Algebraic Expressions and Identities

9.1 What are Expressions?

Examples of expressions are:

etc

You can form many more expressions. As you know expressions are formed from variables and constants. The expression 2y – 5 is formed from the variable y and constants 2 and 5. The expression 4xy + 7 is formed from variables x and y and constants 4 and 7.

Number line and an expression:

Consider the expression x + 5. Let us say the variable x has a position X on the number line;

X may be anywhere on the number line, but it is definite that the value of x + 5 is given by a point P, 5 units to the right of X. Similarly, the value of x – 4 will be 4 units to the left of X and so on.

What about the position of 4x and 4x + 5?

The position of 4x will be point C; the distance of C from the origin will be four times the distance of X from the origin. The position D of 4x + 5 will be 5 units to the right of C.

9.2 Terms, Factors and Coefficients

Take the expression 4x + 5. This expression is made up of two terms, 4x and 5. Terms are added to form expressions. Terms themselves can be formed as the product of factors. The term 4x is the product of its factors 4 and x. The term 5 is made up of just one factor,

i.e., 5.

The expression 7xy – 5x has two terms 7xy and –5x. The term 7xy is a product of factors 7, x and y. The numerical factor of a term is called its numerical coefficient or simply coefficient.

The coefficient in the term 7xy is 7 and the coefficient in the term –5x is –5.

9.3 Monomials, Binomials and Polynomials

● Expression that contains only one term is called a monomial.

● Expression that contains two terms is called a binomial.

● An expression containing three terms is a trinomial and so on.

● In general, an expression containing, one or more terms with non-zero coefficient (with variables having non negative integers as exponents) is called a polynomial.

● A polynomial may contain any number of terms, one or more than one.

Examples of monomials:

, etc.

Examples of binomials:

, etc.

Examples of trinomials:

a + b + c, 2x + 3y – 5, , etc.

Examples of polynomials:

a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.

9.4 Like and Unlike Terms

Look at the following expressions:

7x, 14x, –13x, 5 , 7y, 7xy, –9 , –9 , –5yx

Like terms from these are:

(i) 7x, 14x, –13x are like terms.

(ii) 5 and –9 are like terms.

(iii) 7xy and –5yx are like terms.

Why are 7x and 7y not like?

Why are 7x and 7xy not like?

Why are 7x and not like?

9.5 Addition and Subtraction of Algebraic Expressions

In the earlier classes, we have also learnt how to add and subtract algebraic expressions. For example, to add 7 – 4x + 5 and 9x – 10, we do

Observe how we do the addition. We write each expression to be added in a separate row. While doing so we write like terms one below the other, and add them, as shown.

Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9) x = 5x.

Let us take some more examples.

Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y, –3xz + 5x – 2xy

Solution:

Writing the three expressions in separate rows, with like terms one below the other, we have

Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y.

Note how the terms, – 4y in the second expression and 5x in the third expression, are carried over as they are, since they have no like terms in the other expressions.

Example 2: Subtract 5 – 4 + 6y – 3 from 7 – 4xy + 8 + 5x – 3y.

Solution:

9.6 Multiplication of Algebraic Expressions:

Introduction (i) Look at the following patterns of dots.

9.7 Multiplying a Monomial by a Monomial

9.7.1 Multiplying two monomials

We begin with 4 × x = x + x + x + x = 4x as seen earlier.

Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x

Now, observe the following products.

(i) X × 3y = x × 3 × y = 3 × x × y = 3xy

(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy

(iii) 5x × (–3y) = 5 × x × (–3) × y = 5 × (–3) × x × y = –15xy

(iv) 5x × 4 = (5 × 4) × (x × ) = 20 × = 20

Note that 5 × 4 = 20 i.e., coefficient of product = coefficient of first monomial × coefficient of second monomial; and x × =

i.e., algebraic factor of product = algebraic factor of first monomial × algebraic factor of second monomial.

(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz) = –20 × (x × x × yz) = –20yz

9.7.2 Multiplying three or more monomials

Observe the following examples.

(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz

(ii) 4xy × 5 × 6

= (4xy × 5 ) × 6

= 20 × 6

= 120 ×

= 120 ( × ) × ( × )

= 120 × = 120

It is clear that we first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.

Example 4: Find the volume of each rectangular box with given length, breadth and height.

Solution:

Volume = length × breadth × height Hence,

For (i) volume = (2ax) × (3by) × (5cz) = 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz

(ii) Volume =n ×p ×m = ( × m) × (n × ) × (p × ) = for (iii) volume = 2q × 4 × 8 = 2 × 4 × 8 × q × × = 64

9.8 Multiplying a Monomial by a Polynomial

9.8.1 Multiplying a monomial by a binomial

We commonly use distributive law in our calculations.

For example:

1) 7 × 106 = 7 × (100 + 6)

= 7 × 100 + 7 × 6= 700 + 42 = 742

(Here, we used distributive law)

2) 7 × 38 = 7 × (40 – 2)

= 7 × 40 – 7 × 2 = 280 – 14 = 266

(Here, we used distributive law)

➔ Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) =?

