Algebraic Expressions

Welcome dear students! Today we are going to learn about Algebraic Expressions from Class 7 Maths. We have already come across simple algebraic expressions like x + 3, y – 5, 4x + 5, and 10y – 5. In Class 6, we saw how these expressions are useful in formulating puzzles and problems. We also saw 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. [CHECKPOINT]

Now let us move on to section 10.2, How are expressions formed? We know very well what a variable is. We use letters x, y, l, m, and so on 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, and –17. 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, and 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. [CHECKPOINT]

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, and 4xy + 7. First, the expression x² is obtained by multiplying the variable x by itself. x × x = x². Just as 4 × 4 is written as 4², we write x × x = x². It is commonly read as x square. Later, when you study 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 × x × x = x³. Commonly, x³ is read as x cube. x, x², x³, and so on, are all algebraic expressions obtained from x. [CHECKPOINT]

Second, 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. Third, in 3x² – 5, we first obtain x², and multiply it by 3 to get 3x². From 3x², we subtract 5 to finally arrive at 3x² – 5. Fourth, in xy, we multiply the variable x with another variable y. Thus, x × y = xy. Fifth, in 4xy + 7, we first obtain xy, multiply it by 4 to get 4xy, and add 7 to 4xy to get the expression. [CHECKPOINT]

Let us move to section 10.3, Terms of an Expression. We shall now put in a systematic form what we have learnt 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 3x² + 7y. Here we first formed 3x² separately as a product of 3, x, and x. We then formed 7y separately as a product of 7 and y. Having formed 3x² 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. [CHECKPOINT]

Look at the expression 4x² – 3xy. We say that it has two terms, 4x² 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 4x² and –3xy are added to give the expression 4x² – 3xy. This is because 4x² + (–3xy) = 4x² – 3xy. Note, the minus sign is included in the term. In the expression 4x² – 3xy, we took the term as –3xy and not as 3xy. That is why we do not need to say that terms are added or subtracted to form an expression. Just added is enough. [CHECKPOINT]

Now let us discuss Factors of a term. We saw above that the expression 4x² – 3xy consists of two terms, 4x² and –3xy. The term 4x² is a product of 4, x, and x. We say that 4, x, and x are the factors of the term 4x². 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. In the tree diagram for the expression 4x² – 3xy, we use dotted lines for factors and continuous lines for terms to avoid mixing them. The top of the tree splits into two continuous branches for the terms 4x² and –3xy. From 4x², dotted lines branch down to its factors: 4, x, and x. From –3xy, dotted lines branch down to its factors: –3, x, and y. [CHECKPOINT]

Let us draw a tree diagram for the expression 5xy + 10. The top splits into terms 5xy and 10. From 5xy, dotted lines go to factors 5, x, and y. From 10, a dotted line goes to the factor 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 x³ were a term, it would be written as x × x × x and not x² × x. Also, remember that 1 is not taken as a separate factor. Now, try these on your own. Describe how the following expressions are obtained: 7xy + 5, x²y, and 4x² – 5x. For 7xy + 5, first multiply x and y to get xy, multiply by 7 to get 7xy, then add 5. For x²y, multiply x by x to get x², then multiply by y. For 4x² – 5x, first get x², multiply by 4 to get 4x². Separately, multiply 5 by x to get 5x. Finally, subtract 5x from 4x². [CHECKPOINT]

Next, we learn about 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. 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. 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 –7x²y², –7 is the coefficient of x²y². When the coefficient of a term is +1, it is usually omitted. For example, 1x is written as x. 1x²y² is written as x²y². Also, the coefficient –1 is indicated only by the minus sign. Thus –1 × x is written as –x. 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 10y², and y² is the coefficient of 10x. 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. [CHECKPOINT]

