Algebraic Expressions

12 ALGEBRAIC EXPRESSIONS ALGEBRAIC EXPRESSION CONTENTS • • • • • Constants and variables Any combination of literal numbers or variables and numbers (numerals) connected by +, –, × or ÷ sings is called an algebraic expression. Algebraic Expressions For example, • Addition and subtraction of Algebraic Expressions • • 5, 6x, a + b × c, 4 × m + n, x – y ÷ z are algebraic expressions. The perimeter P of a triangle whose sides are a, b and c is given by P = a + b + c, area of square is x × x i.e., x2 are algebraic expressions. Factors, Coefficients, Terms Like and Unlike Terms Finding the value of an Algebraic Expression A repeated product of a number with itself is written in exponential form. 2 × 2 × 2 × 2 = 24, etc The same is true for literal numbers also. Thus, if x is a literal, then we have x × x × x = x3 (Third power of x or x cube) Types of Brackets x × x × x × x = x4 and so on. Rule for Solving Algebraic Expression Also, 7 × x × x = 7x2 4 × x × x × y × y = 4x2 y2, etc. x × x × x ……, n times = xn and read as n-th power of x. Here x is called base and n, exponent. CONSTANTS AND VARIABLES To deal with the problems of finding unknown numbers we use letters to represent numbers. These letters are called literal numbers. We use letters x, y, z, t,…etc. to denote variables. The values of these letters x, y, z, t,…etc. are not fixed, these can take different values that's why these are called variables. On the other hand, 1, 2, 3, 10, 20, … etc. are called constants as they are fixed, their values can not be changed. For example, EXAMPLES Ex.1 (i) (9p)7 (ii) 9p7 Sol. Ex.2 b = breadth Here, P, l, b are variables and 2 is a constant. Write the following in the exponential form : (i) x × x × x P = 2 (l + b) l = length (i) (9p) 7 = 9p × 9p × 9p × 9p × 9p × 9p × 9p (ii) 9p7 = 9 × p × p × p × p × p × p × p The perimeter of rectangle P = perimeter Write the following in product form : (ii) –2 × 3 × 3 × x × y × y × y × y Sol. (i) x × x × x = x3 (ii) – 2 × 3 × 3 × x × y × y × y × y = – 2 × 32 × x × y4 = – 18xy4 FACTORS When two or more numbers are multiplied together, the numbers themselves are called the factors of the product. The factors of 34 are 2 and 17. The factors of 18x2 are 2, 3 and x, factors of ln are l and n (8 is numerical factor of 8x and x is variable factor of 8x) Note : (i) The words 'mono', 'bi', 'tri' and 'poly' mean one, two, three and many. (ii) A term of an algebraic expression having no literal factor is called a constant term, for example, in x2 + 9x – 8, the constant term is –8. COEFFICIENT Any factor of a product is the coefficient of the remaining factors. In the product of 8 × 5, the number 8 is coefficient of 5 and 5 is coefficient of 8. In the product 5yz, 5 is the (numerical) coefficient of yz, 5y is the coefficient of z and 5z is coefficient of y and yz is the variable coefficient of 5. Note : If a letter has no coefficient written before it, the coefficient 1 is understood. Thus y means 1y and yz means 1yz, similarly, –y means –1(y). TERMS An algebraic expression is a combination of numbers, literals and arithmetical operations. One or more sings (+ and –) separates an expression into several parts. Each part along with its sign is called a term. Type of Algebraic Definition Examples Expression 1. Monomial A monomial is 6, –5xy, –6x2 an expression etc. having one term. 2. Binomial A binomial is 2x – 3y, x – y, an algebraic 3x2 – 6x, expression (x – y)2 + 3xy, having two 3x2 + 5 etc. terms. 