Exam practice: Grade 9 Optional Math Unit 1

a.

Define an ordered pair, with an example.

Answer

An ordered pair is two elements written in a fixed order in round brackets, separated by a comma, where the order is part of the meaning. For example, (2, 4) pairs a number with its double; (4, 2) pairs a number with its half, so it is a different pair.

b.

When are two ordered pairs equal? Give an example.

Answer

When their first elements are equal and their second elements are equal. For example (2 + 3, 1) = (5, 4 − 3), since 2 + 3 = 5 and 1 = 4 − 3.

c.

If (a, b) = (m, n), how are a and m related?

Answer

a = m (and b = n).

a.

(x, 4) = (5, y)

Answer

x = 5, y = 4.

b.

(x − 1, y + 2) = (6, 7)

Answer

x − 1 = 6, so x = 7; y + 2 = 7, so y = 5.

c.

(x − 3, y + 7) = (2, −5)

Answer

x − 3 = 2, so x = 5; y + 7 = −5, so y = −12.

d.

(2x − 5, 4) = (9, y + 4)

Answer

2x − 5 = 9, so 2x = 14 and x = 7; 4 = y + 4, so y = 0.

e.

$\left(\frac{x}{3} + 1, y - \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{2}{3}\right)$

Answer

$\begin{aligned} \frac{x}{3} + 1 &= \frac{5}{3} & y - \frac{2}{3} &= \frac{2}{3} \\ \frac{x}{3} &= \frac{2}{3} & y &= \frac{4}{3} \\ x &= 2 \end{aligned}$
x = 2, $y = \frac{4}{3}$ .

f.

(2x + y, 4) = (6, 3x − y)

Answer

2x + y = 6 … (i) and 3x − y = 4 … (ii). Adding: 5x = 10, so x = 2; then from (i), y = 6 − 4 = 2.
Check: (2 × 2 + 2, 4) = (6, 4) and (6, 3 × 2 − 2) = (6, 4) ✓.

a.

For what values of p and q are the ordered pairs (2p + 4, p − q) and (8, q) equal?

Answer

2p + 4 = 8, so p = 2. Then p − q = q, so 2 = 2q and q = 1.
p = 2, q = 1. Check: (8, 1) = (8, 1) ✓.

b.

For what values of m and n are (3m − 5, 2n + m) and (7, m + 2) equal?

Answer

3m − 5 = 7, so 3m = 12 and m = 4. Then 2n + 4 = 4 + 2 = 6, so 2n = 2 and n = 1.
m = 4, n = 1. Check: (7, 6) = (7, 6) ✓.

a.

(1, …)

Answer

y = 2 × 1 + 4 = 6: (1, 6).

b.

(3, …)

Answer

y = 2 × 3 + 4 = 10: (3, 10).

c.

(…, 0)

Answer

0 = 2x + 4, so x = −2: (−2, 0).

d.

(…, −2)

Answer

−2 = 2x + 4, so 2x = −6 and x = −3: (−3, −2).

a.

The ordered pairs (3, p) and (q, 4) satisfy 4x + 3y = 24. Find the ordered pair (p, q).

Answer

(3, p): 12 + 3p = 24, so p = 4. (q, 4): 4q + 12 = 24, so q = 3.
(p, q) = (4, 3).

b.

The ordered pairs (3, m) and (n, 7) satisfy 7x − 3y = 21. Find m and n.

Answer

(3, m): 21 − 3m = 21, so m = 0. (n, 7): 7n − 21 = 21, so 7n = 42 and n = 6.

a.

What is a Cartesian product? Give an example.

Answer

The Cartesian product A × B of two non-empty sets is the set of all ordered pairs whose first element is from A and second element from B. For example, if A = {1, 2} and B = {x, y}, then A × B = {(1, x), (1, y), (2, x), (2, y)}.

b.

The cardinal numbers of sets A and B are 3 and 2. What is the cardinal number of A × B?

Answer

n(A × B) = 3 × 2 = 6.

c.

If n(A × B) = x, n(A) = y and n(B) = z, how are x, y and z related?

Answer

x = yz, since n(A × B) = n(A) × n(B).

a.

If A = {2, 3} and B = {7}, find A × B and show it on a graph and in a mapping diagram.

Answer

A × B = {(2, 7), (3, 7)}.

Figure 1.6 Both elements of A go to 7.

Figure 1.7 The two points (2, 7) and (3, 7), side by side at height 7.

b.

If A = {2, 3} and B = {4, 5, 6}, find B × A and show it in a table and a mapping diagram.

Answer

B × A = {(4, 2), (4, 3), (5, 2), (5, 3), (6, 2), (6, 3)}.

×	2	3
4	(4, 2)	(4, 3)
5	(5, 2)	(5, 3)
6	(6, 2)	(6, 3)

Figure 1.8 Every element of B to every element of A: 6 arrows.

a.

If A × B = {(a, 1), (a, 5), (a, 2), (b, 2), (b, 1), (b, 5)}, find A, B, n(A), n(B), B × A and n(B × A).

Answer

A = {a, b}, B = {1, 2, 5}, n(A) = 2, n(B) = 3.
B × A = {(1, a), (1, b), (2, a), (2, b), (5, a), (5, b)}, and n(B × A) = 6.

b.

If A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}, find A, B, n(A), n(B), B × A and n(B × A).

Answer

A = {1, 2, 3}, B = {4, 5, 6}, n(A) = n(B) = 3.
B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}, and n(B × A) = 3 × 3 = 9.

a.

If A = {x:x < 4, x ∈ N} and $B = \{x : x^2 - 5x + 6 = 0, \ x \in N\}$ , find A × B and B × A, and show them in a table, a mapping diagram and a tree diagram.

Answer

A = {1, 2, 3} (natural numbers less than 4). $x^2 - 5x + 6 = (x - 2)(x - 3) = 0$ gives B = {2, 3}.
A × B = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

Table of A × B (B × A is the same table with each pair turned round):

×	2	3
1	(1, 2)	(1, 3)
2	(2, 2)	(2, 3)
3	(3, 2)	(3, 3)

Figure 1.9 A × B: each of 1, 2, 3 to each of 2, 3.

Tree of A × B: 1 → 2, 3 gives (1, 2), (1, 3); 2 → 2, 3 gives (2, 2), (2, 3); 3 → 2, 3 gives (3, 2), (3, 3). For B × A: 2 → 1, 2, 3 and 3 → 1, 2, 3.

b.

If M = {x:x ≤ 3, x ∈ N} and N = {y:y = x − 1, x ∈ M}, find M × N and show it on a graph.

Answer

M = {1, 2, 3}, and N = {1 − 1, 2 − 1, 3 − 1} = {0, 1, 2}.
M × N = {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}: 9 pairs.

Figure 1.10 M × N: a 3 by 3 block of points.

a.

Define a relation, with an example.

Answer

A relation from set A to set B is a subset of A × B. For example, with A = {1, 2} and B = {2, 3}, “is less than” is {(1, 2), (1, 3), (2, 3)} ⊆ A × B.

b.

What are the ways of showing a relation?

Answer

As a set of ordered pairs, in set-builder form, in a table, in a mapping (arrow) diagram, and on a graph.

c.

Define, with examples: i. a symmetric relation, ii. a reflexive relation, iii. a transitive relation, iv. an equivalence relation.

Answer

i.Symmetric: whenever (x, y) ∈ R, also (y, x) ∈ R. Example: in {1, 2}, {(1, 2), (2, 1)}; among lines, “is parallel to”.
ii.Reflexive: (x, x) ∈ R for every x in the set. Example: in {1, 2, 3}, {(1, 1), (2, 2), (3, 3)}; among numbers, “is equal to”.
iii.Transitive: whenever (x, y) ∈ R and (y, z) ∈ R, also (x, z) ∈ R. Example: {(1, 2), (2, 3), (1, 3)}; among numbers, “is less than”.
iv.Equivalence: reflexive, symmetric and transitive together. Example: “is congruent to” among triangles; in {1, 2}, the relation {1, 2} × {1, 2}.

d.

In the set {a, b}, R = {(a, a), (b, b)}. What type of relation is R?

Answer

Reflexive: a and b are both related to themselves. It is also symmetric (each pair is its own partner) and transitive (the only chains give back (a, a) and (b, b)), so it is in fact an equivalence relation: “is equal to”.

e.

If R = {(a, b)}, write a symmetric relation that contains R.

Answer

Add the pair turned round: {(a, b), (b, a)}. This is the smallest symmetric relation containing R.

a.

R₁ = {(x, y):x + y = 6}

Answer

R₁ = {(1, 5), (2, 4)}.

Figure 1.14 3 has no partner: 3 + y = 6 needs y = 3, which isn’t in B.

b.

R₂ = {(x, y):x < y}

Answer

Every x (1, 2 or 3) is less than every y (4, 5 or 6), so R₂ = A × B, all nine pairs.

Figure 1.15 Nine arrows: every pair is in R₂.

c.

$R_3 = \{(x, y) : y = x^2\}$

Answer

$1^2 = 1 \notin B$ , $2^2 = 4 \in B$ , $3^2 = 9 \notin B$ : R₃ = {(2, 4)}.

Figure 1.16 Only 2 has its square in B.

a.

is greater than

Answer

A × B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}.
{(x, y):x > y} = {(3, 2)}.

b.

is equal to

Answer

{(x, y):x = y} = {(2, 2), (3, 3)}.

c.

is double of

Answer

{(x, y):x = 2y}: the smallest 2y is 4, more than any x. So it is $\varnothing$ , the empty relation.

d.

is half of

Answer

$\{(x, y) : x = \frac{y}{2}\} =$ {(1, 2), (2, 4)}.

e.

is the square of

Answer

$\{(x, y) : x = y^2\}$ : every $y^2$ is 4 or more, so it is $\varnothing$ .

i.

R₁ = {(x, y):x + y < 12, x ∈ A, y ∈ B}

Answer

a. R₁ = {(6, 2), (6, 4), (7, 2), (7, 4), (8, 2)} (sums 8, 10, 9, 11, 10; every other pair adds to 12 or more).
b.

Figure 1.17 10 has no arrow: 10 + 2 = 12 is already too big.

c.

Figure 1.18 R₁ as points.

d.

x	6	6	7	7	8
y	2	4	2	4	2

ii.

R₂ = {(x, y):2x + y > 10, x ∈ A, y ∈ B}

Answer

a. The smallest 2x + y is 2 × 6 + 2 = 14 > 10, so every pair is in: R₂ = A × B = {(6, 2), (6, 4), (6, 6), (7, 2), (7, 4), (7, 6), (8, 2), (8, 4), (8, 6), (10, 2), (10, 4), (10, 6)}, 12 pairs.
b.

Figure 1.19 Every element of A is joined to every element of B.

c.

Figure 1.20 R₂ as points: all 12 pairs.

d.

x	6	6	6	7	7	7
y	2	4	6	2	4	6

x	8	8	8	10	10	10
y	2	4	6	2	4	6

5.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is a relation in A = {1, 2, 3}. Is it an equivalence relation?

Answer

Reflexive: (1, 1), (2, 2), (3, 3) ∈ R. ✓

Symmetric: (1, 2) ∈ R but (2, 1) ∉ R. ✗

Transitive: (1, 2) and (2, 3) are in R, but (1, 3) ∉ R. ✗

So R is not an equivalence relation: it is reflexive only.

6.

If A = {4, 5, 6}, find a. A × A, b. a reflexive relation R₁ in A, c. a symmetric relation R₂ in A, d. a transitive relation R₃ in A.

Answer

a.A × A = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.
b.For example R₁ = {(4, 4), (5, 5), (6, 6)}: it must contain these three; it may contain more.
c.For example R₂ = {(4, 5), (5, 4), (5, 6), (6, 5)}: every pair with its partner.
d.For example R₃ = {(4, 5), (5, 6), (4, 6)}: the chain 4 → 5 → 6 has its shortcut (4, 6).

Many other answers are right; check yours against the definitions.

a.

Define the domain and the range of a relation, with examples.

Answer

The domain is the set of first elements of the relation’s ordered pairs, the range the set of second elements. For R = {(1, 4), (2, 5), (3, 5)}, the domain is {1, 2, 3} and the range {4, 5}.

b.