Let us solve Example 1. Identify, in the following expressions, terms which are not constants. Give their numerical coefficients. The expressions are xy + 4, 13 – y², 13 – y + 5y², and 4p²q – 3pq² + 5. For xy + 4, the non-constant term is xy, and its coefficient is 1. For 13 – y², the non-constant term is –y², and its coefficient is –1. For 13 – y + 5y², the non-constant terms are –y and 5y². Their coefficients are –1 and 5 respectively. For 4p²q – 3pq² + 5, the non-constant terms are 4p²q and –3pq². Their coefficients are 4 and –3. [CHECKPOINT]

Now, try these. What are the terms in the following expressions? Show how the terms are formed. Draw a tree diagram for each expression: 8y + 3x², 7mn – 4, and 2x²y. For 8y + 3x², the terms are 8y and 3x². 8y is formed by multiplying 8 and y. 3x² is formed by multiplying 3, x, and x. The tree diagram has a top splitting to 8y and 3x². 8y branches to 8 and y. 3x² branches to 3, x, and x. For 7mn – 4, the terms are 7mn and –4. 7mn is formed by multiplying 7, m, and n. The tree splits to 7mn and –4. 7mn branches to 7, m, and n. For 2x²y, there is only one term: 2x²y. It is formed by multiplying 2, x, x, and y. The tree has one branch to 2x²y, which splits to 2, x, x, and y. Second try this: Write three expressions each having 4 terms. For example, 2x + 3y – 4z + 5, a² + b² + c² + d², and 5p – 2q + 3r – 1. [CHECKPOINT]

Identify the coefficients of the terms of the following expressions: 4x – 3y, a + b + 5, 2y + 5, and 2xy. For 4x – 3y, the terms are 4x and –3y. Coefficients are 4 and –3. For a + b + 5, terms are a, b, and 5. Coefficients of a and b are 1. For 2y + 5, terms are 2y and 5. Coefficient of 2y is 2. For 2xy, the single term is 2xy, and its coefficient is 2. [CHECKPOINT]

Let us move to Example 2. Part a: What are the coefficients of x in the following expressions? 4x – 3y, 8 – x + y, y²x – y, and 2z – 5xz. We look for the term with x as a factor. The remaining part is the coefficient. For 4x – 3y, the term is 4x, coefficient is 4. For 8 – x + y, the term is –x, coefficient is –1. For y²x – y, the term is y²x, coefficient is y². For 2z – 5xz, the term is –5xz, coefficient is –5z. Part b: What are the coefficients of y in 4x – 3y, 8 + yz, yz² + 5, and my + m? For 4x – 3y, term is –3y, coefficient is –3. For 8 + yz, term is yz, coefficient is z. For yz² + 5, term is yz², coefficient is z². For my + m, term is my, coefficient is m. [CHECKPOINT]

Now, section 10.4, 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 2xy – 3x + 5xy – 4, look at 2xy and 5xy. The factors of 2xy are 2, x, and y. The factors of 5xy are 5, x, and y. Their algebraic factors are the same, so they are like terms. On the other hand, 2xy and –3x have different algebraic factors. They are unlike terms. Similarly, 2xy and 4 are unlike terms. Also, –3x and 4 are unlike terms. [CHECKPOINT]

Try this: Group the like terms together from: 12x, 12, –25x, –25, –25y, 1, x, 12y, y. The like terms are: 12x, –25x, and x. Next, 12, –25, and 1. Next, –25y and y. And finally, 12y. Try this: Classify the following as monomial, binomial, or trinomial: a, a + b, ab + a + b, ab + a + b – 5, xy, xy + 5, 5x² – x + 2, 4pq – 3q + 5p, 7, 4m – 7n + 10, 4mn + 7. Monomials: a, xy, 7. Binomials: a + b, xy + 5, 4mn + 7. Trinomials: ab + a + b, 5x² – x + 2, 4pq – 3q + 5p, 4m – 7n + 10. The expression ab + a + b – 5 has four terms, so it is a polynomial but not a trinomial. [CHECKPOINT]

