Let’s solve each section step by step. We’ll substitute the given values into the expressions and calculate carefully.
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Section A: Substitute value of a
1.
8a – 9, where
a = 2
→ 8 × 2 = 16
→ 16 – 9 =
7
2.
4 + 2a, where
a = 6
→ 2 × 6 = 12
→ 4 + 12 =
16
3.
a/5 + 20, where
a = 10
→ 10 ÷ 5 = 2
→ 2 + 20 =
22
4.
11 – a/3, where
a = 12
→ 12 ÷ 3 = 4
→ 11 – 4 =
7
5.
8a/10, where
a = 5
→ 8 × 5 = 40
→ 40 ÷ 10 =
4
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Section B: First part — substitute value of b
1.
(14b + 13)/9, where
b = 1
→ 14 × 1 = 14
→ 14 + 13 = 27
→ 27 ÷ 9 =
3
2.
(52 – 2b)/7, where
b = 12
→ 2 × 12 = 24
→ 52 – 24 = 28
→ 28 ÷ 7 =
4
3.
10b – b², where
b = 2
→ 10 × 2 = 20
→ b² = 2 × 2 = 4
→ 20 – 4 =
16
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Section B: Second part — substitute values of a and b
1.
7a + ab, where
a = 2, b = 3
→ 7 × 2 = 14
→ a × b = 2 × 3 = 6
→ 14 + 6 =
20
2.
9b – 6a, where
a = 5, b = 8
→ 9 × 8 = 72
→ 6 × 5 = 30
→ 72 – 30 =
42
3.
a² + b², where
a = 11, b = 4
→ a² = 11 × 11 = 121
→ b² = 4 × 4 = 16
→ 121 + 16 =
137
4.
1 + (5a)/b, where
a = 12, b = 6
→ 5 × 12 = 60
→ 60 ÷ 6 = 10
→ 1 + 10 =
11
5.
(84 – 6a)/b, where
a = 10, b = 3
→ 6 × 10 = 60
→ 84 – 60 = 24
→ 24 ÷ 3 =
8
6.
(a + b)² + 3, where
a = 2, b = 5
→ a + b = 2 + 5 = 7
→ 7² = 49
→ 49 + 3 =
52
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Section C: Substitute values of x and y
1.
(33 – 3x)/(2y), where
x = 7, y = 3
→ 3 × 7 = 21
→ 33 – 21 = 12
→ 2 × 3 = 6
→ 12 ÷ 6 =
2
2.
xy – y², where
x = 10, y = 6
→ x × y = 10 × 6 = 60
→ y² = 6 × 6 = 36
→ 60 – 36 =
24
3.
(xy – 3y)², where
x = 4, y = 8
→ x × y = 4 × 8 = 32
→ 3 × y = 3 × 8 = 24
→ 32 – 24 = 8
→ 8² =
64
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Final Answer:
Section A Answers:
7, 16, 22, 7, 4
Section B (first part) Answers:
3, 4, 16
Section B (second part) Answers:
20, 42, 137, 11, 8, 52
Section C Answers:
2, 24, 64
Parent Tip: Review the logic above to help your child master the concept of algebra substitution worksheet.
