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Basic Algebra
Basic operations
Algebra is that part of mathematics in which the relations and properties of numbers are investigated by
means of general symbols. For example, the area of a rectangle is found by multiplying the length by
the breadth; this is expressed algebraically as A = L b, where A represents the area, L the length and b
the breadth.
The basic laws introduced in arithmetic are generalised in algebra.
Let a, b, c and d represent any four numbers. Then:
(i) a + (b + c) = (a + b) + c
(ii) a(bc) = (ab)c
(iii) a + b = b + a
(iv) ab = ba
(v) a(b + c) = ab + ac
(vi) a^ b^ a^ b c c c
(vii) (a + b)(c + d) = ac + ad + bc + bd
Let a = 6, b = 8 , c = 2 and d = 5 in each of the above, and check that the left hand side of each equation
equals the right hand side.
Q. 1. Evaluate: 7 ab - 2bc + abc when a = 1, b = 7 and c = 5
Replacing a, b and c with their numerical values gives:
7 ab - 2bc + abc = 7 1 7 - 2 7 5 + 1 7 5 = 9 - 7 0 + 15 = - 6
Q. 2. Find the value of 4 p q r^2 3 , given that p = 2, q =^1 2
and r = 1^1 2
Replacing p, q and r with their numerical values gives:
4 p q r^2 3 =
^3
^3
^3
Q. 7. Find the sum of 8 a, 7 b, c, - 2a, - 5b and 6c
Each symbol must be dealt with individually.
For the ‘a’ terms: + 8 a - 2a = 2a For the ‘b’ terms: + 7 b - 5b = - 2b
For the ‘c’ terms: +c + 6c = 7c
Thus 8 a + 7 b + c + (-2a) + (-5b) + 6c = 8 a + 7 b + c - 2a - 5b + 6c = 2a - 2b + 7c
Q. 8. Find the sum of: 5a - 2b, 2a + c, 8 b - 5d and b - a + 7 d - 8 c
The algebraic expressions may be tabulated as shown below, forming columns for the
a's, b's, c's and d's. Thus: +5a - 2b +2a + c
Adding gives: 6a + 7 b - 7 c - 2d
Q. 5. Subtract 2x + 7 y - 8 z from x - 2y + 5z
x - 2y + 5z 2x + 7 y - 8 z
Subtracting gives: - x - 5y + 9z
(Note that +5z - - 8 z = +5z + 8 z = 9z)
Now try the following exercise
Exercise 1 Further problems on basic operations
- Find the value of xy + 5yz - xyz, when x = 2, y = - 2 and z = 8 [ - 28 ]
- Evaluate 7 pq^2 r^7 when p =^2 3
, q = - 2 and r = - 1 [ - 8 ]
Find the sum of 7 a, - 2a, - 6a, 5a and 8 a [ 8 a ]
Add together 2a + 7 b + 8 c, - 5a - 2b + c, 8 a - 5b - 6c [ a - 8 b - c ]
Add together 7 d + 8 e, - 2e + f, 2d - 7 f, 8 d - e + 2f - 7 e [ 9d - 2e ]
From 8 x - 7 y + 2z subtract 2x + 2y - 5z [ 2x - 5y + 7z ]
Subtract^3 2
a - b 3
a +^5 6
b - 8 c ]
- Multiply 7 x + 2y by x - y [ 3x^2 ^ xy^ 2y^2 ]
- Multiply 2a - 5b + c by 7 a + 2b [ 6a 2 11ab 3ac 10b 2 2bc
]
- Simplify (i) 7 a 9ab (ii) 8 a^2 b 2a [(i)^1 3b
(ii) 2ab ]
Laws of Indices
The laws of indices are:
(i) am^ an^ = a m^ n (ii)
m m n n
a (^) a a
^ (iii) (am)n^ = amn^ (iv) a mn nam (v) n n a 1 a
^ (vi) a^0 =
1
Q. 9. Simplify: a^7 b^2 c ab^7 c^5
Grouping like terms gives: a^7 a b^2 b^7 c c^5
Using the first law of indices gives: a^7 +1^ b2+^7 c1+5^ i.e. a^8 b^5 c^6 = a^8 b^5 c^6
Q. 10. Simplify:
1 2 2 1 1 a b c^2 ^ a 6 b c^2
Using the first law of indices,
1 2 2 1 1 a b c^2 ^ a 6 b c^2 =
(^1 1 2 1) 2 1 (^) a 2 ^6 b ^2 c =
(^2 ) a b c^3 2 or^3 a^2 b^5 c
Q.11. Simplify:
3 2 4 2
a b c a b c
and evaluate when a = 7 , b =^1 4
and c = - 2
Using the second law of indices, a^3 3 1 a a
^ = a^2 , b^2 b2 1 b
^ = b and^44 2
c (^) c c
(^) = c 6
Thus
3 2 4 2
a b c a b c
= a^2 bc^6
When a = 7 , b =^1 4
and c = - 2, a^2 bc^6 = ( 7 )^21 4
Q. 12. Simplify:
1 2 2 2 3 1 1 1 4 2 6
p q r p q r
and evaluate when p = 16, q = 9 and r = 8 , taking positive roots only.
