Question

chapter 5: expressions and equations
solving 1-step equations: addition & subtraction
solve each equation.

1. $9 + d = 16$ ____ 2. $y + 3 = 9$ __ 3. $12 + a = 27$ ____
2. $18 - b = 4$ ____ 5. $23 - c = 21$ __ 6. $w - 11 = 11$ ____
3. $n + 8 = 41$ ____ 8. $7 + m = 20$ __ 9. $9 + s = 9$ ____
4. $t - 18 = 5$ ____ 11. $36 - a = 36$ __ 12. $15 - b = 0$ ____

Explanation:

Problem 1: \( 9 + d = 16 \)

Step 1: Subtract 9 from both sides

To isolate \( d \), we subtract 9 from both sides of the equation. This is based on the subtraction property of equality, which states that if \( a = b \), then \( a - c = b - c \).
\( 9 + d - 9 = 16 - 9 \)

Step 2: Simplify both sides

Simplifying the left side, \( 9 - 9 = 0 \), so we are left with \( d \). Simplifying the right side, \( 16 - 9 = 7 \).
\( d = 7 \)

Step 1: Subtract 3 from both sides

Using the subtraction property of equality, we subtract 3 from both sides to isolate \( y \).
\( y + 3 - 3 = 9 - 3 \)

Step 2: Simplify both sides

The left side simplifies to \( y \) (since \( 3 - 3 = 0 \)), and the right side simplifies to \( 6 \) (since \( 9 - 3 = 6 \)).
\( y = 6 \)

Step 1: Subtract 12 from both sides

Applying the subtraction property of equality, we subtract 12 from both sides to solve for \( a \).
\( 12 + a - 12 = 27 - 12 \)

Step 2: Simplify both sides

The left side becomes \( a \) (because \( 12 - 12 = 0 \)), and the right side is \( 15 \) (since \( 27 - 12 = 15 \)).
\( a = 15 \)

Answer:

\( d = 7 \)

Problem 2: \( y + 3 = 9 \)