Let us solve Example 3. State with reasons, which of the following pairs are like or unlike terms. i) 7x, 12y. Factors are 7,x and 12,y. Variables are different. Unlike. ii) 15x, –21x. Factors are 15,x and –21,x. Algebraic factors same. Like. iii) –4ab, 7ba. Factors are –4,a,b and 7,b,a. Same algebraic factors. Remember ab = ba. Like. iv) 3xy, 3x. Factors are 3,x,y and 3,x. Variable y is only in one term. Unlike. v) 6xy², 9x²y. Factors are 6,x,y,y and 9,x,x,y. Variables match but powers do not. Unlike. vi) pq², –4pq². Factors are 1,p,q,q and –4,p,q,q. Same algebraic factors. Like. vii) mn², 10mn. Factors are m,n,n and 10,m,n. Powers of n differ. Unlike. Following simple steps will help you decide. Ignore numerical coefficients. Check variables. They must be same. Check powers of each variable. They must be same. Numerical coefficients and order of multiplication do not matter. [CHECKPOINT]

Now we will solve Exercise 10.1 completely. Question 1: Get the algebraic expressions. i) Subtraction of z from y: y – z. ii) One-half of the sum of x and y: ½(x + y). iii) z multiplied by itself: z². iv) One-fourth of product of p and q: ¼pq. v) x and y both squared and added: x² + y². vi) 5 added to three times product of m and n: 3mn + 5. vii) Product of y and z subtracted from 10: 10 – yz. viii) Sum of a and b subtracted from their product: ab – (a + b). [CHECKPOINT]

Question 2 part i: Identify terms and factors, show tree diagrams. Remember, 1 is not taken as a separate factor. a) x – 3. Terms: x, –3. Factors of x: x. Factors of –3: –3. b) 1 + x + x². Terms: 1, x, x². Factors of 1: 1. Factors of x: x. Factors of x²: x, x. c) y – y³. Terms: y, –y³. Factors of y: y. Factors of –y³: –1, y, y, y. d) 5xy² + 7x²y. Terms: 5xy², 7x²y. Factors: 5, x, y, y and 7, x, x, y. e) –ab + 2b² – 3a². Terms: –ab, 2b², –3a². Factors: –1, a, b; 2, b, b; –3, a, a. Part ii: Identify terms and factors. a) –4x + 5: terms –4x, 5. Factors –4, x and 5. b) –4x + 5y: terms –4x, 5y. Factors –4, x and 5, y. c) 5y + 3y²: terms 5y, 3y². Factors 5, y and 3, y, y. d) xy + 2x²y²: terms xy, 2x²y². Factors x, y and 2, x, x, y, y. e) pq + q: terms pq, q. Factors p, q and q. f) 1.2ab – 2.4b + 3.6a: terms 1.2ab, –2.4b, 3.6a. Factors 1.2, a, b; –2.4, b; 3.6, a. g) ¾x + ¼: terms ¾x, ¼. Factors ¾, x and ¼. h) 0.1p² + 0.2q²: terms 0.1p², 0.2q². Factors 0.1, p, p and 0.2, q, q. [CHECKPOINT]

Question 3: Identify numerical coefficients (other than constants). i) 5 – 3t²: –3. ii) 1 + t + t² + t³: 1, 1, 1. iii) x + 2xy + 3y: 1, 2, 3. iv) 100m + 1000n: 100, 1000. v) –p²q² + 7pq: –1, 7. vi) 1.2a + 0.8b: 1.2, 0.8. vii) 3.14r²: 3.14. viii) 2(l + b) expands to 2l + 2b: 2, 2. ix) 0.1y + 0.01y²: 0.1, 0.01. [CHECKPOINT]

Question 4 part a: Terms containing x and coefficient of x. i) y²x + y: term y²x, coefficient y². ii) 13y² – 8yx: term –8yx, coefficient –8y. iii) x + y + 2: term x, coefficient 1. iv) 5 + z + zx: term zx, coefficient z. v) 1 + x + xy: terms x and xy, coefficients 1 and y. vi) 12xy² + 25: term 12xy², coefficient 12y². vii) 7x + xy²: terms 7x and xy², coefficients 7 and y². Part b: Terms containing y² and coefficient. i) 8 – xy²: term –xy², coefficient –x. ii) 5y² + 7x: term 5y², coefficient 5. iii) 2x²y – 15xy² + 7y²: terms –15xy² and 7y², coefficients –15x and 7. [CHECKPOINT]