3. Trinomial A trinomial is 2a – 3b – 5c, an algebraic 5y2 – 3x + 9, expression a3 + b3 + c3 having three etc. terms. 4.Tetranomial or A quadrinomial a + b + c – 3, Quadrinomial is an algebraic a3 + b3 + c3 + expression 3abc, etc. having four terms. 5. Polynomial Binomial, 2a – 3b, x + y trinomials and all – 3yz + 4x2 – algebraic 6y2 etc. expressions having more than three terms are called polynomials. LIKE (SIMILAR) AND UNLIKE (DISSIMILAR ) TERMS These terms are defined as follows: Like (Similar terms) Unlike (Dissimilar terms) These are terms whose literal (variable) factors are the same. These are terms whose literal (variable) factors are not same. For example, For example, (i) 5x2, – 6x2, + 3x2 (i) 2x and 5y (ii) 2(a + b), – 4(a + b), 6(a + b) (ii) 6xy2 and 8x2y (iii) 6xy2, –8xy2, xy2 (iii) (x + y), (x2 + y), 5(x2 + y2) FINDING THE VALUE OF AN ALGEBRAIC EXPRESSION An algebraic expression contains literal (variable) numbers. If we know the numerical values of these variables and substitute them in the given algebraic expression we get a numerical expression which can be simplified by the methods of arithmetic to get a number, called the value of the algebraic expression. EXAMPLES Ex.3 Evaluate the following expression if a = 2, b = – 1, c = 1: a2 + b2 + c2 – ab – bc – ca. Sol. a2 + b2 + c2 – ab – bc – ca = (2)2 + (–1)2 + (1)2 – 2 × (–1) – (–1) × 1 – 1 × 2 =4+1+1+2+1–2 =9–2=7 So, a2 + b2 + c2 – ab – bc – ca at a = 2, b = – 1, c = 1 has value 7. Ex.4 Get the algebraic expressions in the following cases using variables, constants and arithmetic operations: Ex.7 Identify like and unlike terms in the following : (i) 1, 100 2 (ii) –29x, –29y 2 (iii) 4m p, 4mp (iv) 14xy, 42yx (i) Subtraction of z from y. (ii) One-fourth of the product of numbers p and q. Sol. (i) 1, 100 → like (ii) –29x, –29y → unlike (iii) Number 5 added to three times the product of numbers m and n. (iv) Sum of number a and b subtracted from their product. Sol. (iii) 4m2p, 4mp2 → unlike (iv) 14xy, 42yx → like Ex.8 (i) –xy2, –4yx2, 8x2, 2xy2, 7y, –11x2, –100x, –11yx, 20x2y, –6x2, y, 2xy, 3x. (i) Subtraction of z from y = y – z. (ii) One-fourth of the product of numbers 1 p & q = pq . 4 (iii) Number 5 added to three times the product of numbers m and n = 3mn + 5. (ii) 10pq, 7p, 8q, –p2q2, –7qp, –100q, –23, –12q2p2, –5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2. Sol. (iv) Sum of numbers a and b subtracted from their product = ab – (a + b) Ex.5 Sol. (i) (7y, y), (–100x, 3x), (–11yx, 2xy) terms → Ex.9 –y3 y Factors 1 (ii) (ii) (10pq, –7qp, 78qp), (7p, 2405p), (8q, –100q), (–p2q2, –12q2p2), (–23, 41), (–5p2, 701p2), (13p2q, qp2) (ii) – ab + 2b2 – 3a2 y – y3 (expression) –1 y y y(factors) y – ab + 2b2 – 3a2 (expression) Like terms are : (i) (–xy2, 2xy2), (–4yx2,20x2y), (8x2, –11x2 –6x2), Identify the terms and their factors in the following expressions: (i) y – y3 Identify like terms in the following : Sol. If m = 2, find the value of (i) m – 2 (ii) 9 – 5m (iii) 3m2 – 2m – 7 (iv) 5m −4 2 (i) m –2 = 2 – 2 = 0 (ii) 9 – 5m = 9 – 5 × 2 = 9 – 10 = – 1 terms → Factors –1 Ex.6 Sol. –ab a b2 2b2 –3a2 (iii) 3m2 – 2m – 7 = 3 × (2)2 – 2 × 2 – 7 = 3 × 4 – 4 – 7 = 12 – 11 = 1 b b –3 a a (iv) 5m 5× 2 −4 = −4 = 5 –4 =1 2 2 Classify into monomials, binomials and trinomials : 4y – 7z, y2, x + y – xy, 100, ab – a – b, 5 – 3t, 4p2q – 4pq2, 7mn, z2 – 3z + 8, a3 + b3, z2 + z, 1 + x + x2. Ex.10 If a = 2, b = – 2, find the value of Monomials → y2, 100, 7mn Sol. Binomials → 4y – 7z, 5 –3t, 4p2q – 4pq2, a3 + b3, z2 + z Trinomials → x + y – xy, ab – a – b, z2 –3z + 8, 1 + x + x 2. (i) a2 + b2 (ii) a2 + ab + b2 If a = 2, b = – 2, then (i) a2 + b2 = (2)2 + (–2)2 = 4 + 4 = 8 ⇒ a2 + b 2 = 8 (ii) a2 + ab + b2 = (2)2 + 2 × (–2) + (–2)2 = 4 – 4 + 4 = 4 ⇒ a2 + ab + b2 = 4 Ex.11 Sol. When a = 0, b = –1, find the value of Ex.15 Add –7x, –5x, 8x, 9x. (i) 2a2 + b2 + 1 (ii) 2a2b + 2ab2 + ab Sol. Positive terms are 8x, 9x (iii) a2 + ab + 2 Negative terms are – 7x, – 5x (i) 2a2 + b2 + 1 = 2 (0)2 + (–1)2 + 1 Sum of positive terms = 8x + 9x = 17x =2×0+1+1=0+1+1=2 Sum of negative terms = – 7x + (–5x) ⇒ 2a2 + b2 + 1 = 2 = – 7x – 5x = – 12x (ii) 2a2b + 2ab2 + ab Adding these two terms, we have = 2 (0)2 × (–1) + 2 × 0 × (–1)2 + 0 × (–1) = 0 + 0 + 0 = 0 ⇒ 2a2b + 2ab2 + ab = 0 (iii) a2 + ab + 2 = (0)2 + 0 × (–1) + 2 17x + (–12)x = 17x – 12x = (17 – 12)x = 5x Ex.16 Add 4x + 3y – 5z, –7z + 5x – 8y and – y –3x + 2z. Sol. Column method : = 0 + 0 + 2 = 2 ⇒ a2 + ab + 2 = 2 Ex.12 If z = 10, find the value of z3 – 3(z – 10). Sol. If z = 10, then z3 – 3(z – 10) Re-write the expressions so that their like terms are in a column as 4 x + 3y – 5z = (10) – 3 × (10 – 10) = 1000 – 3 × 0 = 1000 5x − 8 y − 7 z − 3x − y + 2 z Hence, z3 – 3(z – 10) = 1000. 6 x − 6 y − 10z Sum 3 ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS Horizontal Method : Sum = (4x + 3y – 5z) + (–7z + 5x – 8y) Like Terms : + (–y –3x + 2z) (1) The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms. = 4x + 3y – 5z –7z + 5x – 8y – y –3x + 2z = (4x + 5x –3x) + (3y – 8y – y) + ( – 5z – 7z +2z) (2) The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms. EXAMPLES Ex.13 Add 2x2, 5x2 and x2. Sol. The three like terms are 2x2, 5x2 and x2 Adding these terms, we have 2x2 + 5x2 + x2 = (2 + 5 + 1) x2 (Distributive property) = 8x2 Ex.14 Add – xy, –5xy, –2xy. Sol. Adding the given three like terms, we have = – xy + (– 5xy) + (– 2xy) = –xy –5xy – 2xy = (–1 – 5 – 2) xy = – 8xy = (4 + 5 – 3) x + (3 – 8 – 1) y + (–5 –7 + 2)z = 6x – 6y –10z Ex.17 Subtract 10x2 from –8x2. Sol. Minuend → –8x2 Subtrahend → 10x2 (change sign and add) Hence, difference → –18x2 − 8x 2 − 10 x 2 − 18x 2 Sum Ex.18 How much is t2 – 5t + 6 greater than t2 + 5t – 6 ? Sol. minuend → t2 – 5t + 6 Ex.20 Simplify these expressions and find their values if x = 2 : (i) 3(x + 2) + 5x – 7 subtrahend → t2 + 5t – 6 Sol. (i) 3(x + 2) + 5x – 7 (Change the sign of each term and add) t 2 = 3x + 3 × 2 + 5x – 7 − 5t + 6 = 3x + 6 + 5x – 7 − t 2 − 5t + 6 = 3x + 5x + 6 – 7 = 8x – 1 So, 0.t 2 − 10.t + 12 Hence, difference = 16 – 1 = 15 → –10t + 12 Hence, value of given expression at x = 2 is 15. Ex.19 