Let’s solve each section step by step. We’ll plug in the given value for each variable and calculate carefully.
---
Section 1: Evaluate for x = 4
1) x + 8 → 4 + 8 =
12
2) 2x - 1 → 2×4 - 1 = 8 - 1 =
7
3) 10 + x² → 10 + (4)² = 10 + 16 =
26
4) -x - 12 → -4 - 12 =
-16
5) -4 + x² → -4 + 16 =
12
6) x² - 1 → 16 - 1 =
15
---
Section 2: Evaluate for y = -2
1) 3y - 3 → 3×(-2) - 3 = -6 - 3 =
-9
2) -2y - 16 → -2×(-2) - 16 = 4 - 16 =
-12
3) 16 + y² → 16 + (-2)² = 16 + 4 =
20
4) -2y + 20 → -2×(-2) + 20 = 4 + 20 =
24
5) y² + 12 → 4 + 12 =
16
6) -y² - 20 → -( (-2)² ) - 20 = -4 - 20 =
-24
*(Note: In #6, it’s -y², which means “negative of y squared”, not “(-y) squared”. So we square first, then apply the negative.)*
---
Section 3: Evaluate for t = -5
1) t² + 75 → (-5)² + 75 = 25 + 75 =
100
2) -t - 25 → -(-5) - 25 = 5 - 25 =
-20
3) -t + 10 → -(-5) + 10 = 5 + 10 =
15
4) 22 - 2t → 22 - 2×(-5) = 22 + 10 =
32
5) -4t - 5 → -4×(-5) - 5 = 20 - 5 =
15
6) -t - 10 → -(-5) - 10 = 5 - 10 =
-5
---
Section 4: Evaluate for q = -1
1) 12 + q² → 12 + (-1)² = 12 + 1 =
13
2) -q² - q → -[(-1)²] - (-1) = -1 + 1 =
0
3) q + 2q → (-1) + 2×(-1) = -1 - 2 =
-3
8) -2q - 30 → -2×(-1) - 30 = 2 - 30 =
-28
5) -q - 2q → -(-1) - 2×(-1) = 1 + 2 =
3
6) 3q² - 3 → 3×(1) - 3 = 3 - 3 =
0
*(Note: Problem numbering skips from 3 to 8 — probably a typo in the worksheet, but we follow as written.)*
---
Section 5: Evaluate for s = -3
1) s² → (-3)² =
9
2) 2s - 1 → 2×(-3) - 1 = -6 - 1 =
-7
3) s + s³ → (-3) + (-3)³ = -3 + (-27) =
-30
4) s + (-1) → -3 + (-1) =
-4
5) -s + 27 → -(-3) + 27 = 3 + 27 =
30
6) s² - 1 → 9 - 1 =
8
---
Final Answer:
For x = 4:
1) 12
2) 7
3) 26
4) -16
5) 12
6) 15
For y = -2:
1) -9
2) -12
3) 20
4) 24
5) 16
6) -24
For t = -5:
1) 100
2) -20
3) 15
4) 32
5) 15
6) -5
For q = -1:
1) 13
2) 0
3) -3
8) -28
5) 3
6) 0
For s = -3:
1) 9
2) -7
3) -30
4) -4
5) 30
6) 8
Parent Tip: Review the logic above to help your child master the concept of 6th grade expressions and equations worksheet.
