The Lumenary · Grades 6–8

Variables and algebraic expressions

Name: ______________________

Date: ______________________

You will learn to

Write algebraic expressions for situations and evaluate them for given values, including negative ones
Simplify expressions by combining like terms, and expand and factorise using the distributive law
Prove identities such as $(a + b)^2 = a^2 + 2ab + b^2$ from the distributive law, and refute false ones with a counterexample
Use the laws of exponents $a^m a^n = a^{m+n}$ , $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$ for whole-number exponents, and explain why $a^0 = 1$
Explain why testing values can refute an equivalence but can never prove one on its own

1.
Evaluate $2x^2 - 5x$ when $x = -3$ .
2.
Simplify $4x + 3y - x + 5y - 2$ .
- A. $3x + 8y - 2$
- B. $5x + 8y - 2$
- C. $11xy - 2$
- D. $3x + 8y$
3.
Kofi sells bags of oranges at a market in Kumasi for $12$ cedis each. He pays $15$ cedis a day for his stall. Which expression gives his profit, in cedis, on a day when he sells $x$ bags?
- A. $12x - 15$
- B. $15 - 12x$
- C. $12 + x - 15$
- D. $12x + 15$
4.
Expand $(x + 5)(x - 2)$ and write it as $x^2 + bx + c$ . What is $b$ ?
5.
Which is the complete factorisation of $12x + 18$ ?
- A. $6(2x + 3)$
- B. $3(4x + 6)$
- C. $2(6x + 9)$
- D. $6(2x + 18)$
6.
Which of these are equal to $2^{12}$ ? (Choose all that apply.)
- A. $2^5 \cdot 2^7$
- B. $(2^3)^4$
- C. $2^6 + 2^6$
- D. $4^6$
- E. $2^3 \cdot 2^4$
7.
Zainab tests the rule $(a + b)^2 = a^2 + b^2$ with $a = 0, b = 7$ and with $a = 5, b = 0$ . It works both times. What is the right conclusion?
- A.The rule is true for all numbers.
- B.Her tests were unlucky: the rule holds exactly when $a = 0$ or $b = 0$ , and $a = b = 1$ refutes it.
- C.The rule is true for positive numbers.
- D.Two tests are not enough, but ten passing tests would prove it.
8.
Which Lean statement says that $(a + b)^2 = a^2 + 2ab + b^2$ is an identity for the integers?
- A.∀ a b : Int, (a + b) ^ 2 = a ^ 2 + 2 * a * b + b ^ 2
- B.∃ a b : Int, (a + b) ^ 2 = a ^ 2 + 2 * a * b + b ^ 2
- C.(1 + 1) ^ 2 = 1 ^ 2 + 2 * 1 * 1 + 1 ^ 2
- D.∀ a b : Int, (a + b) ^ 2 = a ^ 2 + b ^ 2
9.
Write find_counterexample(f, g, values) that returns the first v in values with f(v) != g(v), or None if f and g agree on all of them. Then think: if it returns None, have you proved that f and g are equivalent?
Computational — Practice
10.
A row of hexagons is made from craft sticks; neighbouring hexagons share one side. How many sticks are needed for a row of $20$ hexagons? (Find an expression for $n$ hexagons first.)

Answer key — Variables and algebraic expressions

1.
$2 \cdot (-3)^2 - 5 \cdot (-3) = 2 \cdot 9 + 15 = 18 + 15 = 33$ .
2.
Group like terms: $(4x - x) + (3y + 5y) - 2 = 3x + 8y - 2$ . This uses the distributive law: $4x - x = (4 - 1)x$ .
3.
Income is $12x$ cedis and the cost is $15$ cedis, so the profit is $12x - 15$ . For example, selling $10$ bags gives $120 - 15 = 105$ cedis.
4.
$(x + 5)(x - 2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10$ , so $b = 3$ (and $c = -10$ ). Check with $x = 1$ : $(6)(-1) = -6$ and $1 + 3 - 10 = -6$ .
5.
The greatest common factor of $12$ and $18$ is $6$ , so $12x + 18 = 6(2x + 3)$ . Expanding gives back $12x + 18$ .
6.
$2^5 \cdot 2^7 = 2^{12}$ , $(2^3)^4 = 2^{12}$ and $4^6 = (2^2)^6 = 2^{12}$ . But $2^6 + 2^6 = 2^7$ and $2^3 \cdot 2^4 = 2^7$ .
7.
Since $(a + b)^2 = a^2 + 2ab + b^2$ , the rule $(a + b)^2 = a^2 + b^2$ holds exactly when $2ab = 0$ , i.e. when $a = 0$ or $b = 0$ — which both of her tests happened to satisfy. A single counterexample, $a = b = 1$ (giving $4 \neq 2$ ), shows the rule is not an identity.
8.
∀ a b : Int, (a + b) ^ 2 = a ^ 2 + 2 * a * b + b ^ 2 — the universal quantifier ∀ expresses "for every pair of integers". This is the statement of square_of_sum in the lesson (written there with a b as inputs, which means the same).
9.
```
def find_counterexample(f, g, values):
    for v in values:
        if f(v) != g(v):
            return v
    return None
```
Returning None proves nothing about values you did not test: the third test shows x + x(x − 1)(x − 2)(x − 3) and x passing four tests, yet x = 4 is a counterexample. A returned counterexample, on the other hand, does prove the two expressions are not equivalent.
10.
The first hexagon needs $6$ sticks and each further one $5$ , so $n$ hexagons need $6 + 5(n - 1) = 5n + 1$ sticks. (Another way: $6n$ sticks, minus the $n - 1$ shared sides counted twice: $6n - (n - 1) = 5n + 1$ .) For $n = 20$ : $5 \cdot 20 + 1 = 101$ .