The relation R is given in a table. Write its domain and range.

x	1	3	5
y	5	7	9

Answer

R = {(1, 5), (3, 7), (5, 9)}: domain = {1, 3, 5}, range = {5, 7, 9}.

c.

If R = {(a, b), (c, d)}, find its domain and range.

Answer

Domain = {a, c}, range = {b, d}.

a.

R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}

Answer

Domain = {1, 2, 3}, range = {3, 5}.

b.

R = {(2, 4), (2, 6), (3, 6), (3, 9), (4, 8), (4, 12)}

Answer

Domain = {2, 3, 4}, range = {4, 6, 8, 9, 12} (6 appears twice but is listed once).

c.

R = {(3, −1), (4, −2), (5, −3), (6, −4)}

Answer

Domain = {3, 4, 5, 6}, range = {−4, −3, −2, −1}.

d.

R = {(4, −2), (4, 2), (1, −1), (1, 1)}

Answer

Domain = {1, 4}, range = {−2, −1, 1, 2}. (This is “y is a square root of x”.)

e.

R = {(5, 8), (6, 9), (7, 10), (8, 11)}

Answer

Domain = {5, 6, 7, 8}, range = {8, 9, 10, 11}.

a.

y = 2x, x ∈ {0, 1, 5}

Answer

2 × 0 = 0, 2 × 1 = 2, 2 × 5 = 10: range = {0, 2, 10}.

b.

y = 2x + 3, x ∈ {−1, 1, 2}

Answer

−2 + 3 = 1, 2 + 3 = 5, 4 + 3 = 7: range = {1, 5, 7}.

c.

$y = 3x^2 - 4$ , −2 < x ≤ 3, x ∈ W

Answer

x is a whole number from −2 (left out) to 3, so x = 0, 1, 2, 3: y = −4, −1, 8, 23.
Range = {−4, −1, 8, 23}.

4.

The figure shows a relation R from A to B.

Figure 1.23

a. Write R as a set of ordered pairs. b. Write its domain and range.

Answer

a.R = {(1, 5), (2, 6), (3, 6), (4, 7)}.
b.Domain = {1, 2, 3, 4}, range = {5, 6, 7}.

a.

R = {(x, y):y = 2x + 1, 0 ≤ x ≤ 3, x ∈ W}

Answer

x = 0, 1, 2, 3: R = {(0, 1), (1, 3), (2, 5), (3, 7)}; domain = {0, 1, 2, 3}, range = {1, 3, 5, 7}.

Figure 1.24

b.

$R = \{(x, y) : y = x^2 - 2, \ 0 \le x \le 3, \ x \in W\}$

Answer

R = {(0, −2), (1, −1), (2, 2), (3, 7)}; domain = {0, 1, 2, 3}, range = {−2, −1, 2, 7}.

Figure 1.25

c.

$R = \{(x, y) : y = 3x^2 - 2x - 1, \ 1 \le x \le 4, \ x \in W\}$

Answer

x = 1: 3 − 2 − 1 = 0; x = 2: 12 − 4 − 1 = 7; x = 3: 27 − 6 − 1 = 20; x = 4: 48 − 8 − 1 = 39.
R = {(1, 0), (2, 7), (3, 20), (4, 39)}; domain = {1, 2, 3, 4}, range = {0, 7, 20, 39}.

Figure 1.26

d.

$R = \{(x, y) : y = x^2 - x, \ 0 \le x \le 3, \ x \in W\}$

Answer

y = 0, 0, 2, 6: R = {(0, 0), (1, 0), (2, 2), (3, 6)}; domain = {0, 1, 2, 3}, range = {0, 2, 6} (0 is the image of both 0 and 1).

Figure 1.27

a.

What is an inverse relation? Give an example.

Answer

The inverse of a relation R is the relation $R^{-1}$ made by swapping the two elements of each of its ordered pairs. For example, if R = {(1, 2), (3, 6)} (“is half of”), then $R^{-1} = \{(2, 1), (6, 3)\}$ (“is double of”).

b.

If R = {(a, b), (c, d)}, write $R^{-1}$ .

Answer

$R^{-1} = \{(b, a), (d, c)\}$ .

a.

R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}

Answer

$R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (3, 3), (5, 3)\}$ .

b.

R = {(1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6)}

Answer

$R^{-1} = \{(3, 1), (6, 1), (3, 2), (6, 2), (3, 3), (6, 3)\}$ .

c.

R = {(3, −1), (4, −2), (5, −3), (6, −4)}

Answer

$R^{-1} = \{(-1, 3), (-2, 4), (-3, 5), (-4, 6)\}$ .

d.

R = {(8, 6), (7, 5), (6, 4), (5, 3), (4, 2), (3, 1)}

Answer

$R^{-1} = \{(6, 8), (5, 7), (4, 6), (3, 5), (2, 4), (1, 3)\}$ .

a.

R = {(x, y):y = 3x, x ∈ {1, 2, 3}}

Answer

R = {(1, 3), (2, 6), (3, 9)}, $R^{-1} = \{(3, 1), (6, 2), (9, 3)\}$ .

b.

R = {(x, y):y = 2x + 3, x ∈ {1, 2, 3}}

Answer

R = {(1, 5), (2, 7), (3, 9)}, $R^{-1} = \{(5, 1), (7, 2), (9, 3)\}$ .

c.

R = {(x, y):y = x − 2, x ∈ {6, 7, 8}}

Answer

R = {(6, 4), (7, 5), (8, 6)}, $R^{-1} = \{(4, 6), (5, 7), (6, 8)\}$ .

d.

$R = \{(x, y) : y = x^2, \ x \in \{0, 1, -1\}\}$

Answer

R = {(0, 0), (1, 1), (−1, 1)}, $R^{-1} = \{(0, 0), (1, 1), (1, -1)\}$ .

4.

The figure shows the relation $R^{-1}$ from B to A.

Figure 1.30

a.Write $R^{-1}$ as ordered pairs, with its domain and range.
b.Write the domain and range of R.
c.Write $R^{-1}$ and R in set-builder form.

Answer

a. $R^{-1} = \{(1, 3), (2, 6), (3, 9)\}$ ; domain {1, 2, 3}, range {3, 6, 9}.
b.R = {(3, 1), (6, 2), (9, 3)}: domain = {3, 6, 9}, range = {1, 2, 3}.
c. $R^{-1} = \{(x, y) : y = 3x\}$ and $R = \{(x, y) : y = \frac{x}{3}\}$ .

5.

In the set A = {2, 3, 4, 5, 6, 7, 8, 9}, R = {(x, y):y is a multiple of x}. Write R as ordered pairs, and find a. its domain and range, b. $R^{-1}$ , c. the domain and range of $R^{-1}$ .

Answer

R = {(2, 2), (2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (3, 9), (4, 4), (4, 8), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)}. (Every number is a multiple of itself, 1 × x.)

a.Domain = range = {2, 3, 4, 5, 6, 7, 8, 9}.
b. $R^{-1} = \{(2, 2), (4, 2), (6, 2), (8, 2), (3, 3), (6, 3), (9, 3), (4, 4), (8, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)\}$ , the relation “y is a factor of x”.
c.Domain = range = {2, 3, 4, 5, 6, 7, 8, 9}.

6.

In A = {2, 3, 4, 5}, R = {(x, y):x is a factor of y}. Write R as ordered pairs and find a. its domain, b. its range, c. $R^{-1}$ , shown in a mapping diagram.

Answer

R = {(2, 2), (2, 4), (3, 3), (4, 4), (5, 5)}.

a.Domain = {2, 3, 4, 5}. b. Range = {2, 3, 4, 5}.
c. $R^{-1} = \{(2, 2), (4, 2), (3, 3), (4, 4), (5, 5)\}$ .

Figure 1.31 4 has two arrows: it is a multiple of 2 and of 4.

a.

{(x, y):y = x − 1, 4 ≤ x ≤ 7, x ∈ N}

Answer

R = {(4, 3), (5, 4), (6, 5), (7, 6)}, so $R^{-1} = \{(3, 4), (4, 5), (5, 6), (6, 7)\}$ , domain {3, 4, 5, 6}, range {4, 5, 6, 7}.

Figure 1.32

b.

$\{(x, y) : y = x^2 - 1, \ 0 \le x \le 3, \ x \in W\}$

Answer

R = {(0, −1), (1, 0), (2, 3), (3, 8)}, so $R^{-1} = \{(-1, 0), (0, 1), (3, 2), (8, 3)\}$ , domain {−1, 0, 3, 8}, range {0, 1, 2, 3} (the mapping diagram is the one in Example 2).

c.

$\{(x, y) : y = 3x^2 - 2x - 1, \ 1 \le x \le 4, \ x \in W\}$

Answer

R = {(1, 0), (2, 7), (3, 20), (4, 39)}, so $R^{-1} = \{(0, 1), (7, 2), (20, 3), (39, 4)\}$ , domain {0, 7, 20, 39}, range {1, 2, 3, 4}.

Figure 1.33

1.

Define a function, with an example from daily life.

Answer

A function from A to B is a relation in which every element of A has exactly one image in B. Example: each student of a class has exactly one roll number, so “roll number of” is a function from the students to the numbers. Another: the price of rice at Rs 90 per kg is y = 90x for x kg.

a.

What does y = f(t) mean?

Answer

y is a function of t: y depends on t, with exactly one y for each t.

b.

D is the depth below the surface of water and P the pressure there. What does P = f(D) mean?

Answer

The pressure P is a function of the depth D: for each depth there is exactly one pressure, and it depends on the depth. (In Grade 9 Science, $P = h \rho g$ : pressure grows in step with depth.)

c.

What does y = f(x) mean?

Answer

y is a function of x: each value of x gives exactly one value of y, by the rule f.

a.

Write the perimeter P of a square as a function of its area A.

Answer

$A = l^2$ , so $l = \sqrt{A}$ , and P = 4l: $P = f(A) = 4\sqrt{A}$ .

b.

Write the area A of a circle as a function of its circumference C.

Answer

$C = 2\pi r$ , so $r = \frac{C}{2\pi}$ . Then
$A = \pi r^2 = \pi \times \frac{C^2}{4\pi^2} = \frac{C^2}{4\pi}$
$A = f(C) = \frac{C^2}{4\pi}$ .

c.

Write the volume V of a cube as a function of its surface area A.

Answer

A cube with edge l has $A = 6l^2$ , so $l = \sqrt{\frac{A}{6}}$ . Then
$V = l^3 = \left(\sqrt{\frac{A}{6}}\right)^3 = \frac{A}{6}\sqrt{\frac{A}{6}} = \sqrt{\frac{A^3}{216}}$
$V = f(A) = \sqrt{\frac{A^3}{216}}$ . Check: a 1 cm cube has A = 6, and $\sqrt{216/216} = 1$ ✓.

a.

Figure 1.43

Answer

A function: 1, 2 and 3 each have exactly one image.

b.

Figure 1.44

Answer

A function: each element of A has one image. Two elements sharing an image, and 9 having none, are both allowed.

c.

Figure 1.45

Answer

Not a function: 3 has no image.

d.

Figure 1.46

Answer

Not a function: 5 has two images, 6 and −1.

e.

Figure 1.47 R₅

Answer

R₅ = {(1, 1), (2, 1), (2, 4), (4, 2), (6, 3)}. Not a function: 2 has two images, 1 and 4 (the points (2, 1) and (2, 4) are one above the other).

f.

Figure 1.48 R₆

Answer

R₆ = {(1, 1), (2, 2), (3, 3), (4, 4)}. A function: no two points share an x.

g.

R₇ = {(2, 4), (3, 6), (4, 8)}

Answer

A function: the first elements 2, 3, 4 are all different.

h.

R₈ = {(3, 6), (2, 4), (3, 9), (4, 16)}

Answer

Not a function: 3 has two images, 6 and 9.

i.

R₉:

x	3	4	5
y	2	3	4

Answer

A function: 3, 4 and 5 each have one image.

j.

$R_{10}$ :

x	2	1	2
y	7	6	3

Answer

Not a function: 2 has two images, 7 and 3.

5.

Write each function of Question 4 as a formula.

Answer

R₁: y = x + 2. R₂: $y = x^2$ . R₆: y = x. R₇: y = 2x. R₉: y = x − 1.

6.

Which of these graphs are graphs of functions? Give reasons.

