Classwork Series and Exercises {Mathematics – JSS1}: ALGEBRAIC SIMPLIFICATION 2: BRACKETS

JSS 1 Mathematics Second Term Week 4

Topic: ALGEBRAIC SIMPLIFICATION 2: BRACKETS

Multiplying and Dividing Algebraic Terms

a. Just as 5a is a short for 5 X a, so ab is short for a X b.

b. Just as 5 X 3 = 3 X 5,

so a X b = b X a.

It is usually to write the letters in alphabetical order, but would be just as correct as ab.

c. Just as 5² is short for 5 X 5, so a² is short for a X a and x³ is shorth for x X x X .

d. 4x + 4x + 4x = 12x

              3 X 4x = 12x

and 3x  + 3x + 3x + 3x = 12x

                            4 X 3x = 12x

Thus: 3 X 4x = 4 X 3x = 12x

The term 3, 4 and x can be multiplied in any order.

3 X 4x = 4 X 3x = 3x X 4 = 4x X 3

            = 4 X x X 3 = x X 3 X 4 = 12x

It is usual to write the numbers before the letters.

Example

We can multiply two algebraic terms to get a product, which is also an algebraic term.

Example :

Evaluate 3pq³× 4qr

Solution:

3pq³× 4qr
= 3 × p × q × q × q × 4 × q × r
= 3 × 4 × p × q × q × q × q × r
= 12 × p × q4 × r
= 12pq4r

Example :

Evaluate –2a³b× 3ab²c

Solution:

–2a³b× 3ab²c
= –2 × 3× a³× a × b × b²× c
= –6× a⁴× b³× c
= –6a⁴b³c

Example

Multiply the following terms.

2x²y • 6x²y

3ab • 5xy²

Answers

First multiply the numerical coefficients, then use the exponent rules.

2x²y • 6x²y = (2 • 6)(x²y • x²y) = 12x²⁺²y¹⁺¹ = 12x⁴y² 

3ab • 5xy² = (3 • 5)abxy² = 15abxy²

Division

In algebra, letters stand for numbers. Just as fractions can be reduced to their lowest terms by equal division of the numerator and denominator so a letter can be divided by the same letter.

Example

x ÷ x = 1, just as 3 ÷ 3 = 1.

Example

Simplify                 Working                                     result

14a/7                    = 7 X 2a/7 =1 X 2a/1                      = 2a

1/3 of 36x           = 36 X x/3 = 3 X 12x/3

                                 = 1 X 12x/1                                         = 12x

1/5 of y               does not simplify                               =1/5y or y/5

Example

Simplify                         working                                result

5ab ÷ a                    = 5 X a X b/a

                                   = 5 X 1 X b/1 = 5 X b                     =  5b

6xy/2y                   = 6 X x X y/2 X y

                                   = 3 X x X 1/1 X 1 =  3 X x             =  3x

X² ÷ x                      = X x X/X = 1 X x/1                        = x

24x²y ÷ 3xy         = 24 X x² X y/3 X x X y

                                    = 3 X 8 X x X 1/3 X 1 X 1              = 8x

                                    = 8 X x

Order of Operation

What is the value of 17 – 5 X 2? It is possible to get two different answers:

a. (17 – 5) X 2 = 12 X 2 = 24

b. 17 – (5 X 2) = 17 – 10 = 7

The answer depends on whether we do the subtraction first or the multiplication first. To avoid confusion, remember the following rules.

a. If there are no brackets do multiplication or division before addition or multiplication.

b. If there are brackets do the operations inside the brackets first.

Usually, operations involving multiplication and division are enclosed in brackets and done before addition and subtraction.

Use the word BODMAS to remember the correct order: Brackets, Of, Division, Multiplication, Addition, Subtraction.

Example

Find the value of 16 X 2 – 3 + 14 ÷ 7.

16 X 2 – 3 + 14 ÷ 7

= (16 X 2) – 3 + (14 ÷ 7)

= 32 – 2 + 2

= 34 – 3

= 31

Example

Simplify

Simplify 7 X 3a – (3a + 5a) X 2

7 X 3a – (3a + 5a) X 2 = 7 X 3a – 8a X 2

                                        = (7 X 3a) – (8a X 2)

                                        =21a – 16a = 5a

Removing Brackets

Always try to simplify the terms inside brackets first. If they will not simplify, remove the brackets.

Positive sign before a bracket

a. 9 + (5 + 2) = 9 + 7 = 16

also 9 + 5 + 2 = 14 + 2 = 16

thus 9 + (5 + 2) = 9 + 5 + 2

Similarly with letters,

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

b. 9 + (5 – 2) = 9 + 3 = 12

also 9 + 5 – 2 = 14 -2 = 12

thus 9 + (5 – 2) = 9 + 5 -2

Similarly with letters,

a + (b – c) = a + b – c

when there is a positive sign before a bracket:

the signs of the terms inside the bracket stay the same when it is removed.

Example

Simplify 7g + (3g + 4h).

7g + (3g + 4h) = 7g + 3g + 4h = 10g + 4h

Example

Simplify 13p + (6p – 3q).

13p + (6p – 3q) = 13p + 6p – 3q = 19p – 3q

Negative sign before a bracket

a. 9 – (5 + 2) = 9 – 7 = 2

The result is the same as first taking away 5, then taking away 2:

i.e. 9 – 5 – 2 = 4 – 2 = 2

thus 9 – (5 + 2) = 9 – 5 – 2 = 2.

Similarly with letters,

a – (b + c) = a – b – c.

b. 9 – (5 – 2) = 9 – 3 = 6

The result is the same as first taking away 5 and then adding 2:

i.e. 9 – 5 +2 = 4 + 2 = 6

thus 9 – (5 – 2) = 9 – 5 + 2.

Similarly with letters,

a – (b – c) = a – b + c.

When there is a negative sign before the bracket:

The signs of the terms inside the bracket are changed when the bracket is removed.

Example

Simplify 5a – (2a + 8).

5a – (2a + 8) = 5a – 2a – 8 = 3a – 8

Example

Simplify 10d – (9c – 4d)

10d – (9c – 4d) = 10d – 9c + 4d = 14d – 9c

Exercise

Simplify the following in two ways. The first example shows you how to do this

1. 9 – (7 – 3)

a. 9 – (7 – 3) = 9 -4 = 5

b. 9 – (7 – 3) = 9 – 7 + 3 = 2 + 3 = 5

2. 9 – (5 + 2)

3. 16 – (4 – 1)

4. 24 – (15 + 8)

5. 12 – (9 – 5)