Simplify, combining like terms : (ii) 4(2x – 1) + 3x + 11 (i) 21b – 32 + 7b – 20b = 4 × 2x – 4 × 1 + 3x + 11 (ii) p – (p – q) – q – (q – p) 2 2 2 2 = 8x – 4 + 3x + 11 2 2 2 2 (iii) 5x y – 5x + 3yx –3y + x –y + 8xy –3y 2 Sol. 3(x + 2) + 5x – 7 = 8x – 1 At x = 2, the value of 8x – 1 = 8 × 2 – 1 → 0.t2 – 10t + 12 difference (ii) 4(2x –1) + 3x + 11 = 11x + 7 2 (iv) (3y + 5y – 4) – (8y – y – 4) So, 4(2x – 1) + 3x + 11 = 11x + 7 (i) 21b – 32 + 7b – 20b at x = 2 the value of 11x + 7 = 11 × 2 + 7 = 29 Hence, value of given expression at x = 2 is 29. = 21b + 7b – 20b – 32 (combining like terms) Ex.21 = 8b – 32 (i) 3x – 5 – x + 9 (ii) p – (p – q) – q – (q – p) (ii) 10 – 3b – 4 – 5b =p–p+q–q–q+p (removing the brackets) =p–p+p+q–q–q (combining like terms) =p–q (iii) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2 = 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2 + 8xy2 (Combining like terms) = x2y(5 + 3) + x2 (–5 + 1) – y2 (3 + 1 + 3) + 8xy2 = 8x2y – 4x – 7y2 + 8xy2 (iv) 3y2 + 5y – 4 – (8y – y2 – 4) = 3y2 + 5y – 4 –8y + y2 + 4 = 3y2 + y2 + 5y – 8y – 4 + 4 = 4y2 – 3y. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2 Sol. (i) 3x – 5 – x + 9 = 3x – x – 5 + 9 = 2x + 4 So, 3x – 5 – x + 9 = 2x + 4 The value of 3x – 5 – x + 9 i.e., 2x + 4 at x = 3 is 2 × 3 + 4 = 6 + 4 = 10. Hence, simplified form is 2x + 4 and its value at x = 3 is 10 (ii) 10 – 3b – 4 – 5b = 10 – 4 – 3b – 5b = 6 – 8b So, 10 – 3b – 4 – 5b = 6 – 8b The value of 6 – 8b at b = – 2 = 6 – 8 × (–2) = 6 + 16 = 22 Hence, simplified form is 6 – 8b and its value at b = – 2 is 22 Ex.22 Subtract : Ex.24 What should be subtracted from 2a + 8b + 10 to get –3a + 7b + 16 ? (i) –5y2 from y2 The answer is obtained by subtracting the second expression –3a + 7b + 16 from the first expression 2a + 8b + 10. Sol. (ii) 6xy from –12xy (iii) a (b – 5) from b (5 – a) (iv) –m2 + 5mn from 4m2 – 3mn + 8 ∴ 2a + 8b + 10 – (–3a + 7b + 16) (v) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq Sol. = 2a + 8b + 10 + 3a – 7b – 16 (i) Required difference = y2 – (–5y2) = y2 + 5y2 = 2a + 3a + 8b – 7b + 10 – 16 = (1 + 5)y2 = (2 + 3)a + (8 – 7)b + 10 –16 = 5a + b – 6 = 6y2 (ii) Required difference = –12xy – (6xy) Ex.25 From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11. Sol. Sum of 3x – y + 11 and – y – 11 = –12xy –6xy = (–12 – 6)xy = –18xy = 3x – y + 11 + (–y –11) (iii) Required difference = b (5 – a) – a (b – 5) = 5b – ba – ba + 5a = 3x – y + 11 – y – 11 = 5b + 5a – ba – ba = 3x – y – y + 11 – 11 = 3x –2y = 5b + 5a – 2ab Now subtracting 3x – y – 11 from 3x –2y (iv) Required difference = 4m2 – 3mn + 8 – (–m2 + 5mn) Required difference = 3x – 2y – (3x – y – 11) = 4m2 – 3mn + 8 + m2 – 5mn = 3x –2y – 3x + y + 11 = 4m2 + m2 – 3mn – 5mn + 8 = 3x – 3x – 2y + y + 11 2 = (4 + 1)m + (–3 –5) mn + 8 = – y + 11 = 5m2 –8mn + 8 Note : (v) Required difference = 5p2 + 3q2 – pq – (4pq –5q2 – 3p2) Unlike terms cannot be added or subtracted as the like terms are added or subtracted. = 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 = 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq = (5 + 3)p2 + (3 + 5)q2 + (–1 – 4) pq ( ) = Parentheses = 8p2 + 8q2 – 5pq [ ] = Square brackets Ex.23 What should be added to x2 + xy + y2 to