Figure 1.49

Answer

Figure 1.50 ✓: every vertical line meets the graph at most once. ✗: the dashed line meets it more than once.

Functions: a, b, d, e, j and l: no vertical line meets them more than once. The path in (l) has corners and a bump, but it never turns back, so it still passes.
Not functions: c, f, g, h, i and k: some vertical line meets each twice (or, for k, three times). The sideways parabola (c), the two curves (f) and the ovals (g, h) have a point above and below the same x; the crossing lines (i) do everywhere except at the crossing; the S-curve (k) doubles back.

a.

f₁ = {(1, 1), (2, 4), (3, 9)}

Answer

x	1	2	3
y	1	4	9

$f_1(x) = x^2$ .

Figure 1.51

b.

f₂ = {(1, 5), (2, 7), (3, 9)}

Answer

x	1	2	3
y	5	7	9

f₂(x) = 2x + 3: each step of 1 in x adds 2 to y, and 2 × 1 + 3 = 5.

Figure 1.52

c.

f₃ = {(3, 1), (5, 3), (7, 5)}

Answer

x	3	5	7
y	1	3	5

f₃(x) = x − 2.

Figure 1.53

d.

f₄ = {(0, 3), (1, 4), (2, 5)}

Answer

x	0	1	2
y	3	4	5

f₄(x) = x + 3.

Figure 1.54

a.

Explain the domain, co-domain and range of a function, with an example.

Answer

For f:A → B, the domain is A, the co-domain is B, and the range is the set of images actually reached. Example: f:{1, 2} → {3, 4, 5}, f(x) = x + 2: domain {1, 2}, co-domain {3, 4, 5}, range {3, 4}.

b.

Explain image and pre-image, with an example.

Answer

If f(x) = y, y is the image of x and x a pre-image of y. For $f(x) = x^2$ , f(3) = 9: 9 is the image of 3, and 3 is a pre-image of 9 (so is −3, if −3 is in the domain).

2.

The table shows a function f.

x	1	2	3	4
y	1	4	9	16

a. Write f as a set of ordered pairs. b. Write its domain and range. c. What is the pre-image of 9? d. Write f as a formula.

Answer

a.f = {(1, 1), (2, 4), (3, 9), (4, 16)}.
b.Domain = {1, 2, 3, 4}, range = {1, 4, 9, 16}.
c.3, since f(3) = 9.
d. $f(x) = x^2$ .

3.

Answer from the mapping diagram.

Figure 1.57

a. Is g a function? Why? b. What is the pre-image of 8? c. Write the domain, co-domain and range of g. d. Find g(5) + g(7). e. Write g as a formula.

Answer

a.Yes: 3, 5 and 7 each have exactly one image.
b.5.
c.Domain = {3, 5, 7}, co-domain = {6, 8, 10, 13}, range = {6, 8, 10}.
d.g(5) + g(7) = 8 + 10 = 18.
e.Each image is 3 more: g(x) = x + 3.

a.

f = {(2, 4), (3, 6), (4, 8)}

Answer

{4, 6, 8}.

b.

g = {(x, y):y = x − 3, 2 < x < 5, x ∈ N}

Answer

x = 3, 4, so y = 0, 1: range = {0, 1}.

c.

$h(x) = x^2 - 2x$ , domain = {0, 1, 2}

Answer

h(0) = 0, h(1) = 1 − 2 = −1, h(2) = 4 − 4 = 0: range = {−1, 0}.

a.

f(x) = 3x + 5, range = {2, 8}

Answer

3x + 5 = 2 gives x = −1; 3x + 5 = 8 gives x = 1. Domain = {−1, 1}.

b.

g(x) = 4x − 5, range = {−1, 7}

Answer

4x − 5 = −1 gives x = 1; 4x − 5 = 7 gives x = 3. Domain = {1, 3}.

c.

$h(x) = x^2$ , range = {9, 16}

Answer

$x^2 = 9$ gives x = ±3; $x^2 = 16$ gives x = ±4. The largest possible domain is {−4, −3, 3, 4}. (Any domain that reaches both 9 and 16 gives this range, {3, 4} for example, but these four are all the pre-images.)

d.

The images of g(x) = 2x + 5 are {7, 11, 15}. Find the set of pre-images.

Answer

2x + 5 = 7, 11, 15 gives x = 1, 3, 5: {1, 3, 5}.

e.

The images of f(x) = 4x − 3 are {1, 5, 9}. Find the set of pre-images.

Answer

4x − 3 = 1, 5, 9 gives x = 1, 2, 3: {1, 2, 3}.

a.

For which element of the domain of $f(x) = \frac{2x - 3}{5}$ is the image 7?

Answer

$\frac{2x - 3}{5} = 7$ , so 2x − 3 = 35, 2x = 38: x = 19.

b.

For which element of the domain of f(x) = 3x + 5 is the image 8?

Answer

3x + 5 = 8, so x = 1.

c.

Which element of the domain has the image $\frac{1}{2}$ under $f(x) = \frac{2}{x - 4}$ ?

Answer

$\frac{2}{x - 4} = \frac{1}{2}$ , so x − 4 = 4: x = 8. Check: $\frac{2}{8 - 4} = \frac{1}{2}$ ✓.

d.

Which element of the domain has the image 9 under f(x) = 2x − 5?

Answer

2x − 5 = 9, so x = 7.

7.

A function f:A → Z, where A = {x:0 ≤ x ≤ 4, x ∈ W}, is f(x) = 2x − 3.

a.Write the domain.
b.Find the range.
c.Draw its arrow diagram and say what type of function it is.

The textbook gives the co-domain as W, the whole numbers, but f(0) = −3 and f(1) = −1 are not whole numbers, so f would not be a function into W. So the book’s numbers don’t work; here the co-domain is Z, the integers.

Answer

a.Domain = {0, 1, 2, 3, 4}.
b.f(0) = −3, f(1) = −1, f(2) = 1, f(3) = 3, f(4) = 5: range = {−3, −1, 1, 3, 5}.

c.

Figure 1.58 Only the integers that are reached are drawn; Z has infinitely many more.

Each element has a different image, so it is one-to-one; most integers (0, 2, 4, …) have no pre-image, so it is into: a one-to-one into function (the types come next).

1.

Define, with examples: a. an onto function, b. an into function, c. a linear function, d. a constant function.

Answer

a.Onto: every element of the co-domain has a pre-image (range = co-domain). Example: {(1, a), (2, b), (3, b)} from {1, 2, 3} to {a, b}.
b.Into: some element of the co-domain has no pre-image. Example: {(1, a), (2, a)} from {1, 2} to {a, b}.
c.Linear: f(x) = mx + c, whose graph is a straight line. Example: f(x) = 2x + 1.
d.Constant: every element has the same image, f(x) = c. Example: f(x) = 5.

a.

Figure 1.68

Answer

One-to-one onto: different elements have different images, and every element of B is reached.

b.

Figure 1.69

Answer

One-to-one onto: a, b, c go to different images, and 1, 2, 3 are all reached.

c.

Figure 1.70

Answer

Many-to-one into: 1 and 2 share a, and c is never reached.

d.

Figure 1.71

Answer

Many-to-one into (in fact constant): 1 and 2 share a, and b, c are never reached.

e.

Figure 1.72

Answer

Many-to-one into: 1 and 3 share a, and b is never reached.

f.

Figure 1.73

Answer

Constant (many-to-one onto): everything goes to d, the only element of B.

3.

State the type of function each graph shows.

Figure 1.74 (a)

Figure 1.75 (b)

Figure 1.76 (c)

Answer

a. Linear (a straight line going up: slope positive). b. Linear (going down: slope negative). c. Constant (horizontal: the same y for every x).
Over the real numbers, (a) and (b) are also one-to-one onto, and (c) is many-to-one.

4.

A = {1, 3} and B = {4, 6, 8}. Make functions from A to B that are a. one-to-one, b. one-to-one into, c. many-to-one into.

Answer

For example:

a.{(1, 4), (3, 6)}.
b.{(1, 6), (3, 8)}: 4 isn’t reached. (With 2 elements going into 3, a one-to-one function is always into.)
c.{(1, 8), (3, 8)}: both go to 8; 4 and 6 aren’t reached.

5.

P = {3, 4, 5} and Q = {4, 5, 6}. Make these functions:

a.a one-to-one onto function from P to Q
b.a constant function from P to Q
c.a many-to-one into function from P to Q
d.the identity function on P

The textbook gives d. as the identity function from P to Q, but the identity function sends each element to itself, and 3 is not in Q. So the book’s numbers don’t work; here it is the identity function on P.

Answer

For example:

a.{(3, 4), (4, 5), (5, 6)}, that is f(x) = x + 1.
b.{(3, 5), (4, 5), (5, 5)}.
c.{(3, 4), (4, 4), (5, 6)}: 3 and 4 share 4, and 5 isn’t reached.
d.{(3, 3), (4, 4), (5, 5)}, from P to P.

a.

f = {(1, 2), (2, 4), (3, 6)}

Answer

Figure 1.77

One-to-one onto (taking the co-domain as {2, 4, 6}).

b.

g = {(1, 1), (−1, 1), (2, 4), (3, 9)}

Answer

Figure 1.78

Many-to-one onto: 1 and −1 share 1 (co-domain {1, 4, 9}).

c.

h = {(1, 0), (2, 0), (3, 0)}

Answer

Figure 1.79

Constant: every element goes to 0.

a.

If f(x) = 4x + 5, find f(2), f(3), f(5).

Answer

8 + 5 = 13, 12 + 5 = 17, 20 + 5 = 25: 13, 17, 25.

b.

If $f(x) = 2x^2 - 1$ , find f(−1), f(0), f(2).

Answer

2 − 1 = 1, 0 − 1 = −1, 8 − 1 = 7: 1, −1, 7.

c.

If $h(x) = 3x^2 + 2x - 1$ , find h(0), h(2), h(4).

Answer

h(0) = −1; h(2) = 12 + 4 − 1 = 15; h(4) = 48 + 8 − 1 = 55: −1, 15, 55.

d.

If $g(x) = x^3 - 2$ , find g(1), g(−1), g(2), g(−2).

Answer

1 − 2 = −1, −1 − 2 = −3, 8 − 2 = 6, −8 − 2 = −10: −1, −3, 6, −10.

e.

If $f(x) = \frac{x^2 + 3x - 2}{x - 2}$ , x ≠ 2, find f(4) − f(1) + f(0).

Answer

$f(4) = \frac{16 + 12 - 2}{2} = 13, \quad f(1) = \frac{1 + 3 - 2}{-1} = -2, \quad f(0) = \frac{-2}{-2} = 1$
f(4) − f(1) + f(0) = 13 + 2 + 1 = 16.

a.

If f(2x + 3) = 2x − 1, find f(x) and f(−2).

Answer

2x − 1 = (2x + 3) − 4, so f(x) = x − 4 and f(−2) = −6.

b.

If f(3x − 4) = 6x − 5, find f(x) and f(2).

Answer

6x − 5 = 2(3x − 4) + 3, so f(x) = 2x + 3 and f(2) = 7.

c.

If f(3x − 5) = 6x − 13, find f(5x) and f(−3).

Answer

6x − 13 = 2(3x − 5) − 3, so f(x) = 2x − 3. Then f(5x) = 10x − 3 and f(−3) = −9.

d.

If f(3x + 2) = 12x − 5, find f(x) and f(6).

Answer

12x − 5 = 4(3x + 2) − 13, so f(x) = 4x − 13 and f(6) = 24 − 13 = 11.
Check: f(6) needs 3x + 2 = 6, $x = \frac{4}{3}$ , and $12 \times \frac{4}{3} - 5 = 16 - 5 = 11$ ✓.

a.

If $f(y) = a^y$ , prove that f(a − b) = f(a) ÷ f(b).

Answer

$\text{L.H.S.} = f(a - b) = a^{a - b} = \frac{a^a}{a^b} = f(a) \div f(b) = \text{R.H.S.}$
Hence proved.

b.

If $f(x) = 5^x$ , prove that f(a + b) = f(a) × f(b).

Answer

$\text{L.H.S.} = f(a + b) = 5^{a + b} = 5^a \times 5^b = f(a) \times f(b) = \text{R.H.S.}$
Hence proved.

c.