Question 5: Classify into monomials, binomials, trinomials. i) 4y – 7z: binomial. ii) y²: monomial. iii) x + y – xy: trinomial. iv) 100: monomial. v) ab – a – b: trinomial. vi) 5 – 3t: binomial. vii) 4p²q – 4pq²: binomial. viii) 7mn: monomial. ix) z² – 3z + 8: trinomial. x) a² + b²: binomial. xi) z² + z: binomial. xii) 1 + x + x²: trinomial. [CHECKPOINT]

Question 6: Like or unlike terms. i) 1, 100: Like (constants). ii) –7x, ⁵⁄₂x: Like. iii) –29x, –29y: Unlike. iv) 14xy, 42yx: Like. v) 4m²p, 4mp²: Unlike. vi) 12xz, 12x²z²: Unlike. Question 7a: Identify like terms. Group 1: –xy² and 2xy². Group 2: –4yx² and 20x²y. Group 3: 8x², –11x², and –6x². Group 4: 7y and y. Group 5: –100x and 3x. Group 6: –11yx and 2xy. Question 7b: Identify like terms. Group 1: 10pq, –7qp, 78qp. Group 2: 7p, 2405p. Group 3: 8q, –100q. Group 4: –p²q² and 12q²p². Group 5: –5p² and 701p². Group 6: –23 and 41. Group 7: 13p²q and qp². [CHECKPOINT]

Now, section 10.6, Finding the Value of an Expression. We know that the value of an algebraic expression depends on the values of the variables. We find values when checking equations or using geometry formulas. For example, area of a square is l². If l equals 5 cm, area is 25 sq cm. If l equals 10 cm, area is 100 sq cm. Let us solve Example 4. Find values for x = 2. i) x + 4: 2 + 4 = 6. ii) 4x – 3: 4 × 2 – 3 = 8 – 3 = 5. iii) 19 – 5x²: 19 – 5 × 2² = 19 – 5 × 4 = 19 – 20 = –1. iv) 100 – 10x³: 100 – 10 × 2³ = 100 – 10 × 8 = 100 – 80 = 20. [CHECKPOINT]

Example 5: Find value when n = –2. i) 5n – 2: 5(–2) – 2 = –10 – 2 = –12. ii) 5n² + 5n – 2: 5(–2)² + 5(–2) – 2 = 5 × 4 – 10 – 2 = 20 – 12 = 8. iii) n³ + 5n² + 5n – 2: (–2)³ + 5(–2)² + 5(–2) – 2 = –8 + 20 – 10 – 2 = –8 + 8 = 0. [CHECKPOINT]

Example 6: Find value for a = 3, b = 2. i) a + b: 3 + 2 = 5. ii) 7a – 4b: 7 × 3 – 4 × 2 = 21 – 8 = 13. iii) a² + 2ab + b²: 3² + 2 × 3 × 2 + 2² = 9 + 12 + 4 = 25. iv) a³ – b³: 3³ – 2³ = 27 – 8 = 19. [CHECKPOINT]

Now, Exercise 10.2. Question 1: m = 2. i) m – 2: 2 – 2 = 0. ii) 3m – 5: 3(2) – 5 = 6 – 5 = 1. iii) 9 – 5m: 9 – 5(2) = 9 – 10 = –1. iv) 3m² – 2m – 7: 3(4) – 2(2) – 7 = 12 – 4 – 7 = 1. v) 5/2 m⁴: 5/2 × 16 = 40. Question 2: p = –2. i) 4p + 7: 4(–2) + 7 = –8 + 7 = –1. ii) –3p² + 4p + 7: –3(4) + 4(–2) + 7 = –12 – 8 + 7 = –13. iii) –2p³ – 3p² + 4p + 7: –2(–8) – 3(4) + 4(–2) + 7 = 16 – 12 – 8 + 7 = 3. [CHECKPOINT]