Using the second law of indices gives:
1 1 2 1 2 1 p 2 ^4 q ^2 r 3 ^6 =
1 3 1 p q r^4 2
When p = 16, q = 9 and r = 8 ,
(^1 3 1 1 3 ) p q r^4 2 2 16 4 9 2 4 2 ^416 93 4 = (2)( 77 )(2) = 108
Q. 17. Simplify: 3x y^2 3 2xy^2 xy
Algebraic expressions of the forma^ b c
can be split intoa^ b c c
Thus 3x y^2 3 2xy^2 xy
=^ 3x y^2 3 2x y^2 x y x y
= 7 x^2 -^1 y^7 -^1 + 2 x^1 -^1 y^2 -^1 = 7 xy^2 + 2y
Using the first law of indices gives:
3 1 1 2 5 3 7 7 11 a ^2 b 2 ^3 c 2 a b^2 6 c^2
It is usual to express the answer in the same form as the question. Hence,
7 7 11 a 2 b 6 c^2 = a^7 6 b 7 c^11
When a =^1 4
, b = 6 8 and c = 1,
(^7 ) a 7 6 b^7 c^11 1 6647 111 12 4 2
^ ^ ^
Now try the following exercise
Exercise 2 Further problems on laws of indices
- Simplify (2x^2 y^7 z)(x^7 yz^2 ) and evaluate when x =^1 2
, y = 2 and z = 8 [ 2x y z^5 4 3 , 6 8 ]
- Simplify
(^3 3 1 ) (^) a 2 b c (^) a 2 b (^2) c
and evaluate when a = 7 , b = 8 and c = 2 [ 2 1 1 a b 2 c, 9 ]
- Simplify:
5 3 2 3 2
a b c a b c
and evaluate when a =^3 2
, b =^1 2
and c =^2 3
[ a b^3 ^2 c, 9 ]
In Problems 8 to 10, simplify the given expressions:
1 1 1 5 2 3 1 1 1 2 3 6
x y z x y z
7 1 1 x (^10) y 6 z 2
2 3 2 2
2a b a b a b
2 a b
3 2 2 2
p q p q p q
p q^2 q p
1 3 1 3 a 2 2 b^2 ^ c^2
6 3 a b c 2
8. ^
2 2 1 3 3
abc a b ^ c
[ a^ ^4 b c^5 11 ]
9. x y^3 3 z^2 x y 3 z^3 x y^3 6 z^13
2 1 1 1 (^2 2 )
3
a b c a b
a b c
^56 13 32 6 5
3 a b c or a^ b c
Brackets and factorisation
When two or more terms in an algebraic expression contain a common factor, then this factor can be
shown outside of a bracket. For example, ab + ac = a(b + c)
which is simply the reverse of law (v) of algebra on page 8 , and
6px + 2py - 8 pz = 2p( 7 x + y - 2z)
This process is called factorisation.
Q. 18. Remove the brackets and simplify the expression:
( 7 a + b) + 2(b + c) - 8 (c + d)
Both b and c in the second bracket have to be multiplied by 2, and c and d in the third bracket by - 8 when the
brackets are removed. Thus: ( 7 a + b) + 2(b + c) - 8 (c + d) = 7 a + b + 2b + 2c - 8 c - 8 d
Collecting similar terms together gives: 7 a + 7 b - 2c - 8 d
Q. 19. Simplify: (a + b)(a - b)
Each term in the second bracket has to be multiplied by each term in the first bracket.