If $f(m) = 7^m$ , prove that $f(xy) = (f(y))^x$ .

Answer

$\text{L.H.S.} = f(xy) = 7^{xy} = (7^y)^x = (f(y))^x = \text{R.H.S.}$
Hence proved.

a.

f:R → R satisfies f(5x) = f(5) + f(x). Find f(1), f(5) and f(25).

Answer

x = 0: f(0) = f(5) + f(0), so f(5) = 0.
x = 1: f(5) = f(5) + f(1), so f(1) = 0.
x = 5: f(25) = f(5) + f(5), so f(25) = 0.

b.

If f(x + 3) = f(3) + f(x), prove that f(0) = 0 and f(−3) = −f(3).

Answer

x = 0: f(3) = f(3) + f(0), so f(0) = 0.
x = −3: f(0) = f(3) + f(−3), so 0 = f(3) + f(−3) and f(−3) = −f(3). Hence proved.

c.

If f(x + 7) = f(10) + f(x), prove that f(3) = 0 and f(−4) = −f(10).

Answer

x = 3: f(10) = f(10) + f(3), so f(3) = 0.
x = −4: f(3) = f(10) + f(−4), so 0 = f(10) + f(−4) and f(−4) = −f(10). Hence proved.

d.

If f(x + a) = f(x) + f(a) for a ∈ R, prove that f(0) = 0 and f(−a) = −f(a).

Answer

x = 0: f(a) = f(0) + f(a), so f(0) = 0.
x = −a: f(0) = f(−a) + f(a), so 0 = f(−a) + f(a) and f(−a) = −f(a). Hence proved.

a.

For which values of x are $f(x) = 2x^2 - 1$ and g(x) = 1 − 3x equal?

Answer

$\begin{aligned} 2x^2 - 1 &= 1 - 3x \\ 2x^2 + 3x - 2 &= 0 \\ (2x - 1)(x + 2) &= 0 \end{aligned}$
$x = \frac{1}{2}$ or x = −2: the domain is $\left\{-2, \frac{1}{2}\right\}$ . Check x = −2: f = 7, g = 7 ✓.

b.

$f(x) = 3x^2 + 4x - 5$ and $g(x) = 2x^2 - 3x - 17$ . Find the values of x for which f(x) = g(x).

Answer

$\begin{aligned} 3x^2 + 4x - 5 &= 2x^2 - 3x - 17 \\ x^2 + 7x + 12 &= 0 \\ (x + 3)(x + 4) &= 0 \end{aligned}$
x = −3 or x = −4. Check x = −3: f = 27 − 12 − 5 = 10, g = 18 + 9 − 17 = 10 ✓.

c.

$h(t) = 5t^2 - 3t + 4$ and $g(t) = t^2 + 4t + 1$ . If h(t) = g(t), find t.

Answer

$\begin{aligned} 5t^2 - 3t + 4 &= t^2 + 4t + 1 \\ 4t^2 - 7t + 3 &= 0 \\ (4t - 3)(t - 1) &= 0 \end{aligned}$
t = 1 or $t = \frac{3}{4}$ .

a.

f(x) = ax + b, with f(2) = 2 and f(−3) = −13. i. Find a and b. ii. Write f(x). iii. Find $f\left(\frac{1}{3}\right)$ .

Answer

i.2a + b = 2 … (i), −3a + b = −13 … (ii). Subtracting: 5a = 15, so a = 3, and b = −4.
ii.f(x) = 3x − 4.
iii. $f\left(\frac{1}{3}\right) = 1 - 4 =$ −3.

b.

f(x) = mx + c, with f(3) = 9 and f(5) = 13. Find m and c.

Answer

3m + c = 9 and 5m + c = 13. Subtracting: 2m = 4, so m = 2, and c = 3: f(x) = 2x + 3.

a.

R = {(1, 2), (3, 5), (5, 7), (7, 9)}

Answer

A function: the first elements 1, 3, 5, 7 are all different.

b.

R = {(1, 2), (2, 3), (3, 5), (3, 6)}

Answer

Not a function: 3 has two images, 5 and 6.

c.

R = {(1, 3), (1, 5), (4, 3)}

Answer

Not a function: 1 has two images, 3 and 5.

a.

Figure 1.85

Answer

A function: each element of A has one image (1 and −1 sharing 1 is allowed).

b.

Figure 1.86

Answer

R = {(1, 2), (2, 1), (3, 3), (4, 1)}: a function, no two points share an x.

c.

Figure 1.87

Answer

R = {(1, 3), (2, 1), (3, 3), (3, 2), (4, 1)}: not a function, 3 has two images, 3 and 2.

d.

x	1	2	3
y	1	4	9

Answer

A function: 1, 2, 3 each have one image ( $y = x^2$ ).

e.

x	1	1	2
y	1	−1	4

Answer

Not a function: 1 has two images, 1 and −1.

f.

Figure 1.88

Answer

Not a function: 1 has two images, 1 and 4. (−1 sharing the image 1, and 2 sharing 4, would be allowed.)

3.

Draw the graphs of a. f(x) = x, b. $f(x) = x^2$ , c. $f(x) = x^3$ .

Answer

Make a table for x = −2, …, 2 and plot, as in the activity of 1.1.10: a. the straight line through (−2, −2), (0, 0), (2, 2); b. the parabola through (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4); c. the cubic curve through (−2, −8), (−1, −1), (0, 0), (1, 1), (2, 8).

a.

Figure 1.89

x	3	−1
y	…	…

Answer

On y = x, y equals x: y = 3 and y = −1.

b.

Figure 1.90

x	…	…
y	4	1

Answer

On $y = x^2$ : y = 4 at x = 2 (and at x = −2); y = 1 at x = 1 (and at x = −1). A horizontal line meets a parabola twice, so each y here has two pre-images.

5.

The graph of a function f(x) is given. Find f(0), f(−1), f(2).

Figure 1.91

Answer

The curve is $y = x^3$ : f(0) = 0, f(−1) = −1, f(2) = 8.

1.

Define a polynomial, with an example.

Answer

A polynomial in x is a sum of terms $c\,x^n$ , each with a constant coefficient c and a whole-number power n. Example: $4x^3 - 2x + 7$ .

2.

Write the coefficients of $x^3$ and $x^2$ in $4x^5 + 2x^3 - 7x + 6$ .

Answer

The coefficient of $x^3$ is 2; there is no $x^2$ term, so its coefficient is 0.

3.

How many terms has $4x^6 - 7x + 6$ ? What is its degree?

Answer

Three terms, degree 6.

a.

$x^2 + \sqrt{5}$

Answer

Yes: the powers are 2 and 0; $\sqrt{5}$ is just a constant.

b.

$\sqrt{y} + 3$

Answer

No: $\sqrt{y} = y^{1/2}$ .

c.

$z + \frac{5}{z}$

Answer

No: $\frac{5}{z} = 5z^{-1}$ .

d.

$\sqrt{y} + y\sqrt{3}$

Answer

No: $\sqrt{y} = y^{1/2}$ . (The second term, $\sqrt{3}\,y$ , would be fine on its own.)

e.

$x^3 - 7x^2 + 2x - 4$

Answer

Yes: powers 3, 2, 1, 0.

f.

$3x^2 - 11x + 6$

Answer

Yes: powers 2, 1, 0.

g.

$4x^3 + \sqrt[3]{x}$

Answer

No: $\sqrt[3]{x} = x^{1/3}$ .

a.

$2x^4 - 5x^2 - 1$

Answer

4.

b.

$x^{10} + 2x - 4 + 3x^3$

Answer

10.

c.

$x^5 + 2x^3 - 7x^7 + 6$

Answer

7 (the highest power, wherever it is written).

d.

$-2x^3 + 5x^2 + 7x + x^5$

Answer

5.

a.

$2x^3 + 5x^2 + 7x + 9x^4$

Answer

$9x^4 + 2x^3 + 5x^2 + 7x$ (the constant term is 0).

b.

$x^4 + 2x + 1 + 3x^3$

Answer

$x^4 + 3x^3 + 0x^2 + 2x + 1$ .

c.

$x^5 + 2x^3 - 7x + 6$

Answer

$x^5 + 0x^4 + 2x^3 + 0x^2 - 7x + 6$ .

d.

$2x^4 + 7x^2 - 3$

Answer

$2x^4 + 0x^3 + 7x^2 + 0x - 3$ .

1.

What is a root of a polynomial? Explain with an example, and give its meaning on a graph.

Answer

A root of p(x) is a value a with p(a) = 0. For p(x) = x − 5, the root is 5, since p(5) = 0. On a graph, a root is where the graph of y = p(x) meets the X-axis: y = x − 5 crosses it at (5, 0).

a.

One root of p(x) is 2. Where does its graph meet the X-axis?

Answer

At (2, 0).

b.

One root of p(x) is −1. Where does its graph meet the X-axis?

Answer

At (−1, 0).

c.

The roots of p(x) are 3 and −1. Where does its graph meet the X-axis?

Answer

At (3, 0) and (−1, 0).

a.

p(x) = x + 3

Answer

x + 3 = 0: x = −3.

b.

$p(x) = x^2 - 9$

Answer

(x − 3)(x + 3) = 0: x = 3 or −3.

c.

$p(x) = x^2 - 1$

Answer

(x − 1)(x + 1) = 0: x = 1 or −1.

d.

$p(x) = 2x^2 - 7x + 6$

Answer

$2x^2 - 4x - 3x + 6 = 2x(x - 2) - 3(x - 2) = (2x - 3)(x - 2) = 0$ : x = 2 or $\frac{3}{2}$ .

e.

$p(x) = x^2 - 5x + 6$

Answer

(x − 2)(x − 3) = 0: x = 2 or 3.

a.

One root of p(x) = kx + 3 is 3. Find k.

Answer

p(3) = 3k + 3 = 0: k = −1.

b.

One root of $p(x) = x^2 - kx + 6$ is 3. Find k.

Answer

p(3) = 9 − 3k + 6 = 0, so 3k = 15: k = 5.

c.

One root of $p(x) = kx^2 - 25$ is 5. Find k.

Answer

p(5) = 25k − 25 = 0: k = 1.

a.

Figure 1.98

Answer

The line crosses the X-axis at x = 1: root 1 (it is y = x − 1).

b.

Figure 1.99

Answer

It crosses at x = 1 and x = 3: roots 1 and 3 (it is $y = x^2 - 4x + 3 = (x - 1)(x - 3)$ ).

c.

Figure 1.100

Answer

It crosses at x = −1: root −1 (it is $y = x^3 + 1$ , and $(-1)^3 + 1 = 0$ ).

1.

What is the degree of the divisor in synthetic division?

Answer

1: synthetic division divides by a linear divisor, x − a (or ax − b).

a.

$(x^3 - 3x^2 + 4x - 3) \div (x - 1)$

Answer

$\begin{array}{r|rrrr} 1 & 1 & -3 & 4 & -3 \\ & & 1 & -2 & 2 \\ \hline & 1 & -2 & 2 & \boxed{-1} \end{array}$
Quotient $x^2 - 2x + 2$ , remainder −1.

b.

$(2x^3 + 4x^2 - 5x + 6) \div (x - 2)$

Answer

$\begin{array}{r|rrrr} 2 & 2 & 4 & -5 & 6 \\ & & 4 & 16 & 22 \\ \hline & 2 & 8 & 11 & \boxed{28} \end{array}$
Quotient $2x^2 + 8x + 11$ , remainder 28.

c.

$(2x^3 - 11x^2 - 19x - 10) \div (x - 5)$

Answer

$\begin{array}{r|rrrr} 5 & 2 & -11 & -19 & -10 \\ & & 10 & -5 & -120 \\ \hline & 2 & -1 & -24 & \boxed{-130} \end{array}$
Quotient $2x^2 - x - 24$ , remainder −130.

d.

$(y^3 - 3y^2 - 4y + 12) \div (y + 2)$

Answer

a = −2:
$\begin{array}{r|rrrr} -2 & 1 & -3 & -4 & 12 \\ & & -2 & 10 & -12 \\ \hline & 1 & -5 & 6 & \boxed{0} \end{array}$
Quotient $y^2 - 5y + 6$ , remainder 0: y + 2 is a factor.

e.