obtain 2x2 + 3xy ? Sol. The answer is obtained by subtracting the first expression x2 + xy + y2 from the second expression 2x2 + 3xy. ∴ 2x2 + 3xy – (x2 + xy + y2) = 2x2 + 3xy – x2 – xy – y2 2 TYPES OF BRACKETS 2 2 { } = Curly brackets or braces — = Vinculum brackets RULE FOR SOLVING ALGEBRAIC EXPRESSIONS BODMAS represents the order Performance of operations namely B = Brackets; 2 2 = 2x – x + 3xy – xy – y = x + 2xy – y O = Of; of D = Division; M = Multiplication; A = Addition; S = Subtraction Alternatively : Sum of 2p2 + 3pq, – p2 – pq – q2 and pq + 2q2 is EXAMPLES Ex.26 Simplify : 3(a + b) – 2(2a – b) + 4a – 7. Sol. 3(a + b) – 2(2a – b) + 4a – 7 2p 2 + 3pq = 3a + 3b – 4a + 2b + 4a – 7 − p 2 − pq − q 2 + pq + 2q 2 = 3a – 4a + 4a + 3b + 2b – 7 p 2 + 3pq + q 2 = (3 – 4 + 4) a + (3 + 2) b – 7 = 3a + 5b – 7 Ex.27 Simplify : 2x – [3y – {2x – (y – x)}]. Sol. We have, 2x – [3y – {2x – (y – x)}] Sum of 3p2 – q2 & – p2 pq + q2 3p 2 − q 2 We first remove the inner most bracket. − p 2 + q 2 + pq 2x – [3y – {2x – y + x}] 2p 2 + 0q 2 + pq Next inner most is the curly bracket. 2x – [3y – 2x + y – x] Now required difference : Now we remove the square bracket. Sum I – Sum II 2x – 3y + 2x – y + x p 2 + 3pq + q 2 = (2x + 2x + x) – 3y – y = (2x + 2x + x) – (3y + y) i.e., = (2 + 2 + 1) x – (3 + 1) y = 5x – 4y Ex.28 From the sum of 2p2 + 3pq, – p2 – pq – q2 and pq + 2q2 subtract the sum of 3p2 – q2 and – p2 + pq + q2. Sol. 2p 2 + pq – – − p 2 + 2pq + q 2 Ex.29 Simplify : 5a – [a 2 – {2a (1 – a + 4a 2 ) –3a (a 2 – 5a – 3)}] – 8a. We have, [(2p2 + 3pq) + (–p2 – pq – q2) + (pq + 2q2)] – [(3p2 – q2) + (–p2 + pq + q2)] = [2p2 + 3pq – p2 – pq – q2 + pq + 2q2) – [3p2 – q2 – p2 + pq + q2] = (2p2 – p2 + 3pq – pq + pq – q2 + 2q2) 2 2 2 2 – (3p – p – q + q + pq) = (p2 + 3pq + q2) – (2p2 + pq) Sol. We first remove the inner most grouping symbol ( ), the { } and then [ ]. Thus we have, 5a – [a2 – {2a (1 – a + 4a2) –3a (a2 – 5a – 3)}] – 8a = 5a –[a2 –{2a –2a2 + 8a3 – 3a3 + 15a2 + 9a}] – 8a = 5a –[a2 – 2a + 2a2 – 8a3 + 3a3 – 15a2 – 9a] – 8a = p2 + 3pq + q2 – 2p2 – pq = 5a – a2 + 2a – 2a2 + 8a3 – 3a3 + 15a2 + 9a – 8a = – p2 + 2pq + q2 = 5a3 + 12a2 + 8a. IMPORTANT POINTS TO BE REMEMBERED (i) A combination of literals numbers or variables and numbers (numerals) connected by +, –, × or ÷ signs is called an algebraic expression. (iv) An expression with only one term is called a monomial, with two terms is called a binomial, with three terms is called a trinomial. In general, an expression is called a polynomial. (ii) Expressions are made up of terms. Terms are added to make an expression. (v) Terms with same algebraic factors are called like terms. (iii) The literals and numbers in a term are called factors. The numerical factor is called the coefficient of the term. In some cases, any one factor is called the coefficient of the remaining part of the term. (vi) To add (or subtract) two algebraic expressions, add (or subtract) the like terms and unlike terms are left as they are. Take care, that the sum of two like terms is a like term with coefficients equal to the sum (or difference) of the coefficients of the two like terms. Also in subtraction do not forget to change the sign of the subtrahend.