["3.30 Chapter 3 \t61.\t Find the square root of a2 1 \u2212 1 + a \u2212 3 . \t\t(a) 3 (b + c) 4 + a2 a 2 4 \t\t(b) 3(b + c \u2212 a) \t\t(a) a \u2212 1 + 1 \t(b) a + 2 \u2212 1 \t\t(c) 3(b + c) 2 a 2 2 a \t\t(d) 3(b + c \u2212 a) \t\t(c) a + 1 \u2212 1 \t(d) a \u2212 2 \u2212 1 6\t 4.\t (a \u2212 b)3 \u2212 (a + b)3 + a(a2 + 3b2 ) = _______ . 2 a 2 2 a 2 2 (x + y )3 + (x \u2212 y )3 \t\t(a) a3 \u2212 b3\t\t (b) (a + b)3 2 \t62.\t \u2212 y(3x 2 + y2 ) = _____ . \t \t (c) a3 + b3\t\t (d) (a \u2212 b)3 \t\t(a) x3 \u2212 y3\t\t (b) (x \u2212 y)3 \t65.\tThe square root of (3a + 2b + 3c)2 \u2212 (2a + 3b + 2c)2 + 5b2 is \t \t (c) 2x3 \u2212 3x2y\t(d) x3 \u2212 6xy2 \t\t(a) 5(a + b + c ) \t(b) 5(a + b) 6\t 3.\tFind the square root of (4a + 5b + 5c)2 \u2212 (5a + 4b + 4c)2 + 9a2. \t\t(c) 5(a + c ) \t(d) 5(a + c \u2212 b) PRACTICE QUESTIONS","Polynomials and Square Roots of Algebraic Expressions 3.31 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t biquadratic 1\t 5.\t f \uf8eb 1 \uf8f6 = 101 \t2.\t 15 \uf8ed\uf8ec 2 \uf8f7\uf8f8 32 \t3.\t False \t4.\t x2 + 10x + 1 \t16.\t a \u2212 b \t5.\t a3 + b3 + c3 = 3abc \t6.\t (x \u2212 y)(x + y)(x2 + y2 \u2212 xy)(x2 + y2 + xy) \t17.\t symmetric m2 n2 \t18.\t a 2 \u22c5 b 2 \t19.\t 20 \t7.\t 8x7y2 2\t 0.\t ab(a2 + b2)(a + b)(a \u2212 b) \t8.\t any real number \t21.\t 49 \t9.\t x + 1 2\t 2.\t 4x5 + x4 \u2212 x3 x \t23.\tm4(m + 1)(m2 \u2212 m + 1) 1\t 0.\t 4x2 \u2212 7 x + 5 \t24.\t 1 (a \u2212 2)(a \u2212 4) 2 2 6 \t11.\t -288. \t25.\t a = \u00b1 3 4 \t12.\t (2x \u2212 1) (3x + 2) \t26.\t (a + b + c)(a2 + b2 + c2 \u2212 ab \u2212 bc \u2212 ca) \t13.\t The LCM of given polynomials = 3 \u00d7 5 \u00d7 x3 2\t 7.\t (a2 + 1)(a + 11) \u00d7 y3 \u00d7 z3 = 15x3y3z3 \t\tThe HCF of given polynomials = 3x2yz2 2\t 8.\t \u2212461700 1\t 4.\tf(2) = 215 \t29.\t \u22128x2 + 6x \u2212 14 \t30.\t 8 Short Answer Type Questions \t31.\t 1, 1 \t39.\t HCF = 9(x + 2)2(x \u2212 1)2(x + 3)4, \t\tLCM = 1260 (x + 2)5(x \u2212 1)5(x + 3)5 \t32.\t a2(b + c)b2(c + a)c2(a + b) 4\t 0.\ta = 6, b = 2 \t41.\t (223 \u2212 1)x + (2 \u2212 223) 3\t 3.\t 3(a \u2212 b)(b \u2212 c)(c \u2212 a) \t43.\t 2c \t44.\t 25abc \t34.\t c2(a2 \u2212 b2) + a2(b2 \u2212 c2) + b2(c2 \u2212 a2) \t45.\t(x + 2)(x + 3)(x + 4) ANSWER KEYS \t35.\t \u2212x3 \u2212 16x2 + 19x \u2212 53 \t36.\t x6 - 4x5 - 2x4 + 4x3 + 5x2 - 1 3\t 7.\t x2 + x \u2212 30 \t38.\t \uf8eb a + x \u2212 1 \uf8f6 \uf8eb a2 + x2 + 1 \u2212 ax + x + a \uf8f6 \uf8ec\uf8ed 4 2 \uf8f7\uf8f8 \uf8ec 16 4 4 8 2 \uf8f7 \uf8ed \uf8f8 Essay Type Questions 4\t 9.\t 3x2 \u2212 2x + 1 \t50.\t (a + b), (b + c), (c + a) \t46.\t (x + 2)(x \u2212 3)(2x + 1)(3x \u2212 1) \t48.