$(m^3 - 7m^2 + 7m + 15) \div (m + 1)$

Answer

a = −1:
$\begin{array}{r|rrrr} -1 & 1 & -7 & 7 & 15 \\ & & -1 & 8 & -15 \\ \hline & 1 & -8 & 15 & \boxed{0} \end{array}$
Quotient $m^2 - 8m + 15$ , remainder 0.

f.

$(5x^2 - 6x - 9) \div (x - 2)$

Answer

$\begin{array}{r|rrr} 2 & 5 & -6 & -9 \\ & & 10 & 8 \\ \hline & 5 & 4 & \boxed{-1} \end{array}$
Quotient 5x + 4, remainder −1.

a.

$(2x^3 + 9x^2 + 7x - 6) \div (2x - 1)$

Answer

$a = \frac{1}{2}$ :
$\begin{array}{r|rrrr} \frac{1}{2} & 2 & 9 & 7 & -6 \\ & & 1 & 5 & 6 \\ \hline & 2 & 10 & 12 & \boxed{0} \end{array}$
Divide $2x^2 + 10x + 12$ by 2: quotient $x^2 + 5x + 6$ , remainder 0.
Check: $(2x - 1)(x^2 + 5x + 6) = 2x^3 + 9x^2 + 7x - 6$ ✓.

b.

$(2x^3 + 3x^2 - 11x - 6) \div (2x + 1)$

Answer

$a = -\frac{1}{2}$ :
$\begin{array}{r|rrrr} -\frac{1}{2} & 2 & 3 & -11 & -6 \\ & & -1 & -1 & 6 \\ \hline & 2 & 2 & -12 & \boxed{0} \end{array}$
Divide by 2: quotient $x^2 + x - 6$ , remainder 0.

c.

$(6x^3 - 5x^2 - 3x + 2) \div (3x + 2)$

Answer

$a = -\frac{2}{3}$ :
$\begin{array}{r|rrrr} -\frac{2}{3} & 6 & -5 & -3 & 2 \\ & & -4 & 6 & -2 \\ \hline & 6 & -9 & 3 & \boxed{0} \end{array}$
Divide by 3: quotient $2x^2 - 3x + 1$ , remainder 0.

d.

$(3x^3 - 13x^2 + 16) \div (3x - 4)$

Answer

Coefficients 3, −13, 0, 16; $a = \frac{4}{3}$ :
$\begin{array}{r|rrrr} \frac{4}{3} & 3 & -13 & 0 & 16 \\ & & 4 & -12 & -16 \\ \hline & 3 & -9 & -12 & \boxed{0} \end{array}$
Divide by 3: quotient $x^2 - 3x - 4$ , remainder 0.

e.

$(6x^3 - 4 + 13x^2) \div (3x + 2)$

Answer

Standard form $6x^3 + 13x^2 + 0x - 4$ ; $a = -\frac{2}{3}$ :
$\begin{array}{r|rrrr} -\frac{2}{3} & 6 & 13 & 0 & -4 \\ & & -4 & -6 & 4 \\ \hline & 6 & 9 & -6 & \boxed{0} \end{array}$
Divide by 3: quotient $2x^2 + 3x - 2$ , remainder 0.

a.

If $2x^3 - 7x^2 + x + 10 = (x - 1) \cdot Q(x) + R$ , find Q(x) and R.

Answer

$\begin{array}{r|rrrr} 1 & 2 & -7 & 1 & 10 \\ & & 2 & -5 & -4 \\ \hline & 2 & -5 & -4 & \boxed{6} \end{array}$
$Q(x) = 2x^2 - 5x - 4$ , R = 6.

b.

If $x^3 - 19x - 30 = (x + 2) \cdot Q(x) + R$ , find Q(x) and R.

Answer

Coefficients 1, 0, −19, −30; a = −2:
$\begin{array}{r|rrrr} -2 & 1 & 0 & -19 & -30 \\ & & -2 & 4 & 30 \\ \hline & 1 & -2 & -15 & \boxed{0} \end{array}$
$Q(x) = x^2 - 2x - 15$ , R = 0.

c.

If $x^3 - 21x - 20 = (x + 1) \cdot Q(x) + R$ , find Q(x) and R.

Answer

Coefficients 1, 0, −21, −20; a = −1:
$\begin{array}{r|rrrr} -1 & 1 & 0 & -21 & -20 \\ & & -1 & 1 & 20 \\ \hline & 1 & -1 & -20 & \boxed{0} \end{array}$
$Q(x) = x^2 - x - 20$ , R = 0.

1.

What is a linear inequality? Give an example.

Answer

An inequality (with <, >, ≤ or ≥) whose expressions are of the first degree, such as 2x + 3 ≤ 11 or x − y > 1.

2.

Show each inequality on a graph and shade its solution set.

a.x ≥ 0 b. x ≤ 0 c. x ≥ 3 d. x > 1 e. x ≥ −2 f. x > −3 g. x < 2
h.y ≥ 0 i. y ≤ 0 j. y ≥ 1 k. y > 3 l. y ≥ −4 m. y > −5 n. y < 2

Answer

An inequality in x alone has a vertical boundary line; one in y alone a horizontal one.

	Boundary line	Drawn	Shade
a.x ≥ 0	x = 0 (the Y-axis)	solid	right of it
b.x ≤ 0	x = 0	solid	left
c.x ≥ 3	x = 3	solid	right
d.x > 1	x = 1	dashed	right
e.x ≥ −2	x = −2	solid	right
f.x > −3	x = −3	dashed	right
g.x < 2	x = 2	dashed	left
h.y ≥ 0	y = 0 (the X-axis)	solid	above it
i.y ≤ 0	y = 0	solid	below
j.y ≥ 1	y = 1	solid	above
k.y > 3	y = 3	dashed	above
l.y ≥ −4	y = −4	solid	above
m.y > −5	y = −5	dashed	above
n.y < 2	y = 2	dashed	below

For example, d and k:

Figure 1.104 d. x > 1

Figure 1.105 k. y > 3

a.

2x + 5 ≤ 15

Answer

2x ≤ 10, so x ≤ 5.

Figure 1.106

b.

2x + 5 ≥ 15

Answer

2x ≥ 10, so x ≥ 5.

Figure 1.107

c.

x + 3 > 7

Answer

x > 4.

Figure 1.108

d.

5x ≤ 20

Answer

x ≤ 4.

Figure 1.109

e.

$\frac{7 - x}{2} \ge 1$

Answer

7 − x ≥ 2, so −x ≥ −5, and multiplying by −1 turns the sign: x ≤ 5.

Figure 1.110

f.

$\frac{2 + x}{3} < 1$

Answer

2 + x < 3, so x < 1.

Figure 1.111

g.

$3x + 1 \ge \frac{2}{3}(4x - 3)$

Answer

Multiply both sides by 3: 9x + 3 ≥ 8x − 6, so x ≥ −9.

Figure 1.112

h.

$\frac{5}{4}(x + 1) \le \frac{3}{2}(x - 1)$

Answer

Multiply by 4: 5x + 5 ≤ 6x − 6, so 11 ≤ x: x ≥ 11.

Figure 1.113

i.

$\frac{1}{3}(x + 2) \ge \frac{1}{5}(x + 6)$

Answer

Multiply by 15: 5x + 10 ≥ 3x + 18, so 2x ≥ 8: x ≥ 4.

Figure 1.114

j.

$\frac{3x + 1}{2} \le \frac{2x - 1}{3}$

Answer

Multiply by 6: 9x + 3 ≤ 4x − 2, so 5x ≤ −5: x ≤ −1.

Figure 1.115

k.

$\frac{4x + 1}{3} > \frac{3x - 1}{2}$

Answer

Multiply by 6: 8x + 2 > 9x − 3, so 5 > x: x < 5.

Figure 1.116

a.

Three times a number, added to 4, is less than 13.

Answer

3x + 4 < 13, so 3x < 9: x < 3.

b.

If 3 is subtracted from twice a number, the result is greater than or equal to 5.

Answer

2x − 3 ≥ 5, so 2x ≥ 8: x ≥ 4.

c.

A motorcycle covers 120 km at a speed of at least 40 km/h and at most 60 km/h. Between what times does the journey take?

Answer

Time $t = \frac{120}{\text{speed}}$ . The fastest, 60 km/h, takes $\frac{120}{60} = 2$ h; the slowest, 40 km/h, takes $\frac{120}{40} = 3$ h. So 2 ≤ t ≤ 3 hours: a faster speed means a shorter time, so the larger speed gives the smaller end.

d.

Sahabuddin scored 72, 85 and 75 in his first three Optional Mathematics exams. He wants an average of at least 80 over four exams. What must he score in the fourth?

Answer

Let the fourth score be x:
$\begin{aligned} \frac{72 + 85 + 75 + x}{4} &\ge 80 \\ 232 + x &\ge 320 \\ x &\ge 88 \end{aligned}$
He needs at least 88. Check: $\frac{232 + 88}{4} = 80$ ✓.

e.

Lakpa has Rs 500. He buys half a dozen pens at Rs 20 each, and then pens of another kind at Rs 40 each. At most how many of the second kind can he buy, and how much money is left?

Answer

Six pens at Rs 20 cost Rs 120. With y pens of the second kind: 120 + 40y ≤ 500, so 40y ≤ 380 and y ≤ 9.5. Pens come whole, so at most 9 pens; they cost Rs 360, and Rs 20 is left (500 − 120 − 360).

f.

Sanam has Rs 2,000 for prize packets. One kind costs Rs 300 a packet and the other Rs 500. She needs 3 packets of the first kind. At most how many of the second kind can she buy, and how much is left?

Answer

3 × 300 + 500y ≤ 2000, so 500y ≤ 1100 and y ≤ 2.2: at most 2 packets. They cost Rs 1,000, so Rs 100 is left (2000 − 900 − 1000).

g.

A triangle’s base is 5 cm, and its area is between 20 cm² and 30 cm². Between what values is its height?

Answer

Area $= \frac{1}{2} \times 5 \times h = 2.5h$ :
$20 < 2.5h < 30 \quad \Rightarrow \quad 8 < h < 12$
The height is between 8 cm and 12 cm.

h.

A square’s perimeter is between 40 cm and 200 cm. What can the length of its side be?

Answer

40 < 4l < 200, so 10 < l < 50 cm.

i.

A batter scored 30 and 85 runs in the first two matches of a T20 series. At least how many must she score in the third match for an average of at least 50?

Answer

$\frac{30 + 85 + x}{3} \ge 50$ , so 115 + x ≥ 150: at least 35 runs.

a.

Figure 1.117

Answer

A solid line at x = 1, shaded to the right: x ≥ 1.

b.

Figure 1.118

Answer

A dashed line at x = −1, shaded to the left: x < −1.

c.

Figure 1.119

Answer

A dashed line at y = 2, shaded above: y > 2.

1.

What is a linear inequality in two variables? Give an example.

Answer

A linear inequality with two variables, such as 2x + y ≤ 8; its solutions are the points of a region of the plane, on one side of the line 2x + y = 8.

a.

y ≤ x + 5

Answer

Boundary y = x + 5 through (0, 5) and (−5, 0), solid. Test (0, 0): 0 ≤ 5, true: shade the origin’s side (below the line).

Figure 1.122

b.

y ≥ 3x + 2

Answer

Boundary through (0, 2) and (−1, −1), solid. Test (0, 0): 0 ≥ 2, false: shade the side away from the origin (above the line).

Figure 1.123

c.

y ≤ 2x − 3

Answer

Boundary through (0, −3) and (2, 1), solid. Test (0, 0): 0 ≤ −3, false: shade away from the origin (below and to the right).

Figure 1.124

d.

2y ≥ x − 4

Answer

Boundary 2y = x − 4 through (0, −2) and (4, 0), solid. Test (0, 0): 0 ≥ −4, true: shade the origin’s side (above).

Figure 1.125

e.

3x + 2y ≥ 6

Answer

Boundary through (2, 0) and (0, 3), solid. Test (0, 0): 0 ≥ 6, false: shade away from the origin.

Figure 1.126

f.

2x − y ≤ 10

Answer

Boundary through (5, 0) and (0, −10), solid. Test (0, 0): 0 ≤ 10, true: shade the origin’s side (above and to the left).

Figure 1.127

g.

