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5. $3(3c + 5) + 1 = 2(c - 20)$ 6. $3 - (4w + 5) = \\frac{1}{2}(8w + 28)…

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:

Problem 5: \( 3(3c + 5) + 1 = 2(c - 20) \)

Step 1: Expand both sides

Expand the left side: \( 3(3c + 5) + 1 = 9c + 15 + 1 = 9c + 16 \)
Expand the 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

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

Step 1: Simplify left side

\( 3 - 4w - 5 = -4w - 2 \)

Step 2: Simplify right side

\( \frac{1}{2}(8w + 28) = 4w + 14 \)
Equation: \( -4w - 2 = 4w + 14 \)

Step 3: Add \( 4w \) to both sides

\( -4w + 4w - 2 = 4w + 4w + 14 \)
\( -2 = 8w + 14 \)

Step 4: Subtract 14 from both sides

\( -2 - 14 = 8w + 14 - 14 \)
\( -16 = 8w \)

Step 5: Divide by 8

\( \frac{-16}{8} = \frac{8w}{8} \)
\( w = -2 \)

Step 1: Simplify left side

\( -17 + 12p \)

Step 2: Expand right side

\( 6(2p - 1) = 12p - 6 \)
Equation: \( -17 + 12p = 12p - 6 \)

Step 3: Subtract \( 12p \) from both sides

\( -17 + 12p - 12p = 12p - 12p - 6 \)
\( -17 = -6 \)

This is a contradiction, so no solution.

Answer:

\( c = -8 \)

Problem 6: \( 3 - (4w + 5) = \frac{1}{2}(8w + 28) \)