➔ Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law, 3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x

➔ Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x and

➔ 5xy × ( + 3) = (5xy × ) + (5xy × 3) = 5x + 15xy

Example 5: Simplify the expressions and evaluate them as directed:

(i) X (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2

Solution:

(i) X (x – 3) + 2 = – 3x + 2

For x = 1, – 3x + 2

= (1)2 – 3 (1) + 2

= 1 – 3 + 2 = 3 – 3 = 0

(ii) 3y (2y – 7) – 3 (y – 4) – 63

= 6 – 21y – 3y + 12 – 63

= 6 – 24y – 51

For y = –2, 6 – 24y – 51

= 6 (–2)2 – 24(–2) – 51

= 6 × 4 + 24 × 2 – 51

= 24 + 48 – 51 = 72 – 51 = 21

Example 6:

Add (i) 5m (3 – m) and 6 – 13m

(ii) 4y (3 + 5y – 7) and 2 (– + 5)

Solution:

(i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5 Now adding the second expression to it, 15m – 5 + 6 – 13m = + 2m (ii) The first expression

= 4y (3 + 5y – 7)

= (4y × 3 ) + (4y × 5y) + (4y × (–7))

= 12 + 20 – 28y

The second expression

= 2 ( – 4 + 5)

= 2 + 2 × (– 4 ) + 2 × 5

= 2– 8 + 10

Example 7: Subtract 3pq (p – q) from 2pq (p + q).

Solution:

We have 3pq (p – q) = 3q – 3p

And

2pq (p + q) = 2q + 2p

9.9 Multiplying a Polynomial by a Polynomial

9.9.1 Multiplying a binomial by a binomial

Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b).

We do this step-by-step, as we did in earlier cases, following the distributive law of multiplication,

(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6+ 9ab + 8ba + 12

= + 17ab + 12 (Since ba = ab)

When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be present. But two of these are like terms, which are combined, and hence we get 3 terms. In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.

Example 8: Multiply

(i) (x – 4) and (2x + 3)

(ii) (x – y) and (3x + 5y)

Solution:

(i) (x – 4) × (2x + 3)

= x × (2x + 3) – 4 × (2x + 3)

= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3)

= 2 + 3x – 8x – 12 = 2 – 5x – 12 (Adding like terms)

(ii) (x – y) × (3x + 5y)

= x × (3x + 5y) – y × (3x + 5y)

= (x × 3x) + (x × 5y) – (y × 3x) – (y × 5y)

= 3x 2 + 5xy – 3yx – 5

= 3 + 2xy – 5 (Adding like terms)

9.9.2 Multiplying a binomial by a trinomial

In this multiplication, we shall have to multiply each of the three terms in the trinomial by each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may reduce to 5 or less, if the term by term multiplication results in like terms

Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.

Solution:

We have

(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)

= 2 – 3ab + ac + 2ab – 3 + bc

= 2a 2 – ab – 3b 2 + bc + ac (Note, –3ab and 2ab are like terms)

and (2a – 3b) c = 2ac – 3bc

Therefore,

(a + b) (2a – 3b + c) – (2a – 3b) c

= 2 – ab – 3 + bc + ac – (2ac – 3bc)

= 2 – ab – 3 + bc + ac – 2ac + 3bc

= 2 – ab – 3 + (bc + 3bc) + (ac – 2ac)

= 2 – 3 – ab + 4bc – ac

9.10 What is an Identity?

Consider the equality (a + 1) (a +2) = + 3a + 2

We shall evaluate both sides of this equality for some value of a, say a = 10.

For a = 10,

LHS = (a + 1) (a + 2)

= (10 + 1) (10 + 2)

= 11 × 12 = 132

RHS = + 3a + 2

= 102 + 3 × 10 + 2

= 100 + 30 + 2

= 132

Thus, the values of the two sides of the equality are equal for a = 10.

Let us now take a = –5

LHS = (a + 1) (a + 2)

= (–5 + 1) (–5 + 2)

= (– 4) × (–3)

= 12 RHS

= + 3a + 2

= (–5)2 + 3 (–5) + 2

= 25 – 15 + 2

= 10 + 2 = 12

Thus, for a = –5, also LHS = RHS.

We shall find that for any value of a, LHS = RHS. Such an equality, true for every value of the variable in it, is called an identity.

Thus, (a + 1) (a + 2) = + 3a + 2 is an identity.

An equation is true for only certain values of the variable in it. It is not true for all values of the variable.

9.11 Standard Identities

We shall now study three identities which are very useful in our work. These identities are obtained by multiplying a binomial by another binomial.

Let us first consider the product (a + b) (a + b) or (a + b .

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

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

= + ab + ba + b 2

= + 2ab + (since ab = ba)

Thus (a + b = + 2ab + ….(I)

Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication. One may verify that for any value of a and any value of b, the values of the two sides are equal.

• Next we consider (a – b = (a – b) (a – b) = a (a – b) – b (a – b)

We have = – ab – ba +

= – 2ab + or (a – b

= – 2ab +…. (II)

• Finally, consider (a + b) (a – b).

We have (a + b) (a – b)

= a (a – b) + b (a – b)

= – ab + ba – = – (since ab = ba) or (a + b) (a – b)

= – ... (III)

The identities (I), (II) and (III) are known as standard identities.

Example 12:

Using Identity (II), find (i) (4p – 3q (ii) (4.9

Solution:

(i) (4p – 3q = (4p – 2 (4p) (3q) + (3q [Using the Identity (II)]

= 16 – 24pq + 9

Do you agree that for squaring (4p – 3q the method of identities is quicker than the direct method?

(ii) (4.9 = (5.0 – 0.1

= (5.0 – 2 (5.0) (0.1) + (0.1

= 25.00 – 1.00 + 0.01 = 24.01

Is it not that, squaring 4.9 using Identity (II) is much less tedious than squaring it by direct multiplication?

Example 13: Using Identity (III), find (i) (m+n) (m-n)

Solution:

(i) (m+n) (m-n)

= (m- (n

=-