Question

1. $3(3c + 5) + 1 = 2(c - 20)$
2. $3 - (4w + 5) = \frac{1}{2}(8w + 28)$
3. $-13 + 12p - 4 = 6(2p - 1)$
4. $-7(m - 5) = 4(4 - m) + 1$
5. $2(8r + 5) - 3 = 4(4r - 1) + 11$
6. $12 - 4(2x + 9) = -8(x + 3)$
7. $3(8k - 3) = -6(7 - 4k)$
8. $7v - (2v - 16) = 5(v + 4)$

Explanation:

Let's solve one of the equations, say problem 5: \( 3(3c + 5) + 1 = 2(c - 20) \)

Step 1: Expand both sides

Left side: \( 3(3c + 5) + 1 = 9c + 15 + 1 = 9c + 16 \)
Right side: \( 2(c - 20) = 2c - 40 \)
So the equation becomes \( 9c + 16 = 2c - 40 \)

Step 2: Subtract \( 2c \) from both sides

\( 9c - 2c + 16 = 2c - 2c - 40 \)
\( 7c + 16 = -40 \)

Step 3: Subtract 16 from both sides

\( 7c + 16 - 16 = -40 - 16 \)
\( 7c = -56 \)

Step 4: Divide both sides by 7

\( \frac{7c}{7} = \frac{-56}{7} \)
\( c = -8 \)

Answer:

\( c = -8 \)