Question

name
date \t\t\tperiod

1. $6 = 1 - 2n - 5$
2. $a + 5 = -5a + 5$
3. $p - 1 = 5p + 3p - 8$
4. $p - 4 = -9 + p$
5. $12 = -4(-6x - 3)$
6. $-(7 - 4x) = 9$
7. $5n + 34 = -2(1 - 7n)$
8. $3n - 5 = -8(6 + 5n)$
9. $-3(4x + 3) + 4(6x + 1) = 43$
10. $-5(1 - 5x) + 5(-8x - 2) = -4x - 8x$

Explanation:

Let's solve each equation one by one:

Problem 2: \( 6 = 1 - 2n - 5 \)

Step 1: Simplify the right side

Combine like terms: \( 1 - 5 = -4 \), so the equation becomes \( 6 = -4 - 2n \)

Step 2: Add \( 4 \) to both sides

\( 6 + 4 = -2n \) which simplifies to \( 10 = -2n \)

Step 3: Divide both sides by \( -2 \)

\( n = \frac{10}{-2} = -5 \)

Step 1: Subtract \( 5 \) from both sides

\( a = -5a \)

Step 2: Add \( 5a \) to both sides

\( a + 5a = 0 \) which simplifies to \( 6a = 0 \)

Step 3: Divide by \( 6 \)

\( a = 0 \)

Step 1: Combine like terms on the right

\( 5p + 3p = 8p \), so equation is \( p - 1 = 8p - 8 \)

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

\( -1 = 7p - 8 \)

Step 3: Add \( 8 \) to both sides

\( 7 = 7p \)

Step 4: Divide by \( 7 \)

\( p = 1 \)

Answer:

\( n = -5 \)

Problem 4: \( a + 5 = -5a + 5 \)