Question

1st semester credit recovery packet
must complete packet (show work) and then come in and pass the credit recovery test.
unit 1 review
find the solution to each equation.

1. $-3p - 4(4p - 8) = 3(-8p - 1)$
2. $3x - 2(6x - 10) = -16 + 3(5x + 4)$
3. $4(2x + 1.5) = 4x - 7$
4. $9x - (3x - 6) = 2(x + 1)$
5. $3(6 - 2x) = -2(4x - 5)$
6. $2(x + 2) = 6(x - 12)$

Explanation:

Problem 1: $-3p - 4(4p - 8) = 3(-8p - 1)$

Step 1: Distribute the terms

First, distribute the -4 on the left side and the 3 on the right side:
$-3p - 16p + 32 = -24p - 3$

Step 2: Combine like terms

Combine the $p$ terms on the left side:
$-19p + 32 = -24p - 3$

Step 3: Add $24p$ to both sides

$-19p + 24p + 32 = -3$
$5p + 32 = -3$

Step 4: Subtract 32 from both sides

$5p = -3 - 32$
$5p = -35$

Step 5: Divide both sides by 5

$p = \frac{-35}{5}$
$p = -7$

Step 1: Distribute the terms

Distribute -2 on the left and 3 on the right:
$3x - 12x + 20 = -16 + 15x + 12$

Step 2: Combine like terms

Left side: $-9x + 20$
Right side: $15x - 4$
So, $-9x + 20 = 15x - 4$

Step 3: Add $9x$ to both sides

$20 = 24x - 4$

Step 4: Add 4 to both sides

$24 = 24x$

Step 5: Divide both sides by 24

$x = 1$

Step 1: Distribute the 4 on the left side

$8x + 6 = 4x - 7$

Step 2: Subtract $4x$ from both sides

$8x - 4x + 6 = -7$
$4x + 6 = -7$

Step 3: Subtract 6 from both sides

$4x = -7 - 6$
$4x = -13$

Step 4: Divide both sides by 4

$x = \frac{-13}{4} = -3.25$

Answer:

$p = -7$

Problem 2: $3x - 2(6x - 10) = -16 + 3(5x + 4)$