**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 5^{2 }is short for 5 X 5, so a^{2 }is short for a X a and x^{3 }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 3*pq*^{3}*×* 4*qr*

**Solution:**

3*pq*^{3}*×* 4*qr
*= 3

*× p × q × q × q ×*4

*× q × r*

= 3 × 4 × p × q × q × q × q × r

= 12 × p × q4 × r

= 12pq4r

**Example :**

Evaluate –2a^{3}b× 3ab^{2}c

**Solution:**

–2a^{3}b× 3ab^{2}c

= –2 × 3× a^{3}× a × b × b^{2}× c

= –6× a^{4}× b^{3}× c

= –6a^{4}b^{3}c

**Example**

Multiply the following terms.

2*x*^{2}*y* • 6*x*^{2}*y*

3*ab* • 5*xy*^{2}

**Answers**

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

*2x ^{2}y • 6x^{2}y =* (2 • 6)(

*x*

^{2}

*y*•

*x*

^{2}

*y*) = 12

*x*

^{2+2}

*y*

^{1+1}= 12

*x*

^{4}

*y*

^{2 }

*3ab • 5xy ^{2} =* (3 • 5)

*abxy*

^{2}= 15

*abxy*

^{2}

**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

X^{2 }÷ x = X x X/X = 1 X x/1 = x

24x^{2}y ÷ 3xy = 24 X x^{2 }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: **B**rackets, **O**f, **D**ivision, **M**ultiplication, **A**ddition, **S**ubtraction.

**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)