3x − 5y + 15 ≥ 0

Answer

Boundary through (−5, 0) and (0, 3), solid. Test (0, 0): 15 ≥ 0, true: shade the origin’s side (below the line).

Figure 1.128

h.

3x + 4y ≤ 24

Answer

Boundary through (8, 0) and (0, 6), solid. Test (0, 0): 0 ≤ 24, true: shade the origin’s side.

Figure 1.129

a.

Raman has Rs 5,000 for pants at Rs 1,000 each and T-shirts at Rs 500 each. Write the possible numbers he can buy as an inequality.

Answer

With x pants and y T-shirts: 1000x + 500y ≤ 5000, which simplifies (dividing by 500) to 2x + y ≤ 10.

b.

Shristi has Rs 1,000 for two kinds of exercise books, at Rs 50 and Rs 80 each. Write the possible numbers she can buy as an inequality.

Answer

50x + 80y ≤ 1000, that is 5x + 8y ≤ 100.

c.

3 kg of apples and 2 kg of pomegranates together cost less than Rs 1,200. Write the cost as an inequality.

Answer

With apples at Rs x per kg and pomegranates at Rs y per kg: 3x + 2y < 1200.

1.

Define a surd, with an example.

Answer

A surd is a root $\sqrt[n]{a}$ of a positive rational number a whose value is not rational. Example: $\sqrt{3}$ (= 1.732…) or $\sqrt[3]{5}$ .

a.

$\sqrt{6}$

Answer

2.

b.

$\sqrt[3]{9}$

Answer

3.

c.

$3\sqrt[7]{6}$

Answer

7 (the coefficient 3 doesn’t change the order).

d.

$\sqrt[n]{a}$

Answer

n.

a.

For what value of m are $x^3$ and $\sqrt[3]{x^{m - 1}}$ equal?

Answer

$\sqrt[3]{x^{m - 1}} = x^{\frac{m - 1}{3}}$ . Equal to $x^3$ when $\frac{m - 1}{3} = 3$ , so m − 1 = 9: m = 10.

b.

If $y^4$ and $\sqrt[4]{y^{2x}}$ are equal, find x.

Answer

$\sqrt[4]{y^{2x}} = y^{\frac{2x}{4}} = y^{\frac{x}{2}}$ . Equal to $y^4$ when $\frac{x}{2} = 4$ : x = 8.

a.

$\sqrt{3} \times \sqrt{5}$

Answer

$\sqrt{15}$ .

b.

$\sqrt{2} \times \sqrt{8}$

Answer

$\sqrt{16} =$ 4.

c.

$\dfrac{\sqrt{15}}{\sqrt{5}}$

Answer

$\sqrt{\frac{15}{5}} =$ $\sqrt{3}$ .

d.

$\dfrac{\sqrt[3]{27}}{\sqrt{36}}$

Answer

$\frac{3}{6} =$ $\frac{1}{2}$ .

e.

$\sqrt[3]{40} \times \sqrt[3]{25}$

Answer

$\sqrt[3]{1000} =$ 10.

f.

$\dfrac{\sqrt[3]{64}}{\sqrt{16}}$

Answer

$\frac{4}{4} =$ 1.

a.

$4\sqrt[3]{7} + 2\sqrt[3]{7} - \sqrt[3]{7}$

Answer

$(4 + 2 - 1)\sqrt[3]{7} =$ $5\sqrt[3]{7}$ .

b.

$\dfrac{24}{\sqrt{3}} - \dfrac{18}{\sqrt{3}}$

Answer

$\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} =$ $2\sqrt{3}$ .

c.

$3\sqrt{3} + \dfrac{6}{\sqrt{3}}$

Answer

$\frac{6}{\sqrt{3}} = 2\sqrt{3}$ , so $3\sqrt{3} + 2\sqrt{3} =$ $5\sqrt{3}$ .

d.

$\sqrt[4]{16807}$

Answer

$16807 = 7^5 = 7^4 \times 7$ , so $\sqrt[4]{16807} =$ $7\sqrt[4]{7}$ .

e.

$\sqrt[3]{250}$

Answer

250 = 125 × 2: $5\sqrt[3]{2}$ .

f.

$3\sqrt{2} + \sqrt[4]{2500} + \sqrt[4]{64} + 6\sqrt{8}$

Answer

$\sqrt[4]{2500} = \sqrt[4]{625 \times 4} = 5\sqrt[4]{4} = 5\sqrt{2}$ (since $\sqrt[4]{4} = \sqrt[4]{2^2} = \sqrt{2}$ ); $\sqrt[4]{64} = \sqrt[4]{16 \times 4} = 2\sqrt{2}$ ; $6\sqrt{8} = 12\sqrt{2}$ .
$3\sqrt{2} + 5\sqrt{2} + 2\sqrt{2} + 12\sqrt{2} =$ $22\sqrt{2}$ .

g.

$\sqrt{50} + \sqrt{18} - 8\sqrt{2}$

Answer

$5\sqrt{2} + 3\sqrt{2} - 8\sqrt{2} =$ 0.

h.

$4\sqrt[3]{250} - 8\sqrt[3]{128} + 4\sqrt[3]{54}$

Answer

$\sqrt[3]{250} = 5\sqrt[3]{2}$ , $\sqrt[3]{128} = 4\sqrt[3]{2}$ , $\sqrt[3]{54} = 3\sqrt[3]{2}$ :
$20\sqrt[3]{2} - 32\sqrt[3]{2} + 12\sqrt[3]{2} =$ 0.

i.

$4\sqrt[3]{24} - 11\sqrt[3]{192} + 6\sqrt[3]{648}$

Answer

$\sqrt[3]{24} = 2\sqrt[3]{3}$ , $\sqrt[3]{192} = \sqrt[3]{64 \times 3} = 4\sqrt[3]{3}$ , $\sqrt[3]{648} = \sqrt[3]{216 \times 3} = 6\sqrt[3]{3}$ :
$8\sqrt[3]{3} - 44\sqrt[3]{3} + 36\sqrt[3]{3} =$ 0.

j.

$\sqrt{125} - \sqrt{45} + \sqrt{5}$

Answer

$5\sqrt{5} - 3\sqrt{5} + \sqrt{5} =$ $3\sqrt{5}$ .

a.

$\sqrt{5} \times \sqrt{2} \times \sqrt{10}$

Answer

$\sqrt{100} =$ 10.

b.

$\sqrt[3]{54} \times \sqrt[3]{4}$

Answer

$\sqrt[3]{216} =$ 6.

c.

$6\sqrt[3]{4} \times 7\sqrt[3]{12}$

Answer

$42\sqrt[3]{48} = 42\sqrt[3]{8 \times 6} = 42 \times 2\sqrt[3]{6} =$ $84\sqrt[3]{6}$ .

d.

$\sqrt{40} \times \sqrt{18}$

Answer

$\sqrt{720} = \sqrt{144 \times 5} =$ $12\sqrt{5}$ .

e.

$(2\sqrt{5} + 3\sqrt{2})(3\sqrt{5} - 4\sqrt{2})$

Answer

$\begin{aligned} &= 6 \times 5 - 8\sqrt{10} + 9\sqrt{10} - 12 \times 2 \\ &= 30 + \sqrt{10} - 24 = 6 + \sqrt{10} \end{aligned}$
$6 + \sqrt{10}$ .

f.

$(\sqrt{2} + 4\sqrt{3})(\sqrt{12} - 2\sqrt{3})$

Answer

$\sqrt{12} = 2\sqrt{3}$ , so the second bracket is $2\sqrt{3} - 2\sqrt{3} = 0$ , and the product is 0.

g.

$(3 - 2\sqrt{3})^2$

Answer

$9 - 12\sqrt{3} + 4 \times 3 =$ $21 - 12\sqrt{3}$ .

h.

$(5 + 2\sqrt{2})^2$

Answer

$25 + 20\sqrt{2} + 8 =$ $33 + 20\sqrt{2}$ .

i.

$(7\sqrt{5} - 2\sqrt{7})^2$

Answer

$49 \times 5 - 28\sqrt{35} + 4 \times 7 = 245 - 28\sqrt{35} + 28 =$ $273 - 28\sqrt{35}$ .

j.

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})$

Answer

a − b.

k.

$(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})$

Answer

5 − 3 = 2.

l.

$\dfrac{\sqrt[3]{128} - \sqrt[3]{16}}{2\sqrt[3]{2}}$

Answer

$\sqrt[3]{128} = 4\sqrt[3]{2}$ , $\sqrt[3]{16} = 2\sqrt[3]{2}$ : $\frac{2\sqrt[3]{2}}{2\sqrt[3]{2}} =$ 1.

m.

$\dfrac{\sqrt{12} + \sqrt{27}}{\sqrt{3}}$

Answer

$\frac{2\sqrt{3} + 3\sqrt{3}}{\sqrt{3}} =$ 5.

n.

$\dfrac{6\sqrt{45} - 2\sqrt{80} - 3\sqrt{20}}{2\sqrt{180}}$

Answer

$6\sqrt{45} = 18\sqrt{5}$ , $2\sqrt{80} = 8\sqrt{5}$ , $3\sqrt{20} = 6\sqrt{5}$ , $2\sqrt{180} = 12\sqrt{5}$ :
$\frac{18 - 8 - 6}{12} = \frac{4}{12} =$ $\frac{1}{3}$ .

a.

$\dfrac{12}{\sqrt{7} - \sqrt{5}}$

Answer

$\frac{12(\sqrt{7} + \sqrt{5})}{7 - 5} =$ $6(\sqrt{7} + \sqrt{5})$ .

b.

$\dfrac{2 + \sqrt{7}}{2 - \sqrt{7}}$

Answer

$\frac{(2 + \sqrt{7})^2}{4 - 7} = \frac{4 + 4\sqrt{7} + 7}{-3} = -\frac{11 + 4\sqrt{7}}{3}$
$-\dfrac{11 + 4\sqrt{7}}{3}$ .

c.

$\dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} - \sqrt{18}}$

Answer

$\sqrt{48} - \sqrt{18} = 4\sqrt{3} - 3\sqrt{2}$ . Multiply top and bottom by $4\sqrt{3} + 3\sqrt{2}$ :
$\begin{aligned} \text{top} &= 28 \times 3 + 21\sqrt{6} - 20\sqrt{6} - 15 \times 2 = 54 + \sqrt{6} \\ \text{bottom} &= 48 - 18 = 30 \end{aligned}$
$\dfrac{54 + \sqrt{6}}{30}$ .

d.

$\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$

Answer

$\frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} =$ $2 + \sqrt{3}$ .

e.

$\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

Answer

$\frac{(\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2}{5 - 3} = \frac{(8 + 2\sqrt{15}) + (8 - 2\sqrt{15})}{2} = \frac{16}{2}$
8.

f.

$\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Answer

$\frac{(5 + 2\sqrt{6}) + (5 - 2\sqrt{6})}{3 - 2} =$ 10.

g.

$\dfrac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}$

Answer

Multiply top and bottom by $x - \sqrt{x^2 - 1}$ . The bottom becomes $x^2 - (x^2 - 1) = 1$ :
$(x - \sqrt{x^2 - 1})^2 = x^2 - 2x\sqrt{x^2 - 1} + x^2 - 1$
$2x^2 - 1 - 2x\sqrt{x^2 - 1}$ .

h.

$\dfrac{\sqrt{x} + \sqrt{a}}{\sqrt{x} - \sqrt{a}} + \dfrac{\sqrt{x} - \sqrt{a}}{\sqrt{x} + \sqrt{a}}$

Answer

$\frac{(x + 2\sqrt{xa} + a) + (x - 2\sqrt{xa} + a)}{x - a} = \frac{2(x + a)}{x - a}$
$\dfrac{2(x + a)}{x - a}$ .

i.

$\dfrac{1}{x + \sqrt{x^2 - 1}} + \dfrac{1}{x - \sqrt{x^2 - 1}}$

Answer

The common denominator is $x^2 - (x^2 - 1) = 1$ :
$(x - \sqrt{x^2 - 1}) + (x + \sqrt{x^2 - 1}) =$ 2x.

j.

$\dfrac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} - \dfrac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}$

Answer

The common denominator is 1, and by $(p + q)^2 - (p - q)^2 = 4pq$ :
$4x\sqrt{x^2 - 1}$ .