\t p = -154, q = 121","3.32 Chapter 3 CONCEPT APPLICATION Level 1 \t1.\u2002(d)\t 2.\u2002 (d)\t3.\u2002(a)\t 4.\u2002 (c)\t5.\u2002(c)\t 6.\u2002 (a)\t7.\u2002(c)\t 8.\u2002 (c)\t9.\u2002(d)\t 10.\u2002 (d) \t11.\u2002 (c)\t 12.\u2002 (a)\t13.\u2002(c)\t 14.\u2002 (d)\t15.\u2002(d)\t 16.\u2002 (d)\t17.\u2002(b)\t 18.\u2002 (b)\t19.\u2002(d)\t 20.\u2002 (b) \t21.\u2002 (b)\t 22.\u2002 (c)\t23.\u2002(a)\t 24.\u2002 (c)\t25.\u2002(c)\t 26.\u2002 (a)\t27.\u2002(b)\t 28.\u2002 (d)\t29.\u2002(c)\t 30.\u2002 (a) Level 2 \t31.\u2002 (a)\t 32.\u2002 (a)\t33.\u2002(b)\t 34.\u2002 (a)\t35.\u2002(d)\t 36.\u2002 (b)\t37.\u2002(c)\t 38.\u2002 (d)\t39.\u2002(c)\t 40.\u2002 (d) \t41.\u2002 (b)\t 42.\u2002 (d)\t43.\u2002(a)\t 44.\u2002 (d)\t45.\u2002(d)\t 46.\u2002 (c)\t47.\u2002(c)\t 48.\u2002 (b)\t49.\u2002(b)\t 50.\u2002 (a) \t51.\u2002 (c)\t 52.\u2002 (d)\t53.\u2002(d)\t 54.\u2002 (a)\t55.\u2002(b) Level 3 58.\u2002 (a)\t59.\u2002(a)\t 60.\u2002 (c)\t61.\u2002(a)\t 62.\u2002 (b)\t63.\u2002(c)\t 64.\u2002 (d)\t65.\u2002(c) \t56.\u2002 (c)\t57.\u2002(c)\t ANSWER KEYS","Polynomials and Square Roots of Algebraic Expressions 3.33 CONCEPT APPLICATION Level 1 \t1.\t Degree of AB = degree of A + degree of B. \t18.\t \u2009\u2009(i)\tUse factor theorem. Hints and Explanation \t2.\t Factorize the given expression. \t\t(ii)\tPut x = -1. \t3.\t Use the formula, HCF \u00d7 LCM = product of \t\t(iii)\tC\u0007 heck for what values of n, (-1)n + 1 is divis- polynomials. ible by x + 1. \t4.\t Use remainder theorem. \t5.\t Degree of every term should be same. \t19.\t Use division method to find p and q. \t6.\t Use factorization concept. \t7.\t Use remainder theorem. 2\t 0.\t Apply the division method to find the square root. \t8.\t Find the common factors with the least exponents. \t9.\t Use factor theorem. 2\t 1.\t Recall the concept of finding the square root of \t10.\t Take the terms in common and factorize. monomial. \t11.\t An expression is symmetric if all the coefficients 2\t 2.\t Use algebraic identities. are equal. 1\t 2.\t Use a2 + b2 + 2ab = (a + b)2 identity. \t23.\t Use the concept of polynomials multiplication. \t13.\t Use the concept of multiplication of polynomials. \t14.\t Factorize the given polynomials. \t24.\t Use the formula (LCM) (HCF) = f(x) \u22c5 g(x) \t15.\t Use the addition and subtraction concept of \t25.\t Use factor theorem. polynomials. \t16.\t The degree of the homogeneous expression is 3. \t26.\t Use factor theorem. \t17.\t (i)\tFactorize the given polynomials. \t\t(ii)\t\u0007Factorize the middle term such that product 2\t 7.