic fraction problems to avoid mistakes.

Solve 6a: Simplify the given algebraic fraction and state restrictions on the variable.

Learn how to simplify a rational expression and determine the restrictions on the variable with a step-by-step approach. The problem involves factoring and division of fractions. Suitable for students in Grades 9-11.

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