1.

What is a closed interval? Give an example.

Answer

The set of all real numbers from a to b, both ends included: [a, b] = {x:a ≤ x ≤ b}. Example: [0, 1], all real numbers from 0 to 1, with 0 and 1.

2.

What is an open interval? Give an example.

Answer

The set of all real numbers between a and b, without the ends: (a, b) = {x:a < x < b}. Example: (0, 1), all real numbers strictly between 0 and 1.

a.

[1, 3]

Answer

All real numbers from 1 to 3, both included: 1 ≤ x ≤ 3.

Figure 1.137

b.

(−2, 4)

Answer

All real numbers between −2 and 4, neither included: −2 < x < 4.

Figure 1.138

c.

(−3, 0)

Answer

All real numbers between −3 and 0, neither included: −3 < x < 0.

Figure 1.139

d.

(0, 5)

Answer

All real numbers between 0 and 5, neither included: 0 < x < 5.

Figure 1.140

a.

Figure 1.141

Answer

Open dots at −1 and 2: (−1, 2).

b.

Figure 1.142

Answer

An open dot at −1 and a filled dot at 3: (−1, 3].

c.

Figure 1.143

Answer

A filled dot at 0 and an open dot at 2: [0, 2).

d.

Figure 1.144

Answer

Filled dots at 1 and 3: [1, 3].

a.

What is a matrix? Give an example, and say how a matrix is named.

Answer

An arrangement of numbers in rows and columns inside brackets, named by a capital letter. Example: $A = \begin{pmatrix} 2 & 5 \\ 1 & 0 \end{pmatrix}$ , with 2 rows and 2 columns.

b.

What is meant by the order of a matrix? Explain with an example.

Answer

The number of its rows × the number of its columns, rows first. A matrix with m rows and n columns is of order m × n. Example: $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ has 2 rows and 3 columns, so its order is 2 × 3.

a.

$P = \begin{pmatrix} 1 & 4 & 2 \end{pmatrix}$

Answer

1 row and 3 columns: 1 × 3.

b.

$X = \begin{pmatrix} 0 & 4 \\ -7 & 0 \\ 2 & -3 \\ 1 & -5 \end{pmatrix}$

Answer

4 rows and 2 columns: 4 × 2.

c.

$M = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ -3 & -2 \end{pmatrix}$

Answer

3 rows and 2 columns: 3 × 2.

a.

Find $a_{32}$ .

Answer

Row 3, column 2: $a_{32} =$ 0.

b.

Find $(a_{32} + a_{21})^2$ .

Answer

$a_{32} = 0$ , $a_{21} = 2$ : $(0 + 2)^2 =$ 4.

c.

Find $a_{31} - a_{22} + a_{13}$ .

Answer

$a_{31} = 2$ , $a_{22} = 1$ , $a_{13} = 3$ : 2 − 1 + 3 = 4.

a.

$A = [a_{ij}]_{2 \times 3}$ , where $a_{ij} = 3i + 2j$ .

Answer

Row 1: 3 + 2 = 5, 3 + 4 = 7, 3 + 6 = 9. Row 2: 6 + 2 = 8, 10, 12.
$A = \begin{pmatrix} 5 & 7 & 9 \\ 8 & 10 & 12 \end{pmatrix}$

b.

$B = [b_{ij}]_{3 \times 2}$ , where $b_{ij} = 2i^2 - j$ .

Answer

Row 1: 2 − 1 = 1, 2 − 2 = 0. Row 2: 8 − 1 = 7, 8 − 2 = 6. Row 3: 18 − 1 = 17, 18 − 2 = 16.
$B = \begin{pmatrix} 1 & 0 \\ 7 & 6 \\ 17 & 16 \end{pmatrix}$

c.

$X = [x_{ij}]_{3 \times 3}$ , where $x_{ij} = |2i - j|$ .

Answer

Row 1: |2 − 1| = 1, |2 − 2| = 0, |2 − 3| = 1. Row 2: |4 − 1| = 3, 2, 1. Row 3: |6 − 1| = 5, 4, 3.
$X = \begin{pmatrix} 1 & 0 & 1 \\ 3 & 2 & 1 \\ 5 & 4 & 3 \end{pmatrix}$
The bars make every element 0 or more: |2 − 3| = | − 1| = 1.

d.

$M = [m_{ij}]_{2 \times 4}$ , where $m_{ij} = \dfrac{(i - j)^2}{2}$ .

Answer

Row 1: $\frac{0}{2} = 0$ , $\frac{1}{2}$ , $\frac{4}{2} = 2$ , $\frac{9}{2}$ . Row 2: $\frac{1}{2}$ , 0, $\frac{1}{2}$ , 2.
$M = \begin{pmatrix} 0 & \frac{1}{2} & 2 & \frac{9}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} & 2 \end{pmatrix}$

e.

$P = [p_{ij}]_{2 \times 3}$ , where $p_{ij} = (i \times j)^2$ .

Answer

Row 1: $1^2 = 1$ , $2^2 = 4$ , $3^2 = 9$ . Row 2: $2^2 = 4$ , $4^2 = 16$ , $6^2 = 36$ .
$P = \begin{pmatrix} 1 & 4 & 9 \\ 4 & 16 & 36 \end{pmatrix}$

b.

Define each type of matrix with an example: (i) square matrix, (ii) diagonal matrix, (iii) scalar matrix, (iv) symmetric matrix.

Answer

i.Square: as many rows as columns, $\begin{pmatrix} 1 & 4 \\ 2 & 0 \end{pmatrix}$ .
ii.Diagonal: a square matrix with every element off the main diagonal 0, $\begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}$ .
iii.Scalar: a diagonal matrix whose diagonal elements are all equal, $\begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}$ .
iv.Symmetric: a square matrix with $a_{ij} = a_{ji}$ for all i, j, $\begin{pmatrix} 1 & 7 \\ 7 & 2 \end{pmatrix}$ .

c.

Compare an identity matrix and a scalar matrix.

Answer

Both are diagonal matrices with all their diagonal elements equal. In an identity matrix that equal element must be 1; in a scalar matrix it can be any number. So every identity matrix is a scalar matrix, but not every scalar matrix is an identity matrix: $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is both, $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ is scalar only.

a.

$\begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$

Answer

Scalar matrix: diagonal, both diagonal elements −3. (It is also diagonal and symmetric; “scalar” is the most exact name.)

b.

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Answer

Identity matrix I₃.

c.

$\begin{pmatrix} 2 & 0 & 0 \\ 3 & 3 & 0 \\ 0 & 2 & 6 \end{pmatrix}$

Answer

Lower triangular matrix: every element above the main diagonal is 0.

d.

$\begin{pmatrix} 1 & 2 & 3 & 5 \end{pmatrix}$

Answer

Row matrix, of order 1 × 4.

e.

$\begin{pmatrix} 5 & 2 \\ 2 & 5 \end{pmatrix}$

Answer

Symmetric matrix: $a_{12} = a_{21} = 2$ .

f.

$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}$

Answer

Diagonal matrix: zeros off the main diagonal, and the diagonal elements 2, 3, 6 not all equal, so not scalar.

a.

$\begin{pmatrix} x - 1 & 2 \\ 2y & z \end{pmatrix}$ and $\begin{pmatrix} 7 & 2 \\ 10 & 5 \end{pmatrix}$ are equal matrices.

Answer

x − 1 = 7, so x = 8; 2y = 10, so y = 5; z = 5.
x = 8, y = 5, z = 5.

b.

$\begin{pmatrix} 5x - 1 & 0 \\ 2y & 3z + 5 \end{pmatrix}$ is an identity matrix.

Answer

An identity matrix has 1 on the diagonal and 0 elsewhere, so it is $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ .
5x − 1 = 1 gives $x = \frac{2}{5}$ ; 2y = 0 gives y = 0; 3z + 5 = 1 gives 3z = −4, $z = -\frac{4}{3}$ .
$x = \frac{2}{5}$ , y = 0, $z = -\frac{4}{3}$ .

c.

$\begin{pmatrix} 4 & x^2 \\ 5 & xy \end{pmatrix} = \begin{pmatrix} 4 & 16 \\ 2z - 3 & 8 \end{pmatrix}$

Answer

2z − 3 = 5, so z = 4. $x^2 = 16$ , so x = ±4. xy = 8: if x = 4, y = 2; if x = −4, y = −2.
z = 4, and x = 4, y = 2 or x = −4, y = −2. (x and y have the same sign, because their product 8 is positive.)

d.

$\begin{pmatrix} 3x + 1 & 0 \\ 0 & \frac{z}{3} \end{pmatrix}$ is a scalar matrix with 7 on its main diagonal.

Answer

Both diagonal elements are 7: 3x + 1 = 7, so x = 2; $\frac{z}{3} = 7$ , so z = 21.
x = 2, z = 21.

e.

$\begin{pmatrix} 5 & y + 5 \\ 2y & 4 \end{pmatrix}$ is a symmetric matrix.

Answer

Symmetric means $a_{12} = a_{21}$ : y + 5 = 2y, so y = 5. Check: both become 10.

f.

$\begin{pmatrix} x + y & 5 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 2 & x - y \end{pmatrix}$

Answer

x + y = 4 and x − y = 4. Adding: 2x = 8, x = 4; then y = 0.
x = 4, y = 0.

g.

$\begin{pmatrix} 2x + y & x - 3y \\ 2 & 2z - 6 \end{pmatrix} = \begin{pmatrix} 10 & 12 \\ 2 & 5z + 3 \end{pmatrix}$

Answer

2x + y = 10 … (i) and x − 3y = 12 … (ii). From (i), y = 10 − 2x; in (ii), x − 30 + 6x = 12, 7x = 42, x = 6, so y = 10 − 12 = −2.
2z − 6 = 5z + 3 gives −3z = 9, z = −3.
x = 6, y = −2, z = −3.

h.

$\begin{pmatrix} \sin x & \cos y \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 1 \\ 3 & \tan z \end{pmatrix}$ , where $0^\circ \le x, y, z \le 90^\circ$

Answer

$\sin x = \frac{1}{2}$ , so $x = 30^\circ$ ; cos y = 1, so $y = 0^\circ$ ; tan z = 1, so $z = 45^\circ$ .
$x = 30^\circ$ , $y = 0^\circ$ , $z = 45^\circ$ . (These use the values of sine, cosine and tangent at 0°, 30°, 45°, 60° and 90° from Grade 9 Mathematics’ trigonometry, which Unit 2 uses again.)

4.

If $\begin{pmatrix} a - b & 5 \\ 3 & 8 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 3 & ab \end{pmatrix}$ , find the possible values of a and b.

Answer

a − b = 2 and ab = 8. From the first, a = b + 2; then (b + 2)b = 8, $b^2 + 2b - 8 = 0$ , (b + 4)(b − 2) = 0, so b = 2 or b = −4.
b = 2 gives a = 4; b = −4 gives a = −2.
a = 4, b = 2 or a = −2, b = −4. Check: 4 × 2 = 8 and (−2)(−4) = 8. ✓

5.

Matrix A is of order (2x − 3) × (y + 5) and matrix B of order (x + 2) × (3y − 1). If A and B are equal, find the order of A.

Answer

Equal matrices have the same order, so the rows match and the columns match:
2x − 3 = x + 2, so x = 5; y + 5 = 3y − 1, so 2y = 6, y = 3.
Rows: 2 × 5 − 3 = 7; columns: 3 + 5 = 8. The order of A is 7 × 8.

1.

What are the conditions for adding and subtracting matrices?

Answer

The matrices must be of the same order; then the corresponding elements (in the same places) are added or subtracted.

a.

What are the properties of matrix addition? Write them.

Answer

For matrices A, B, C of the same order: closure (A + B is a matrix of the same order); commutative (A + B = B + A); associative ((A + B) + C = A + (B + C)); additive identity (A + O = O + A = A); additive inverse (A + (−A) = (−A) + A = O).

c.

Write the associative property of matrix addition.

Answer

If A, B and C are matrices of the same order, (A + B) + C = A + (B + C).

3.

