Algebraic Expressions and Identities

Addition and Subtraction of Algebraic Expressions

In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are. Examples of expressions are:

                      x + 3, 2y – 5, 3x ² , 4xy + 7 etc.

In the earlier classes, we have also learnt how to add and subtract algebraic expressions.

For example, to add 7x ² – 4x + 5 and 9x – 10, we do

                                                    7x ² – 4x + 5

                                                                + 9x – 10

                                                         7x ² + 5x – 5

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

                                            7xy + 5yz –3zx

                                    + 4yz + 9zx – 4y

                          + –2xy – 3zx + 5x (Note xz is same as zx)

                               5xy + 9yz +3zx + 5x – 4y

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

Solution:                 

                                          7x ² – 4xy + 8y ² + 5x – 3y

                        5x ²          – 4y ²         + 6y – 3

                        (-)              (+)             (-)   (+)

                       2x ² – 4xy + 12y ² +5x – 9y + 3

Note that subtraction of a number is the same as addition of its additive inverse. Thus subtracting –3 is the same as adding +3. Similarly, subtracting 6y is the same as adding – 6y; subtracting – 4y ² is the same as adding 4y ² and so on. The signs in the third row written below each term in the second row help us in knowing which operation has to be performed.

Algebraic Expressions

Multiplication of Algebraic Expressions:Introduction
(i) Look at the following patterns of dots

(ii) Can you now think of similar other situations in which two algebraic expressions have to be multiplied? Ameena gets up. She says, “We can think of area of a rectangle.” The area of a rectangle is l × b, where l is the length, and b is breadth. If the length of the rectangle is increased by 5 units, i.e., (l + 5) and breadth is decreased by 3 units , i.e., (b – 3) units, the area of the new rectangle will be (l + 5) × (b – 3).

(iii) Can you think about volume? (The volume of a rectangular box is given by the product of its length, breadth and height).

(iv) Sarita points out that when we buy things, we have to carry out multiplication. For example, if price of bananas per dozen = ₹p

and for the school picnic bananas needed = z dozens,

then we have to pay = ₹p × z Suppose, the price per dozen was less by  ₹2 and the bananas needed were less by 4 dozens.

Then, price of bananas per dozen = ₹ (p – 2)

and bananas needed = (z – 4) dozens,

Therefore, we would have to pay = ₹ (p – 2) × (z – 4)

Algebraic Expressions

Multiplying a Monomial by a Monomial

Expression that contains only one term is called a monomial.

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

Some more useful examples follow.

(iv) 5x × 4x ² = (5 × 4) × (x × x ² ) = 20 × x ³ = 20x ³

(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz) = –20 × (x × x × yz) = –20x 2yz

Observe how we collect the powers of different variables in the algebraic parts of the two monomials. While doing so, we use the rules of exponents and powers.

Multiplying three or more monomials:

Observe the following examples.

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

(ii) 4xy × 5x ²y ² × 6x ³y ³ = (4xy × 5x 2y ² ) × 6x ³y ³= 20x 3y ³ × 6x ³y ³ = 120x ³y ³ × x ³y ³

                                                                                                             = 120 (x ³ × x ³ ) × (y ³ × y ³ ) = 120x ⁶ × y ⁶ = 120x ⁶ y ⁶

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.

We can find the product in other way also.

4xy × 5x ²y ² × 6x ³ y ³ = (4 × 5 × 6) × (x × x ² × x ³ ) × (y × y ² × y ³ ) = 120 x 6y

TRY THESE

Find 4x × 5y × 7z

First find 4x × 5y and multiply it by 7z; or

first find 5y × 7z and multiply it by 4x.

Is the result the same? What do you observe?

Does the order in which you carry out the multiplication matter?

Example 3: Complete the table for area of a rectangle with given length and breadth.

Solution: 

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

for (ii) volume = m²n × n ²p × p ²m

                     = (m² × m) × (n × n ² ) × (p × p ² ) = m³n ³p ³

for (iii) volume = 2q × 4q ² × 8q ³

                      = 2 × 4 × 8 × q × q ² × q ³ = 64q ⁶

Algebraic expressions and Identities

Multiplying a Monomial by a Polynomial

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.

Multiplying a monomial by a binomial:

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 × (y 2 + 3) = (5xy × y 2 ) + (5xy × 3) = 5xy3 + 15xy.

What about a binomial × monomial? For example, (5y + 2) × 3x = ?

We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a

Similarly,   (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.

Multiplying a monomial by a trinomial:

Consider 3p × (4p ² + 5p + 7).

As in the earlier case, we use distributive law; 3p × (4p ² + 5p + 7) = (3p × 4p ² ) + (3p × 5p) + (3p × 7) = 12p ³ + 15p ² + 21p

Multiply each term of the trinomial by the monomial and add products.

Observe, by using the distributive law, we are able to carry out the multiplication term by term.

TRY THESE

Find the product: (4p ² + 5p + 7) × 3p

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 = x ² – 3x + 2

For x = 1, x ² – 3x + 2 = (1)² – 3 (1) + 2 = 1 – 3 + 2 = 3 – 3 = 0

(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y ² – 21y – 3y + 12 – 63

                                                     = 6y ² – 24y – 51

For y = –2, 6y ² – 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 6m² – 13m (ii) 4y (3y ² + 5y – 7) and 2 (y ³ – 4y ² + 5)

Solution: (i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m²

Now adding the second expression to it,15m – 5m² + 6m² – 13m = m² + 2m

(ii) The first expression = 4y (3y ² + 5y – 7) = (4y × 3y ² ) + (4y × 5y) + (4y × (–7)) = 12y ³ + 20y ² – 28y

The second expression= 2 (y ³ – 4y ² + 5)= 2y ³ + 2 × (– 4y ² ) + 2 × 5

                                                             = 2y ³ – 8y ² + 10

Adding the two expressions,

                       12y ³ + 20y ² – 28y

                      + 2y ³ – 8y ²          + 10

                                14y ³ + 12y ² – 28y + 10  

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

Solution: We have 3pq (p – q) = 3p ²q – 3pq² and

2pq (p + q) = 2p ²q + 2pq²

Subtracting, 2p ²q + 2pq²

                  3p ²q – 3pq²

                   –        +

                         -p²q + 5pq²

Algebraic Expressions

Multiplying a Polynomial by a Polynomial

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)

= 6a ² + 9ab + 8ba + 12b ²

= 6a ² + 17ab + 12b ² (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 : 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)

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

Example : Multiply (i) (a + 7) and (b – 5) (ii) (a 2 + 2b 2 ) and (5a – 3b)

Solution: (i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5) = ab – 5a + 7b – 35

Note that there are no like terms involved in this multiplication.

(ii) (a ² + 2b² ) × (5a – 3b) = a ² (5a – 3b) + 2b ² × (5a – 3b)

                                               = 5a ³ – 3a ²b + 10ab² – 6b ³

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. Consider

( a+7) (binomial) \({{a}^{2}}+3a+5\)(trinomial)= a × (a ² + 3a + 5) + 7 × (a ² + 3a + 5) [using the distributive law]

= a ³ + 3a ² + 5a + 7a ² + 21a + 35

= a ³ + (3a ² + 7a ² ) + (5a + 21a) + 35

= a ³ + 10a ² + 26a + 35      (Why are there only 4 terms in the final result?)

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)

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

                                                                  = 2a ² – ab – 3b ² + bc + ac – 2ac + 3bc

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

                                                                   = 2a ² – 3b ² – ab + 4bc – ac