\t Use addition and subtraction concept of polynomials. obtained is 22 and sum obtained is \u221212 2. \t28.\t Find the square root. \t29.\t (i)\tx\u0007n + yn is always disable by x + y if n is odd. \t\t(ii)\tx\u0007n - yn is divisible by x - y, if n is odd number. \t30.\t (i)\t\u0007The product of four consecutive numbers added to 1 is a perfect square. \t\t(ii)\t\u0007The continued product of 4 consecutive inte- gers added by 1 is always a perfect square. Level 2 \t31.\t \u2009\u2009(i)\tDivide A by B. \t\t(ii)\t\u0007(x \u2212 3) is a factor of 5x2 \u2212 ax + 3 and 2x -1 is a factor of 2x2 + x + b. \t\t(ii)\tD\u0007 ivide the polynomial A by B and then write the remainder. \t\t(iii)\t\u0007Substitute x = 1 in 5x2 \u2212 ax + 3 and x = 1 in 2 \t32.\t \u2009\u2009\u2009\u2009(i)\tTake common terms and factorize. 2x2 + x + b, then obtain the values of a and b. \t\t(ii)\tF\u0007 rom the first two terms take x3 common and 3\t 5.\t \u2009\u2009\u2009\u2009(i)\tUse division algorithm. from last two terms take y3 common. \t\t(ii)\tf\u0007(x) = g(x)Q(x) + [ax + b] where Q(x) is quo- \t\t(iii)\tAgain take x2 \u2212 a2 common in the product. tient and (ax + b) is the remainder. \t\t\u2009(iv)\tNow write the factors of a2 \u2212 b2 and a3 + b3. \t\t(iii)\t\u0007Put x = 1 and x = \u22121, then find the values of a 3\t 3.\t \u2009\u2009(i)\tFactorize the polynomials. and b. \t\t(ii)\tFind the HCF of the first two polynomials. \t36.\t \u2009\u2009\u2009\u2009(i)\tFactorize the cyclic expression. \t\t(iii)\tN\u0007 ow find the HCF of the third polynomial \t\t(ii)\t\u0007Put b = c; the expression become zero, i.e., and HCF of the first two polynomials. b \u2212 c is a factor of the expression. \t34.\t \u2009\u2009\u2009\u2009(i)\t\u0007Factorize the polynomials and use factor theorem.","3.34 Chapter 3 \t\t(iii)\t\u0007Similarly (a \u2212 b) and (c \u2212 a) are also the factors \t\t(ii)\tAdd and subtract 4 to the expression. of the expression. \t\t(iii)\tExpress the expression of the form a2 \u2212 b2. \t46.\t BCAD is the required sequential order of steps in \t\t(iv)\tS\u0007ince the degree of the expression is 4, the fourth factor is k(a + b + c). solving the given problem. \t47.\t CBDAE is the required sequential order. 3\t 7.\t \u2009\u2009\u2009\u2009(i)\tF\u0007ind a factor by hit and trial method and \t48.\t Given polynomials are 8x3ya and 12xby2 HCF = remaining factors by division method. 4xayb \t\t(ii)\tB\u0007 y observation, sum of the coefficients of odd terms = Sum of the coefficients of even terms. \t\t \u21d2\u2003 a \u2264 3 or a \u2264 b and b \u2264 a or b \u2264 2 \u21d2 a = b \t\t\u2234 Each of a and b is less than or equal to 2. \t\t(iii)\tN\u0007 ow divide the expression by y + 1. \t\t\u2234 The maximum value of a + b is 2 + 2, i.e., 4. \t\t(iv)\tF\u0007 ind the roots of quotient by trial and error \t49.