Algebraic Expressions

INTRODUCTION

We have already come across simple algebraic expressions like x + 3, y – 5, 4x + 5, 10y – 5 and so on. In Class VI, we have seen how these expressions are useful in formulating puzzles and problems. We have also seen examples of several expressions in the chapter on simple equations.

Expressions are a central concept in algebra. This Chapter is devoted to algebraic expressions. When you have studied this Chapter, you will know how algebraic expressions are formed, how they can be combined, how we can find their values and how they can be used.

HOW ARE EXPRESSIONS FORMED?

 We now know very well what a variable is. We use letters x, y, l, m, ... etc. to denote variables. A variable can take various values. Its value is not fixed. On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, –17, etc.

We combine variables and constants to make algebraic expressions. For this, we use the operations of addition, subtraction, multiplication and division. We have already come across expressions like 4x + 5, 10y – 20. The expression 4x + 5 is obtained from the variable x, first by multiplying x by the constant 4 and then adding the constant 5 to the product. Similarly, 10y – 20 is obtained by first multiplying y by 10 and then subtracting 20 from the product.

The above expressions were obtained by combining variables with constants. We can also obtain expressions by combining variables with themselves or with other variables. Look at how the following expressions are obtained: x² , 2y² , 3x² – 5, xy, 4xy + 7

(i) The expression x\(^2\) is obtained by multiplying the variable x by itself;

\(x\times x={{x}^{2}}\)

 Just as 4 × 4 is written as 4\(^2\) , we write \(x\times x={{x}^{2}}\). It is commonly read as x squared.

(Later, when you study the chapter ‘Exponents and Powers’ you will realise that x² may also be read as x raised to the power 2).

In the same manner, we can write \(x\times x\times x={{x}^{3}}\)

Commonly, x³ is read as ‘x cubed’. Later, you will realise that x³ may also be read as x raised to the power 3.

x, x² , x³ , ... are all algebraic expressions obtained from x.

(ii) The expression 2y² is obtained from y: 2y² = 2 × y × y Here by multiplying y with y we obtain y² and then we multiply y² by the constant 2.

(iii) In (3x ² – 5) we first obtain x ² , and multiply it by 3 to get 3x² . From 3x ² , we subtract 5 to finally arrive at 3x² – 5.

(iv) In xy, we multiply the variable x with another variable y. Thus, x × y = xy.

 (v) In 4xy + 7, we first obtain xy, multiply it by 4 to get 4xy and add 7 to 4xy to get the expression

Algebraic Expressions

TERMS OF AN EXPRESSION

We shall now put in a systematic form what we have learnt above about how expressions are formed. For this purpose, we need to understand what terms of an expression and their factors are. Consider the expression (4x + 5). In forming this expression, we first formed 4x separately as a product of 4 and x and then added 5 to it. Similarly consider the expression (3x ² + 7y). Here we first formed 3x ² separately as a product of 3, x and x. We then formed 7y separately as a product of 7 and y. Having formed 3x 2 and 7y separately, we added them to get the expression. You will find that the expressions we deal with can always be seen this way. They have parts which are formed separately and then added. Such parts of an expression which are formed separately first and then added are known as terms. Look at the expression (4x ² – 3xy). We say that it has two terms, 4 ² and –3xy. The term 4x² is a product of 4, x and x, and the term (–3xy) is a product of (–3), x and y.

Terms are added to form expressions. Just as the terms 4x and 5 are added to form the expression (4x + 5), the terms

4x ² and (–3xy) are added to give the expression (4x ² – 3xy). This is because 4x ² + (–3xy) = 4x² – 3xy

Factors of a term

We saw above that the expression (4x ² – 3xy) consists of two terms 4x ² and –3xy. The term 4x 2 is a product of 4, x and x; we say that 4, x and x are the factors of the term 4x ² . A term is a product of its factors. The term –3xy is a product of the factors –3, x and y.

We can represent the terms and factors of the terms of an expression conveniently and elegantly by a tree diagram. The tree for the expression (4x ² – 3xy) is as shown in the adjacent figure.

