To remove the fractions in each equation, we multiply through by the least common denominator (LCD) of all the denominators in the equation. Here's how you approach each part:

a.

$\frac{3x - 4}{8} = \frac{x + 2}{3}$

• The denominators are $8$ and $3$.
• The LCD of $8$ and $3$ is $24$.
• Multiply both sides by $24$:
  $24 \cdot \frac{3x - 4}{8} = 24 \cdot \frac{x + 2}{3}.$
  This eliminates the fractions.

b.

$\frac{3(2 - r)}{4} = \frac{3 + r}{6}$

• The denominators are $4$ and $6$.
• The LCD of $4$ and $6$ is $12$.
• Multiply both sides by $12$:
  $12 \cdot \frac{3(2 - r)}{4} = 12 \cdot \frac{3 + r}{6}.$
  This eliminates the fractions.

c.

$\frac{4p + 3}{8} = \frac{p + 2}{4}$

• The denominators are $8$ and $4$.
• The LCD of $8$ and $4$ is $8$.
• Multiply both sides by $8$:
  $8 \cdot \frac{4p + 3}{8} = 8 \cdot \frac{p + 2}{4}.$
  This eliminates the fractions.

d.

$\frac{2(a - 7)}{15} = \frac{a + 4}{6}$

• The denominators are $15$ and $6$.
• The LCD of $15$ and $6$ is $30$.
• Multiply both sides by $30$:
  $30 \cdot \frac{2(a - 7)}{15} = 30 \cdot \frac{a + 4}{6}.$
  This eliminates the fractions.

Summary of Multiplications:

• a. Multiply through by $24$.
• b. Multiply through by $12$.
• c. Multiply through by $8$.
• d. Multiply through by $30$.

Would you like detailed steps on any of these or further practice solving such equations?

Related Questions:

1. How do you determine the LCD of multiple denominators?
2. Why does multiplying by the LCD eliminate fractions?
3. Solve one of these equations fully.
4. How do you handle equations with three or more fractions?
5. What happens if the fractions have variables in the denominator?

Tip: Always check for the LCD of all fractions in an equation before multiplying. It simplifies the work significantly!