6th Grade Algebraic Expressions Worksheets - Math Monks - Free Printable

Sure! Let’s evaluate each algebraic expression step by step using the given values.

---

1) \( p^2 + m \); where \( m = 2, p = 1 \)
→ Substitute: \( (1)^2 + 2 = 1 + 2 = \boxed{3} \)

---

2) \( x - x^2 + 1 + y \); where \( x = 1, y = -1 \)
→ Substitute: \( 1 - (1)^2 + 1 + (-1) = 1 - 1 + 1 - 1 = \boxed{0} \)

---

3) \( p(r + q) \); where \( p = 3, q = 6, r = 4 \)
→ First, inside parentheses: \( r + q = 4 + 6 = 10 \)
→ Then: \( p \cdot 10 = 3 \cdot 10 = \boxed{30} \)

---

4) \( y - (x - x) \); where \( x = 6, y = 2 \)
→ Inside parentheses: \( x - x = 6 - 6 = 0 \)
→ Then: \( y - 0 = 2 - 0 = \boxed{2} \)

---

5) \( p(p + m) \); where \( m = 1, p = 6 \)
→ Inside: \( p + m = 6 + 1 = 7 \)
→ Then: \( p \cdot 7 = 6 \cdot 7 = \boxed{42} \)

---

6) \( j^2 h \); where \( h = 2, j = 4 \)
→ First: \( j^2 = 4^2 = 16 \)
→ Then: \( 16 \cdot 2 = \boxed{32} \)

---

7) \( xy \div 6 \); where \( x = 5, y = 6 \)
→ First: \( xy = 5 \cdot 6 = 30 \)
→ Then: \( 30 \div 6 = \boxed{5} \)

---

8) \( 2p - q \); where \( p = 4, q = -2 \)
→ Substitute: \( 2(4) - (-2) = 8 + 2 = \boxed{10} \)
*(Note: Subtracting a negative is adding.)*

---

9) \( y^2 - x \); where \( x = 1, y = 5 \)
→ \( y^2 = 25 \), so \( 25 - 1 = \boxed{24} \)

---

10) \( 6 - (p - m) \); where \( m = 1, p = -5 \)
→ Inside: \( p - m = -5 - 1 = -6 \)
→ Then: \( 6 - (-6) = 6 + 6 = \boxed{12} \)

---

11) \( (x - y) \div 2 \); where \( x = 3, y = 1 \)
→ Inside: \( 3 - 1 = 2 \)
→ Then: \( 2 \div 2 = \boxed{1} \)

---

12) \( (x + y)^2 \); where \( x = 1, y = 1 \)
→ Inside: \( 1 + 1 = 2 \)
→ Then: \( 2^2 = \boxed{4} \)

---

✔ Final Answers:

1) 3
2) 0
3) 30
4) 2
5) 42
6) 32
7) 5
8) 10
9) 24
10) 12
11) 1
12) 4

Let me know if you’d like to see any step explained further!