Let us draw a tree diagram for the expression 5xy + 10. The factors are such that they cannot be further factorised. Thus we do not write 5xy as

 5 × xy, because xy can be further factorised. Similarly, if x ³ were a term, it would be written as x\({{x}^{2}}\times x\) and not \( {{x}^{2}}\times x\). Also, remember that 1 is not taken as a separate factor.

Coefficients

We have learnt how to write a term as a product of factors. One of these factors may be numerical and the others algebraic (i.e., they contain variables). The numerical factor is said to be the numerical coefficient or simply the coefficient of the term. It is also said to be the coefficient of the rest of the term (which is obviously the product of algebraic factors of the term). Thus in 5xy, 5 is the coefficient of the term. It is also the coefficient of xy. In the term 10xyz, 10 is the coefficient of xyz, in the term –7x ²y ² , –7 is the coefficient of x ²y ² .

When the coefficient of a term is +1, it is usually omitted. For example, 1x is written as x; 1 x² y ² is written as x ² y ² and so on. Also, the coefficient (–1) is indicated only by the minus sign. Thus (–1) \(x\) is written as – x; (–1) x ² y ² is written as – x² y ² and so on.

Sometimes, the word ‘coefficient’ is used in a more general way. Thus we say that in the term 5xy, 5 is the coefficient of xy, x is the coefficient of 5y and y is the coefficient of 5x. In 10xy² , 10 is the coefficient of xy² , x is the coefficient of 10y ² and

y ² is the coefficient of 10x. Thus, in this more general way, a coefficient may be either a numerical factor or an algebraic factor or a product of two or more factors. It is said to be the coefficient of the product of the remaining factors.

EXAMPLE :Identify, in the following expressions, terms which are not constants. Give their numerical coefficients: xy + 4, 13 – y ² , 13 – y + 5y ² , 4p 2q – 3pq² + 5

SOLUTION

EXAMPLE : (a) What are the coefficients of x in the following expressions? 4x – 3y, 8 – x + y, y 2 x – y,2z – 5xz

(b) What are the coefficients of y in the following expressions? 4x – 3y, 8 + yz, yz² + 5, my + m

SOLUTION (a) In each expression we look for a term with x as a factor. The remaining part of that term is the coefficient of x

(b) The method is similar to that in (a) above

Algebraic Expressions

LIKE AND UNLIKE TERMS

When terms have the same algebraic factors, they are like terms. When terms have different algebraic factors, they are unlike terms. For example, in the expression 2xy – 3x + 5xy – 4, look at the terms 2xy and 5xy. The factors of 2xy are 2, x and y. The factors of 5xy are 5, x and y. Thus their algebraic (i.e., those which contain variables) factors are the same and hence they are like terms. On the other hand the terms 2xy and –3x, have different algebraic factors. They are unlike terms. Similarly, the terms, 2xy and 4, are unlike terms. Also, the terms –3x and 4 are unlike terms.

Algebraic Expressions

MONOMIALS, BINOMIALS, TRINOMIALS AND POLYNOMIALS

 An expression with only one term is called a monomial; for example, 7xy, – 5m, 3z 2 , 4 etc. An expression which contains two unlike terms is called a binomial; for example, x + y, m – 5, mn + 4m,

a ² – b ² are binomials. The expression 10pq is not a binomial; it is a monomial. The expression (a + b + 5) is not a binomial. It contains three terms

An expression which contains three terms is called a trinomial; for example, the expressions x + y + 7,

ab + a +b, 3x 2 – 5x + 2, m + n + 10 are trinomials. The expression ab + a + b + 5 is, however not a trinomial; it contains four terms and not three. The expression x + y + 5x is not a trinomial as the terms x and 5x are like terms.

In general, an expression with one or more terms is called a polynomial. Thus a monomial, a binomial and a trinomial are all polynomials.

