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Let's solve each problem step-by-step.

### Problem 1
\[
5x = 35
\]
To solve for \( x \), divide both sides by 5:
\[
x = \frac{35}{5} = 7
\]
**Answer:** \( x = 7 \)

---

### Problem 2
\[
24 = -3x
\]
To solve for \( x \), divide both sides by -3:
\[
x = \frac{24}{-3} = -8
\]
**Answer:** \( x = -8 \)

---

### Problem 3
\[
14c + 15 = 43 + 7c
\]
1. Move \( 7c \) to the left side by subtracting \( 7c \) from both sides:
   \[
   14c - 7c + 15 = 43
   \]
   \[
   7c + 15 = 43
   \]
2. Subtract 15 from both sides:
   \[
   7c = 28
   \]
3. Divide by 7:
   \[
   c = 4
   \]
**Answer:** \( c = 4 \)

---

### Problem 4
\[
5(2y - 1) = 2(4y - 3)
\]
1. Expand both sides:
   \[
   10y - 5 = 8y - 6
   \]
2. Move \( 8y \) to the left side by subtracting \( 8y \) from both sides:
   \[
   10y - 8y - 5 = -6
   \]
   \[
   2y - 5 = -6
   \]
3. Add 5 to both sides:
   \[
   2y = -1
   \]
4. Divide by 2:
   \[
   y = -\frac{1}{2}
   \]
**Answer:** \( y = -\frac{1}{2} \)

---

### Problem 5
\[
5p + 6(p + 7) = 3(p + 2)
\]
1. Expand both sides:
   \[
   5p + 6p + 42 = 3p + 6
   \]
   \[
   11p + 42 = 3p + 6
   \]
2. Move \( 3p \) to the left side by subtracting \( 3p \) from both sides:
   \[
   11p - 3p + 42 = 6
   \]
   \[
   8p + 42 = 6
   \]
3. Subtract 42 from both sides:
   \[
   8p = -36
   \]
4. Divide by 8:
   \[
   p = -\frac{36}{8} = -\frac{9}{2}
   \]
**Answer:** \( p = -\frac{9}{2} \)

---

### Problem 6
\[
32 - [4 + 6x - 5(x + 4)] = 4(3x + 4) - [6(3x - 4) + 7 - 4x]
\]
1. Start by expanding within the brackets on both sides.

For the left side:
   \[
   32 - [4 + 6x - 5x - 20]
   \]
   \[
   = 32 - [ -16 + x]
   \]
   \[
   =

Detailed solutions for solving various linear equations, covering concepts such as the distributive property, combining like terms, and basic equation manipulation. Perfect for students in Grades 8-10.

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