\t Let f(x) = 5x5 \u2212 3x3 + 2x2 \u2212 k method. \t\tf(\u22121) = 1 (given) \t\t\u21d2 5(\u22121)5 \u22123(\u22121)3 + 2(\u22121)2 \u2212 k = 1 3\t 8.\t \u2009\u2009\u2009\u2009(i)\tL\u0007 CM is the product of the all the factors with \t\t\u22125 + 3 + 2 \u2212 k = 1 highest powers. \t\t\\\\ k = \u22121. \t\t(ii)\tU\u0007 se the LCM concept and find the values of a 5\t 0.\t a3 + b3 + 3ab \u2212 1 and b. \t\t= a3 + b3 + (\u22121)3 \u2212 3 \u00b7 a \u00b7 b(\u22121) \t\t(iii)\ta + b + c is least when c = 0. \t\t= (a + b \u2212 1)(a2 + b2 + 1 \u2212 ab + b + a). 3\t 9.\t \u2009\u2009\u2009\u2009(i)\tF\u0007ind the common and uncommon factors 5\t 1.\t Degree of f = 3 with highest powers. Hints and Explanation \t\tDegree of g = 4 \t\t(ii)\tFind LCM of 195 and 221. \t\t \u21d2 Degree of (\u2009\u2009f - g) = 4, since the term of degree \t\t(iii)\tFind the LCM of the expressions given. 4 cannot be vanished. \t40.\t \u2009\u2009\u2009\u2009(i)\tUse factor theorem. 5\t 2.\t x2 9 x 3 5 9 + 4x2 \u2212 3 + 2x \u2212 4 \t\t(ii)\tIf x \u2212 a is HCF of f(x), then f(a) = 0. x x2 9 x 3 5 x \u2212 1 + 3 \t\t(iii)\t\u0007Substitute x = 2 in f(x) or g(x) and obtained the 3 9 + 4x2 \u2212 3 \u2212 2x + 4 3 2 2x value of k. 2x \u2212 1 9 x35 \t41.\t \u2009\u2009\u2009\u2009(i)\tH\u0007 CF is the product of all common factors 3 2 4x2 \u2212 3 \u2212 2x + 4 with least exponents. \u2212 x + 1 \t\t(ii)\tU\u0007 se the HCF concept and find the values of a, 3 4 b and c. +- \t42.\t \u2009\u2009\u2009\u2009(i)\tUse remainder theorem. 2x \u22121+ 3 9 \u2212 3 +1 \t\t(ii)\t\u0007Put x = 0, x = \u22121, and x = \u22122 in the given 3 2x 4x2 2x expression, and obtain the corresponding reminders. 9 \u2212 3 +1 4x2 2x 4\t 3.\t \u2009\u2009\u2009\u2009(i)\tF\u0007 ind the LCM of denominations and add the polynomials. 0 \t\t(ii)\tTake LCM. \t\t\u2234 x2 + 9 \u2212 x \u2212 3 + 5 = x \u2212 1 + 3 . 9 4x2 3 2x 4 3 2 2x \t\t(iii)\tFind the factors and simplify. 4\t 4.\t \u2009\u2009\u2009\u2009(i)\tU\u0007 se algebraic identities to factorize the given polynomial. \t\t(ii)\tMultiply the first three terms with xyz. \t\t(iii)\t\u0007Now express the expression in (a + b + c)2 form. \t45.\t \u2009\u2009\u2009\u2009(i)\tU\u0007 se algebraic identities to factorize the given polynomial.","Polynomials and Square Roots of Algebraic Expressions 3.35 \t53.