Chapter 10

Further algebra

10.1 Introduction

In this chapter, the use of brackets and factorization

with algebra is explained, together with further practice

with the laws of precedence. Understanding of these

topics is often necessary when solving and transposing

equations.

10.2 Brackets

With algebra

(a) 2(a+b)= 2 a+ 2 b

(b) (a+b)(c+d)=a(c+d)+b(c+d)

=ac+ad+bc+bd

Here are some worked examples to help understanding

of brackets with algebra.

Problem 1. Determine 2b(a− 5 b)

2 b(a− 5 b)= 2 b×a+ 2 b×− 5 b

= 2 ba− 10 b^2

= 2 ab− 10 b^2 (note that 2bais the

same as 2ab)

Problem 2. Determine( 3 x+ 4 y)(x−y)

( 3 x+ 4 y)(x−y)= 3 x(x−y)+ 4 y(x−y)

= 3 x^2 − 3 xy+ 4 yx− 4 y^2

= 3 x^2 − 3 xy+ 4 xy− 4 y^2

(note that 4yxis the same as 4xy)

= 3 x^2 +xy− 4 y^2

Problem 3. Simplify 3( 2 x− 3 y)−( 3 x−y)

3 ( 2 x− 3 y)−( 3 x−y)= 3 × 2 x− 3 × 3 y− 3 x−−y

(Note that−( 3 x−y)=− 1 ( 3 x−y)and the

−1 multipliesbothterms in the bracket)

= 6 x− 9 y− 3 x+y

(Note:−×−=+)

= 6 x− 3 x+y− 9 y

= 3 x− 8 y

Problem 4. Remove the brackets and simplify the

expression(a− 2 b)+ 5 (b−c)− 3 (c+ 2 d)

(a− 2 b)+ 5 (b−c)− 3 (c+ 2 d)

=a− 2 b+ 5 ×b+ 5 ×−c− 3 ×c− 3 × 2 d

=a− 2 b+ 5 b− 5 c− 3 c− 6 d

=a+ 3 b− 8 c− 6 d

Problem 5. Simplify(p+q)(p−q)

(p+q)(p−q)=p(p−q)+q(p−q)

=p^2 −pq+qp−q^2

=p^2 −q^2

Problem 6. Simplify( 2 x− 3 y)^2

( 2 x− 3 y)^2 =( 2 x− 3 y)( 2 x− 3 y)

= 2 x( 2 x− 3 y)− 3 y( 2 x− 3 y)

= 2 x× 2 x+ 2 x×− 3 y− 3 y× 2 x

− 3 y×− 3 y

= 4 x^2 − 6 xy− 6 xy+ 9 y^2

(Note:+×−=−and−×−=+)

= 4 x^2 − 12 xy+ 9 y^2

DOI: 10.1016/B978-1-85617-697-2.00010-7