BASIC ALGEBRA I
Q.1. Express the following in their simplest form:
(a) 2a + 3b + 6a – 8b (b) 2mn + 3pq – 3mn – pq (c) 11x + 2y –z – 2y + x
(d) x2 + y2 + 3xy - 4 + 2xy + 8 (e) 2cd – 2ef + 5dc – 2fe + cd2
Q.2. Simplify: (a) g3 x g5 (b) m6 ÷ m4 (c) (k2)4 (d) p2 ÷ p5
(e) (2g)3 (f) (pq)2 (g) pq4 ÷ q2 (h) (k2)-2 (i) r3s2 x (rs)2
Q.3. Simplify the
following: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
Q.4. Expand the following:
(a) (2x – y)(2x + y) (b) (3b – a)2 (c) (x + 5)(x – 3) (d) 2b(x – 4) -6(x – 4)
(e) (2x – 4)(3x + 6) (f) (x – 3)(x2 + 3x + 9) (g) (2xy + z)(4xy – z)
Q.5. Factorise the following:
(a) (a2 – 4) (b) (4m2 – 9n2) (c) k2 + 2k + 1 (d) 4g2 – 12gh + 9h2
(e) 8p3 + q3 (f) p2 + 9p + 14 (g) k2 – 3k – 18 (h) 2g2 + 9g – 18
Q.6. Solve the following equations:
(a) 5x = 15 (b) 2x + 4 = 12 (c) 4g – 6 = 2g + 4 (d) 3m + 4 = m – 8
(e) 2k – 3 = 7k + 7 (f) 2 – p = 6 – 2p (g) 6q + 18 + 3q = 3 – q
Q.7. An apricot costs 10 cents more than a peach. If 2 peaches plus 3 apricots cost $2.80, what is the cost of a peach and what is the cost of an apricot?
Answers:
Q.1. (a) 6a – 5b (b) 2pq – mn (c) 12x – z (d) x2 + y2 + 5xy + 4
(e) 7cd – 4ef + cd2
Q.2. (a) g8 (b) m2 (c) k8 (d) p-3 or 
(e) 8g3 (f) p2q2 (g) pq2 (h) k-4 or 
(i) r5s4
Q.3. (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) ab5 (h) 
Q.4. (a) 4x2 – y2 (b) 9b2 – 6ab + a2 (c) x2 + 2x – 15 (d) 2bx – 8b – 6x + 24
(e) 6x2 – 24 (f) x3 – 27 (g) 8x2y2 + 2xyz – z2
Q.5. (a) (a + 2)(a – 2) (b) (2m + 3n)(2m – 3n) (c) (k + 1)2 (d) (2g – 3h)2
(e) (2p + q)(4p2 – 2pq + q2) (f) (p + 7)( p + 2) (g) (k – 6)(k + 3)
(h) (2g – 3)(g + 6)
Q.6. (a) x = 3 (b) x = 4 (c) g = 5 (d) m = -6 (e) k = -2 (f) p = 4 (g) q = -1½
Q.7. Let cost of peach be
x. 
cost
of apricot = (x + 10)
2x + 3(x + 10) = 280 2x + 3x + 30 = 280 5x = 250 x = 50
peach = 50 cents, apricot = 50 + 10 = 60 cents
BASIC ALGEBRA II
Q.1. Express the following in their simplest form:
(a) 5d + 16e – 6d + 4e (b) 4ab + 6cd – 2ab – 3c (c) 2p + 3q – r – 2q + 2r
(d) x2 + y2 - 4xy + 6 + 2xy + 3 (e) 5pq – 2rs – 3qp – 2sr + pq + 4rs
Q.2. Simplify: (a) p4 x p7 (b) r8 ÷ r5 (c) (b4)3 (d) g3 ÷ g7
(e) (4d)2 (f) (2ab)3 (g) x2y4 ÷ y3 (h) (m3)-2 (i) (mn)3 ÷ m2n
Q.3. Simplify the
following: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
Q.4. Expand the following:
(a) (m + 3n)(m – 3n) (b) (g + 2h)2 (c) (x + 4)(x – 5) (d) 3k(j + 3) -4(j + 3)
(e) (4p - 2)(3p + 4) (f) (r + 2)(r2 – 2r + 4) (g) (3pq + r)(2pq – r)
Q.5. Factorise the following:
(a) (4p2 – 9) (b) (2g2 – 18) (c) m2 – 2m + 1 (d) 16r2 + 24rs + 9s2
(e) g3 – 27h3 (f) m2 - m - 12 (g) 2q2 + 3q – 20 (h) 3n2 - 2n – 21
Q.6. Solve the following equations:
(a) 6x = 24 (b) 3x + 2 = 4x - 6 (c) 7g – 5 = 2g + 10 (d) 5k + 9 = 23 – 2k
(e) 2m + 6 + m = 5m - 4 (f) 5 – g + 10 – 3g = 12 + 4g - 5
(g) 2q + 3 = 7q + 8 – q + 5
Q.7. A custard tart costs 80 cents more than a lamington. If 2 lamingtons plus a custard tart cost $3.20, what is the cost of a custard tart, and what is the cost of a lamington?
Answers:
Q.1. (a) 20e - d (b) 2ab – 3cd (c) 2p + q + r (d) x2 + y2 - 2xy + 9 (e) 3pq
Q.2. (a) p11 (b) r3 (c) b12 (d) g-4 or 
(e) 16d2 (f) 8a3b3 (g) x2y (h) m-6 or 
(i) mn2
Q.3. (a) 
(b) gjk (c) 
(d) 
(e) 
(f) 
(g) f 2 e3 (h) 
Q.4. (a) m2 – 9n2 (b) g2 + 4gh + 4h2 (c) x2 - x – 20 (d) 3jk – 4j + 9k - 12
(e) 12p2 + 10p - 8 (f) r3 + 8 (g) 6p2q2 - rpq – r2
Q.5. (a) (2p + 3)(2p – 3) (b) 2(g + 3)(g – 3) (c) (m - 1)2 (d) (4g + 3s)2
(e) (g - 3h)(g2 + 3gh + 9h2) (f) (m - 4)( m + 3) (g) (q + 4)(2q - 5)
(h) (3n + 7)(n - 3)
Q.6. (a) x = 4 (b) x = 8 (c) g = 3 (d) k = 2 (e) m = 5 (f) g = -1 (g) q = -2½
Q.7. Lamington = 80 cents, Custard tart = $1.60