Numerical and Algebraic Expressions: practice solutions, Grade 6 – download the PDF

1 Reading exponents ★★★

1. \(3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81\).
2. \(2^6 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 64\).
3. \(10^5 = 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 100{,}000\), a 1 followed by five zeros.
4. \(7^2 = 7 \cdot 7 = 49\).

2 Order of operations ★★★

1. Multiply first: \(8 + 18 = 26\).
2. Parentheses first: \(14 \cdot 3 = 42\).
3. Divide first: \(30 - 3 = 27\).
4. Exponent first: \(5 \cdot 4 = 20\).

Notice that (a) and (b) use the same numbers but give different values because of the parentheses.

3 Words to expressions ★★★

1. \(n + 9\)
2. \(4y\)
3. \(k - 6\)
4. \(\dfrac{m}{5}\), or \(m \div 5\).

4 Evaluate with one variable ★★★

1. \(6 + 9 = 15\).
2. \(4 \cdot 6 = 24\).
3. \(30 - 6 = 24\).
4. \(6 \div 3 = 2\).

5 Name the parts ★★★

1. Three terms, separated by plus signs: \(7a\), \(3b\), and \(12\).
2. The coefficient of \(a\) is 7 and the coefficient of \(b\) is 3.
3. The constant is 12, because it has no variable.

6 Sort the like terms ★★★

The \(x\)-terms: \(5x\) and \(2x\). The \(y\)-terms: \(3y\) and \(9y\). The constants: \(8\) and \(4\).

Terms are like terms when they have exactly the same variable part, or when none of them has a variable.

7 True or false? ★★★

1. False. \(2^3 = 2 \cdot 2 \cdot 2 = 8\).
2. False. Multiplication comes first: \(4 + 6 = 10\).
3. True. Both sides equal 15, and this is the distributive property.
4. False. The left side is \(8 - 2 = 6\), but the right side is \(12 - 2 = 10\). Subtraction goes from left to right.

8 Several operations ★★★

1. \(3 + 2 \cdot 3^2 = 3 + 2 \cdot 9 = 3 + 18 = 21\).
2. \(48 \div 8 + 20 = 6 + 20 = 26\).
3. \(8^2 \div 16 = 64 \div 16 = 4\).

9 Pizza party ★★★

1. Each pizza gives 8 slices, so the expression is \(8p\).
2. \(8 \cdot 3 = 24\), \(8 \cdot 5 = 40\), and \(8 \cdot 12 = 96\) slices.
3. For \(p = 8\): \(8 \cdot 8 = 64\), too few. For \(p = 9\): \(8 \cdot 9 = 72\). The party should order 9 pizzas.

10 Saving money ★★★

1. She saves \(15w\) dollars and starts with 40, so the expression is \(15w + 40\).
2. For \(w = 8\): \(15 \cdot 8 + 40 = 120 + 40 = 160\) dollars. For \(w = 12\): \(180 + 40 = 220\) dollars.
3. The coefficient of \(w\) is 15 (dollars saved per week) and the constant is 40 (dollars at the start).

11 Two variables ★★★

1. \(2(5) + 4(3) = 10 + 12 = 22\).
2. \(5^2 - 3^2 = 25 - 9 = 16\).
3. \(3(5 + 3) = 3 \cdot 8 = 24\).

12 Expand with the distributive property ★★★

1. \(6 \cdot x + 6 \cdot 4 = 6x + 24\).
2. \(3 \cdot 2y + 3 \cdot 5 = 6y + 15\).
3. \(7 \cdot n - 7 \cdot 2 = 7n - 14\).

13 Combine like terms ★★★

1. \((7 + 3)x = 10x\).
2. \((9 - 4)y + 2 = 5y + 2\).
3. \((5a + 3a) + (2b + b) = 8a + 3b\), because \(b = 1b\).

14 Equivalent or not? ★★★

1. \(3(6) = 18\) but \(3(4) + 2 = 14\). The values differ, so they are not equivalent. (The 3 must multiply the 2 as well.)
2. Both give 15. Since \(2x + x = (2 + 1)x = 3x\), they are equivalent.
3. Both give 44. By the distributive property \(4(x + 1) = 4x + 4\), so they are equivalent.

One matching value is not a proof; the property is.

15 Factor out the common factor ★★★

1. The GCF of 12 and 18 is 6: \(12x + 18 = 6(2x + 3)\).
2. The GCF of 20 and 30 is 10: \(20y + 30 = 10(2y + 3)\).
3. The GCF of 35 and 14 is 7: \(35 + 14 = 7(5 + 2) = 7 \cdot 7 = 49\).

16 Phone plan ★★★

1. The expression is \(20 + 4g\).
2. \(g = 0\): 20 dollars. \(g = 3\): \(20 + 12 = 32\) dollars. \(g = 6\): \(20 + 24 = 44\) dollars.
3. The cost grows by 4 dollars per gigabyte. From 44 dollars at \(g = 6\) we need 4 more dollars, so try \(g = 7\): \(20 + 4 \cdot 7 = 20 + 28 = 48\). The plan used 7 extra gigabytes.

17 Expand, then simplify ★★★

Distribute: \(5(x + 2) = 5x + 10\), so the expression becomes \(5x + 10 + 3x - 4\). Combine the like terms: \(5x + 3x = 8x\) and \(10 - 4 = 6\). The simplified expression is \(8x + 6\).

Check with \(x = 3\): the original gives \(5(5) + 9 - 4 = 30\), and \(8(3) + 6 = 30\). The results match.

18 Area and perimeter of a rectangle ★★★

1. Area: \(3(x + 5) = 3x + 15\) square feet. Perimeter: \(2(x + 5) + 2 \cdot 3 = 2x + 10 + 6 = 2x + 16\) feet.
2. For \(x = 4\) the length is 9 feet. Area: \(3 \cdot 9 = 27\) and \(3(4) + 15 = 27\) square feet. Perimeter: \(2 \cdot 9 + 6 = 24\) and \(2(4) + 16 = 24\) feet.

19 Find the errors ★★★

1. He forgot to multiply the 3 by 4. Correct: \(4(x + 3) = 4x + 12\).
2. He multiplied the base by the exponent. Correct: \(2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16\).
3. He combined unlike terms. With \(x = 2\), \(5 + 2x = 9\) but \(7x = 14\). The expression \(5 + 2x\) cannot be simplified.

20 Prove an equivalence ★★★

Table: for \(x = 1\) the first expression is \(2(7) - 2 = 12\) and the second is \(4 + 8 = 12\). For \(x = 2\): \(2(10) - 4 = 16\) and \(8 + 8 = 16\). For \(x = 3\): \(2(13) - 6 = 20\) and \(12 + 8 = 20\).

Reason: \(2(3x + 4) - 2x = 6x + 8 - 2x = 4x + 8\). The expressions are equivalent.

21 Complex phrases ★★★

1. \(3(n + 8) - 5\). For \(n = 4\): \(3 \cdot 12 - 5 = 36 - 5 = 31\).
2. \(m^2 + 2m\). For \(m = 5\): \(25 + 10 = 35\).
3. \(\dfrac{a + 10}{2}\). For \(a = 6\): \(16 \div 2 = 8\).