Which of these matrices can be added? Give a reason.
$A = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}$ , $P = \begin{pmatrix} 1 & 1 \\ 2 & 0 \\ -4 & 2 \end{pmatrix}$ , $Q = \begin{pmatrix} 1 & 2 & -2 \end{pmatrix}$ , $R = \begin{pmatrix} 3 & 7 \\ 2 & 5 \\ -4 & 2 \end{pmatrix}$ , $X = \begin{pmatrix} 1 & 5 \\ 7 & 0 \end{pmatrix}$

Answer

Write each order: A 2 × 2, P 3 × 2, Q 1 × 3, R 3 × 2, X 2 × 2.
A and X can be added (both 2 × 2), and P and R can be added (both 3 × 2). Q is the only 1 × 3 matrix, so it can’t be added to any of the others.

a.

X + Y

Answer

$X + Y = \begin{pmatrix} 4+1 & 6+5 \\ 8+7 & 0+0 \end{pmatrix} = \begin{pmatrix} 5 & 11 \\ 15 & 0 \end{pmatrix}$

b.

X − Y

Answer

$X - Y = \begin{pmatrix} 4-1 & 6-5 \\ 8-7 & 0-0 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 1 & 0 \end{pmatrix}$

c.

2X + 3Y

Answer

$2X + 3Y = \begin{pmatrix} 8 & 12 \\ 16 & 0 \end{pmatrix} + \begin{pmatrix} 3 & 15 \\ 21 & 0 \end{pmatrix} = \begin{pmatrix} 11 & 27 \\ 37 & 0 \end{pmatrix}$

d.

$\frac{1}{2}X - 2Y$

Answer

$\tfrac{1}{2}X - 2Y = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} - \begin{pmatrix} 2 & 10 \\ 14 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -7 \\ -10 & 0 \end{pmatrix}$

a.

If $X = \begin{pmatrix} -2 & 2 \\ 8 & 6 \end{pmatrix}$ and $Y = \begin{pmatrix} -4 & -5 \\ 7 & 0 \end{pmatrix}$ , verify the commutative property of matrix addition. Does it hold for subtraction too? Write and verify your conclusion.

Answer

$X + Y = \begin{pmatrix} -2-4 & 2-5 \\ 8+7 & 6+0 \end{pmatrix} = \begin{pmatrix} -6 & -3 \\ 15 & 6 \end{pmatrix}, \quad Y + X = \begin{pmatrix} -4-2 & -5+2 \\ 7+8 & 0+6 \end{pmatrix} = \begin{pmatrix} -6 & -3 \\ 15 & 6 \end{pmatrix}$
So X + Y = Y + X: verified.
$X - Y = \begin{pmatrix} 2 & 7 \\ 1 & 6 \end{pmatrix}, \quad Y - X = \begin{pmatrix} -2 & -7 \\ -1 & -6 \end{pmatrix}$
X − Y ≠ Y − X: the commutative property does not hold for subtraction. (Y − X is always −(X − Y).)

b.

If $P = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}$ , $Q = \begin{pmatrix} 4 & 5 \\ 6 & 1 \end{pmatrix}$ and $R = \begin{pmatrix} 4 & 8 \\ 3 & 0 \end{pmatrix}$ , verify the associative property of matrix addition.

Answer

$(P + Q) + R = \begin{pmatrix} 8 & 7 \\ 14 & 7 \end{pmatrix} + \begin{pmatrix} 4 & 8 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 12 & 15 \\ 17 & 7 \end{pmatrix}$
$P + (Q + R) = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix} + \begin{pmatrix} 8 & 13 \\ 9 & 1 \end{pmatrix} = \begin{pmatrix} 12 & 15 \\ 17 & 7 \end{pmatrix}$
Both are equal, so (P + Q) + R = P + (Q + R): verified.

c.

If $M = N = H = \begin{pmatrix} 1 & 0 & -1 \\ 3 & 2 & 5 \end{pmatrix}$ , prove: (i) M + H = H + M, (ii) (M + H) + N = M + (H + N), (iii) M + (−M) = O, where O is the zero matrix of the same order.

Answer

(The three matrices are the same, so this checks the laws on one matrix; the working is still the same as for any three.)

i. $M + H = \begin{pmatrix} 2 & 0 & -2 \\ 6 & 4 & 10 \end{pmatrix}$ and $H + M = \begin{pmatrix} 2 & 0 & -2 \\ 6 & 4 & 10 \end{pmatrix}$ , so M + H = H + M.
ii. $(M + H) + N = \begin{pmatrix} 2 & 0 & -2 \\ 6 & 4 & 10 \end{pmatrix} + \begin{pmatrix} 1 & 0 & -1 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} 3 & 0 & -3 \\ 9 & 6 & 15 \end{pmatrix}$ ; $H + N = \begin{pmatrix} 2 & 0 & -2 \\ 6 & 4 & 10 \end{pmatrix}$ , so $M + (H + N) = \begin{pmatrix} 3 & 0 & -3 \\ 9 & 6 & 15 \end{pmatrix}$ . Equal.
iii. $-M = \begin{pmatrix} -1 & 0 & 1 \\ -3 & -2 & -5 \end{pmatrix}$ , and $M + (-M) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = O$ .

a.

$\begin{pmatrix} 1 & 4x \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 4 & 2 \\ 4 & 6 \end{pmatrix} = \begin{pmatrix} 5 & 8 \\ 4 & 3y \end{pmatrix}$

Answer

The left side is $\begin{pmatrix} 5 & 4x + 2 \\ 4 & 9 \end{pmatrix}$ . So 4x + 2 = 8, $x = \frac{3}{2}$ ; and 3y = 9, y = 3.
$x = \frac{3}{2}$ , y = 3.

b.

$\begin{pmatrix} 1 & 3x \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 0 & 2 \\ 0 & y \end{pmatrix} = I$ , the identity matrix of the same order.

The textbook gives the second matrix as $\begin{pmatrix} 4 & 2 \\ 0 & y \end{pmatrix}$ , but then the top left of the sum is 1 + 4 = 5, and an identity matrix has 1 there, whatever x and y are. So the book’s numbers don’t work; with 0 in that place they give the book’s answer.

Answer

The sum is $\begin{pmatrix} 1 & 3x + 2 \\ 0 & 3 + y \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ .
3x + 2 = 0, so $x = -\frac{2}{3}$ ; 3 + y = 1, so y = −2.
$x = -\frac{2}{3}$ , y = −2.

c.

The additive inverse of $\begin{pmatrix} x + 5 & 2 \\ 0 & 2y - 3 \end{pmatrix}$ is $\begin{pmatrix} 4 & -2 \\ 0 & -7 \end{pmatrix}$ .

The textbook gives the inverse as $\begin{pmatrix} 4 & 2 \\ 0 & -7 \end{pmatrix}$ , but the additive inverse changes the sign of every element, so the 2 must become −2. So the book’s numbers don’t work; with −2 in that place they give the book’s answer.

Answer

The additive inverse is $\begin{pmatrix} -x - 5 & -2 \\ 0 & 3 - 2y \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ 0 & -7 \end{pmatrix}$ .
−x − 5 = 4, so x = −9; 3 − 2y = −7, so 2y = 10, y = 5.
x = −9, y = 5.

d.

$2\begin{pmatrix} 5 & 4x \\ 7 & y - 3 \end{pmatrix} + 3\begin{pmatrix} 4 & 1 \\ -2 & 6 \end{pmatrix} = \begin{pmatrix} 22 & 8 \\ 8 & 3y \end{pmatrix}$

The textbook gives the right side as $\begin{pmatrix} 5 & 8 \\ 4 & 3y \end{pmatrix}$ , but the left side’s first column is 2 × 5 + 3 × 4 = 22 and 2 × 7 + 3 × (−2) = 8, not 5 and 4. So the book’s numbers don’t work; with 22 and 8 there they give the book’s answer.

Answer

The left side is $\begin{pmatrix} 10 & 8x \\ 14 & 2y - 6 \end{pmatrix} + \begin{pmatrix} 12 & 3 \\ -6 & 18 \end{pmatrix} = \begin{pmatrix} 22 & 8x + 3 \\ 8 & 2y + 12 \end{pmatrix}$ .
8x + 3 = 8, so $x = \frac{5}{8}$ ; 2y + 12 = 3y, so y = 12.
$x = \frac{5}{8}$ , y = 12.

e.

$2\begin{pmatrix} 3x & 0 \\ 0 & y - 3 \end{pmatrix} + 3\begin{pmatrix} 4 & 0 \\ 0 & 6 \end{pmatrix}$ is the zero matrix of the same order.

Answer

The sum is $\begin{pmatrix} 6x + 12 & 0 \\ 0 & 2y - 6 + 18 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ .
6x + 12 = 0, so x = −2; 2y + 12 = 0, so y = −6.
x = −2, y = −6.

a.

If $P = \begin{pmatrix} 3 & 0 \\ 4 & -2 \end{pmatrix}$ and $3P + X = \begin{pmatrix} 8 & 6 \\ 9 & 6 \end{pmatrix}$ , find the matrix X.

Answer

$X = \begin{pmatrix} 8 & 6 \\ 9 & 6 \end{pmatrix} - 3P = \begin{pmatrix} 8 & 6 \\ 9 & 6 \end{pmatrix} - \begin{pmatrix} 9 & 0 \\ 12 & -6 \end{pmatrix} = \begin{pmatrix} -1 & 6 \\ -3 & 12 \end{pmatrix}$ .
Check: $3P + X = \begin{pmatrix} 9 - 1 & 0 + 6 \\ 12 - 3 & -6 + 12 \end{pmatrix} = \begin{pmatrix} 8 & 6 \\ 9 & 6 \end{pmatrix}$ . ✓

b.

If $P = \begin{pmatrix} 1 & 0 \\ 3 & -2 \end{pmatrix}$ and $R = \begin{pmatrix} 4 & 8 \\ 3 & 0 \end{pmatrix}$ , find the matrix X such that 2P − 3X = 3R.

Answer

Move 3X and 3R across, as with numbers: 3X = 2P − 3R.
$2P - 3R = \begin{pmatrix} 2 & 0 \\ 6 & -4 \end{pmatrix} - \begin{pmatrix} 12 & 24 \\ 9 & 0 \end{pmatrix} = \begin{pmatrix} -10 & -24 \\ -3 & -4 \end{pmatrix}$
$X = \frac{1}{3}\begin{pmatrix} -10 & -24 \\ -3 & -4 \end{pmatrix} = \begin{pmatrix} -\frac{10}{3} & -8 \\ -1 & -\frac{4}{3} \end{pmatrix}$

8.

If $P + Q = \begin{pmatrix} 5 & 0 \\ 3 & -2 \end{pmatrix}$ and $P - Q = \begin{pmatrix} 8 & 4 \\ 2 & 6 \end{pmatrix}$ , find P and Q.

Answer

Add the two equations, as with numbers: $2P = \begin{pmatrix} 13 & 4 \\ 5 & 4 \end{pmatrix}$ , so $P = \begin{pmatrix} \frac{13}{2} & 2 \\ \frac{5}{2} & 2 \end{pmatrix}$ .
Subtract them: $2Q = \begin{pmatrix} 5 - 8 & 0 - 4 \\ 3 - 2 & -2 - 6 \end{pmatrix} = \begin{pmatrix} -3 & -4 \\ 1 & -8 \end{pmatrix}$ , so $Q = \begin{pmatrix} -\frac{3}{2} & -2 \\ \frac{1}{2} & -4 \end{pmatrix}$ .
Check: $P + Q = \begin{pmatrix} 5 & 0 \\ 3 & -2 \end{pmatrix}$ . ✓

9.

If $X = \begin{pmatrix} -3 & -1 \\ 6 & 7 \end{pmatrix}$ , find the additive inverse of X.

Answer

Change the sign of every element: $-X = \begin{pmatrix} 3 & 1 \\ -6 & -7 \end{pmatrix}$ . Check: X + (−X) = O. ✓

Exam practice

Unit 1 · Algebra

Exercise 1.1 (A)

Question 1

Question 2

Question 3

Question 4

Question 5

Exercise 1.1 (B)

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Exercise 1.1 (C)

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Exercise 1.1 (D)

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Exercise 1.1 (E)

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Exercise 1.1 (F)

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Exercise 1.1 (G)

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Exercise 1.1 (H): types of function

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Exercise 1.1 (H): functional values

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Exercise 1.1 (I)

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Exercise 1.2 (A)

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Exercise 1.2 (B)

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