Question 3: x = –1. i) 2x – 7: 2(–1) – 7 = –2 – 7 = –9. ii) –x + 2: –(–1) + 2 = 1 + 2 = 3. iii) x² + 2x + 1: (–1)² + 2(–1) + 1 = 1 – 2 + 1 = 0. iv) 2x² – x – 2: 2(–1)² – (–1) – 2 = 2 + 1 – 2 = 1. Question 4: a = 2, b = –2. i) a² + b²: 2² + (–2)² = 4 + 4 = 8. ii) a² + ab + b²: 4 + 2(–2) + 4 = 4 – 4 + 4 = 4. iii) a² – b²: 4 – 4 = 0. Question 5: a = 0, b = –1. i) 2a + 2b: 0 + 2(–1) = –2. ii) 2a² + b² + 1: 0 + 1 + 1 = 2. iii) 2a²b + 2ab² + ab: 0 + 0 + 0 = 0. iv) a² + ab + 2: 0 + 0 + 2 = 2. [CHECKPOINT]

Question 6: Simplify and find value for x = 2. i) x + 7 + 4(x – 5): x + 7 + 4x – 20 = 5x – 13. For x = 2: 5(2) – 13 = 10 – 13 = –3. ii) 3(x + 2) + 5x – 7: 3x + 6 + 5x – 7 = 8x – 1. For x = 2: 8(2) – 1 = 16 – 1 = 15. iii) 6x + 5(x – 2): 6x + 5x – 10 = 11x – 10. For x = 2: 11(2) – 10 = 22 – 10 = 12. iv) 4(2x – 1) + 3x + 11: 8x – 4 + 3x + 11 = 11x + 7. For x = 2: 11(2) + 7 = 22 + 7 = 29. [CHECKPOINT]

Question 7: Simplify and find value for x = 3, a = –1, b = –2. i) 3x – 5 – x + 9: 2x + 4. For x = 3: 2(3) + 4 = 10. ii) 2 – 8x + 4x + 4: –4x + 6. For x = 3: –4(3) + 6 = –6. iii) 3a + 5 – 8a + 1: –5a + 6. For a = –1: –5(–1) + 6 = 11. iv) 10 – 3b – 4 – 5b: –8b + 6. For b = –2: –8(–2) + 6 = 22. v) 2a – 2b – 4 – 5 + a: 3a – 2b – 9. For a = –1, b = –2: 3(–1) – 2(–2) – 9 = –3 + 4 – 9 = –8. [CHECKPOINT]

Question 8: i) z = 10, find z³ – 3(z – 10): 10³ – 3(10 – 10) = 1000 – 0 = 1000. ii) p = –10, find p² – 2p – 100: (–10)² – 2(–10) – 100 = 100 + 20 – 100 = 20. Question 9: Value of 2x² + x – a equals 5 when x = 0. Substitute x = 0: 0 + 0 – a = 5. So –a = 5, which means a = –5. Question 10: Simplify 2(a² + ab) + 3 – ab. This is 2a² + 2ab + 3 – ab, which simplifies to 2a² + ab + 3. For a = 5, b = –3: 2(25) + 5(–3) + 3 = 50 – 15 + 3 = 38. [CHECKPOINT]

Finally, let us review what we have discussed. Algebraic expressions are formed from variables and constants. We use the operations of addition, subtraction, multiplication and division on the variables and constants to form expressions. Expressions are made up of terms. Terms are added to make an expression. A term is a product of factors. The coefficient is the numerical factor in the term. Any expression with one or more terms is called a polynomial. Specifically a one term expression is called a monomial; a two-term expression is called a binomial; and a three-term expression is called a trinomial. Terms which have the same algebraic factors are like terms. Terms which have different algebraic factors are unlike terms. In situations such as solving an equation and using a formula, we have to find the value of an expression. The value of the expression depends on the value of the variable from which the expression is formed. Thus, the value of 7x – 3 for x = 5 is 32, since 7(5) – 3 = 35 – 3 = 32. [CHECKPOINT]

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]