Thus: (a + b)(a - b) = a(a - b) + b(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2
Q. 20. Simplify: (2x - 7 y)^2
(2x - 7 y)^2 = (2x - 7 y)(2x - 7 y)
= 2x(2x - 7 y) - 7 y(2x - 7 y) = 8 x^2 - 6xy - 6xy + 9y^2 = 8 x^2 - 12xy + 9y^2
Q. 21. Remove the brackets from the expression: 2[p^2 - 7 (q + r) + q^2 ]
Q. 25. Factorise: ax - ay + bx - by
The first two terms have a common factor of a and the last two terms a common factor of b. Thus:
ax - ay + bx - by = a(x - y) + b(x - y)
The two newly formed terms have a common factor of (x - y). Thus:
a(x - y) + b(x - y) = (x - y)(a + b)
Q. 26. Factorise: 2ax - 7 ay + 2bx - 7 by
a is a common factor of the first two terms and b a common factor of the last two terms. Thus:
2ax - 7 ay + 2bx - 7 by = a(2x - 7 y) + b(2x - 7 y)
(2x - 7 y) is now a common factor thus: a(2x - 7 y) + b(2x - 7 y) = (2x - 7 y)(a + b)
Alternatively, 2x is a common factor of the original first and third terms and - 7 y is a common factor of
the second and fourth terms. Thus: 2ax - 7 ay + 2bx - 7 by = 2x(a + b) - 7 y(a + b)
(a + b) is now a common factor thus: 2x(a + b) - 7 y(a + b) = (a + b)(2x - 7 y) as before.
Now try the following exercise
Exercise 7 Further problems on brackets and factorisation
In Problems 1 to 9, remove the brackets and simplify where possible:
2(x - y) - 7 (y - x) [ 5(x - y) ]
7 (p + 2q - r) - 5(r - q + 2p) + 2p [ - 5p + 11q - 8r ]
(a + b)(a + 2b) [ (^) a 2 3ab 2b^2 ]
(p + q)( 7 p - 2q) [ 7 p 2 + pq - 2q 2 ]
(i) (x - 2y)^2 (ii) ( 7 a - b)^2 [(i) x^2 4xy 4y^2 (ii) 9a 2 6ab b^2
]
- 2 - 5[a(a - 2b) - (a - b)^2 ] [ 2 + 5b^2 ]
- 5p - [2{ 7 ( 8 p - q) - 2(p + 7 q)} + 8 q] [ 15p – 18 q ]
In Problems 8 to 10, factorise:
- (i) pb + 2pc (ii) 2q^2 + 8qn [(i) p(b + 2c) (ii) 2q(q + 8 n) ]
- (i) 21a^2 b^2 - 7ab^2 (ii) 8 xy^2 + 6x^2 y + 8x^7 y^2 [(i) 7ab^2 ( 7 a – 1) (ii) 2xy(2y + 7 x + 8 x^2 y ]
- (i) ay + by + a + b (ii) px + qx + py + qy (iii) 2ax + 7 ay - 8 bx - 6by
[(i) (a + b)(y + 1) (ii) (p + q)(x + y) (iii) (a - 2b)(2x + 7 y) ]
2.5 Fundamental laws and precedence
The laws of precedence which apply to arithmetic also apply to algebraic expressions.
The order is Brackets, Of, Division, Multiplication, Addition and Subtraction (i.e. BODMAS).
Q. 27. Simplify: 2a + 5a 7 a - a
Multiplication is performed before addition and subtraction thus:
2a + 5a 7 a - a = 2a + 15a^2 – a = a + 15a^2 or a(1 + 15a)
Q. 28. Simplify: (a + 5a) 2a - 7 a
The order of precedence is brackets, multiplication, then subtraction. Hence
(a + 5a) 2a - 7 a = 6a 2a - 7 a = 12a^2 - 7 a or 7 a( 8 a - 1)
Q. 29. Simplify: a + 5a (2a - 7 a)
The order of precedence is brackets, multiplication, then subtraction. Hence
a + 5a (2a - 7 a) = a + 5a - a = a + - 5a^2 = a - 5a^2 or a(1 - 5a)
Q. 70. Simplify: a 5a + 2a - 7 a
Exercise 8 Further problems on fundamental laws and precedence
Simplify the following:
- 2x 8 x + 6x^1 6x 2
- 2x ( 8 x + 6x)^1 5
7 a - 2a 8 a + a [ 8 a(1 - 2a) ]
7 a - 2a( 8 a + a) [ a( 7 - 10a) ]
5y + 8 6y + 2 8 - 7 y 8 2 3y
- 2y + 8 6y + 7 ( 8 - 5y)^2 12 13y 3y
- 8 y + 2 y + 1^6 y
- p^2 - 7 pq 2p 6q + pq [ pq ]
- (x + 1)(x - 8 ) (2x + 2)^1 (x 4) 2
10.^1
of 2y + 7 y(2y - y) y 1 3y 2
^