EXAMPLE:State with reasons, which of the following pairs of terms are of like terms and which are of unlike terms: (i) 7x, 12y (ii) 15x, –21x (iii) – 4ab, 7ba (iv) 3xy, 3x (v) 6xy² , 9x ² y (vi) pq² , – 4pq²

(vii) mn² , 10mn

SOLUTION

Following simple steps will help you to decide whether the given terms are like or unlike terms: (i) Ignore the numerical coefficients. Concentrate on the algebraic part of the terms. (ii) Check the variables in the terms. They must be the same. (iii) Next, check the powers of each variable in the terms. They must be the same. Note that in deciding like terms, two things do not matter (1) the numerical coefficients of the terms and (2) the order in which the variables are multiplied in the terms

Algebraic Expressions

FINDING THE VALUE OF AN EXPRESSION

We know that the value of an algebraic expression depends on the values of the variables forming the expression. There are a number of situations in which we need to find the value of an expression, such as when we wish to check whether a particular value of a variable satisfies a given equation or not. We find values of expressions, also, when we use formulas from geometry and from everyday mathematics. For example, the area of a square is  \(l\)² , where l is the length of a side of the square. If \(l\) = 5 cm., the area is 52 cm² or 25 cm² ; if the side is 10 cm, the area is 10² cm² or 100 cm² and so on. We shall see more such examples in the next section.

EXAMPLE : Find the values of the following expressions for x = 2. (i) x + 4 (ii) 4x – 3 (iii) 19 – 5x ²

(iv) 100 – 10x ³

SOLUTION :Putting x = 2

 (i) In x + 4, we get the value of x + 4,

i.e., x + 4 = 2 + 4 = 6

(ii) In 4x – 3, we get 4x – 3 = (4 × 2) – 3 = 8 – 3 = 5

(iii) In 19 – 5x ² , we get 19 – 5x ² = 19 – (5 × 2² ) = 19 – (5 × 4) = 19 – 20 = – 1

 (iv) In 100 – 10x ³ , we get 100 – 10x ³ = 100 – (10 × 23 ) = 100 – (10 × 8)

 (Note 2³ = 8) = 100 – 80 = 20

EXAMPLE :Find the value of the following expressions when n = – 2. (i) 5n – 2 (ii) 5n ² + 5n – 2

 (iii) n ³ + 5n ² + 5n – 2

 SOLUTION (i) Putting the value of n = – 2, in 5n – 2, we get, 5(– 2) – 2 = – 10 – 2 = – 12

 (ii) In 5n ² + 5n – 2, we have, for n = –2, 5n – 2 = –12 and 5n ² = 5 × (– 2)² = 5 × 4 = 20 [as (– 2)2 = 4] Combining

5n ² + 5n – 2 = 20 – 12 = 8

 (iii) Now, for n = – 2, 5n ² + 5n – 2 = 8 and n ³ = (–2)³ = (–2) × (–2) × (–2) = – 8

Combining,

n ³ + 5n ² + 5n – 2 = – 8 + 8 = 0

We shall now consider expressions of two variables, for example, x + y, xy. To work out the numerical value of an expression of two variables, we need to give the values of both variables. For example, the value of (x + y), for x = 3 and y = 5, is

3 + 5 = 8.

EXAMPLE :Find the value of the following expressions for a = 3, b = 2. (i) a + b (ii) 7a – 4 (iii) a ² + 2ab + b ² (iv) a ³ – b ³

 SOLUTION :Substituting a = 3 and b = 2 in

 (i) a + b, we get a + b = 3 + 2 = 5

ii) 7a – 4b, we get 7a – 4b = 7 × 3 – 4 × 2 = 21 – 8 = 13.

(iii) a ² + 2ab + b ² , we get a ² + 2ab + b 2 = 3² + 2 × 3 × 2 + 2² = 9 + 2 × 6 + 4 = 9 + 12 + 4 = 25

 (iv) a ³ – b ³ , we get a ³ – b ³ = 3³ – 2³ = 3 × 3 × 3 – 2 × 2 × 2 = 9 × 3 – 4 × 2 = 27 – 8 = 19