\t (xy + xz \u2212 yz)2 \u2212 4xyz(x \u2212 y) \t\t= x\u0007 2 + 2x + 1 + y2 + 2y + 1 + z2 + 2z + 1 \u2212 x2 - y2 \u2212 z2 \u2212 2xy \u2212 2yz \u2212 2zx \u2212 3 \t\t= (xy + z(x \u2212 y))2 \u22124(xy) [z(x \u2212 y)] \t\t= 2x + 2y + 2z \u2212 2xy \u2212 2yz \u2212 2zx \t\t= [xy \u2212 z(x \u2212 y)]2 \uf8eb \uf8f6 \t\t[ (a + b)2 - 4ab\t = (a - b)2] \t\t= 2 \u2211x \u2212 2 \u2211 xy = 2\uf8ec \u2211 x \u2212 \u2211 xy \uf8f7 x,y,z x,y,z \uf8ed x,y,z x,y,z \uf8f8 \t\t\t = (xy \u2212 zx + yz)2 \uf8f62 \uf8f7\uf8f7\uf8f8 \t\t\t = (xy + yz \u2212 zx)2 \u2211 \u221155.\t\uf8eb x \u2212 \uf8eb x2 \uf8f6 \uf8ed\uf8ec\uf8ec x,y,z \uf8ec\uf8ed\uf8ec \uf8f7\uf8f7\uf8f8 \t\t\u2234 The square root of the given expression is x,y,z (xy + yz \u2212 zx). \t\t = (x + y + z)2 \u2212 (x2 + y2 + z2) \uf8eb \uf8f62 \t\t = x2 + y2 + z2 + 2xy + 2yz + 2zx \u2212 x2 \u2212 y2 \u2212 z2 \uf8ec\uf8ec\uf8ed x,y,z \uf8f7\uf8f7\uf8f8 \u2211 \u22115\t 4.\t (x + 1)2 \u2212 x \u2212 3 \t\t = 2xy + 2yz + 2zx x,y,z \uf8eb\uf8f6 \t\t= (x + 1)2 + (y + 1)2 + (z + 1)2 - (x + y + z)2 - 3 \u2211\t\t= 2 \uf8ec\uf8ed\uf8ec x,y,z xy \uf8f8\uf8f7\uf8f7. Level 3 \t56.\t (i)\t\u0007Find the square root and equate it to ax2 \t62.\t (x + y)3 + (x \u2212 y)3 \u2212 (y3 + 3x2y) + bx + c. 2 \t\t(ii)\t\u0007Square on both sides and obtain the values of = x3 + y3 + 3x 2 y + 3xy2 + x3 \u2212 y3 \u2212 3x 2 y + 3xy2 Hints and Explanation a, b and c. 2 \t\t \u2212 y3 \u2212 3x2y \t\t(iii)\tVerify the relation between a, b and c. 5\t 7.\t \u2009\u2009\u2009\u2009(i)\tF\u0007ind average of the given polynomials and then apply the division method. \t\t= 2x3 + 6xy2 \u2212 y3 \u2212 3x2y 2 \t\t(ii)\tAverage of a, b and c is a + b + c . 3 \t\t= x3 \u2212 3x2y + 3xy2 \u2212 y3 \t\t(iii)\t\u0007Express the numerator in the form of (a + b + c)2. \t\t= (x \u2212 y)3. 6\t 1.\t a a2 + 1 \u2212 1 + a \u2212 3 \u2212 a2 a + 1 \u2212 1 \t63.\t (4a + 5b + 5c)2 \u2212 (5a + 4b + 4c)2 + 9a2 2 4 a2 a 2 4 4 2 2 a \t\t = \u0007(4a + 5b + 5c + 5a + 4b + 4c) (4a + 5b + 5c \u2212 5a 2a + 1 a2 + 1 \u2212 1 + a \u2212 3 \u2212 a2 \u2212 4b \u2212 4c) + 9a2 2 2 4 a2 a 2 4 4 \t\t = (9a + 9b + 9c)(\u2212a + b + c) + 9a2 \t\t = 9(a + b + c)(\u2212a + b + c) + 9a2 a + 1 \u2212 1 1 \u2212 1 \u22121 \t\t = 9(b + c)2 \u2212 9a2 + 9a2 a a2 a \t\t = 9(b + c)2. 1 \u2212 1 \u22121 a2 a \t\t\\\\\t The square root of the given expression is 0 9(b + c )2 = 3(b + c ). \t\t\u2234 The square root of the given expression is \u2212 b)3 + b)3 a \u2212 1 + 1 . 64.\t (a \u2212 (a + a(a2 + 3b2 ) 2 a 2 2","3.36 Chapter 3 = a3 \u2212 3a2b + 3ab2 \u2212 b3 \u2212 a3 \u2212 3a2b \u2212 3ab2 \u2212 b3 \t\t = (5a + 5b + 5c)(a \u2212 b + c) + 5b2 2 \t\t = 5(a + c + b)(a + c \u2212 b) + 5b2 \t\t + a3 + 3ab2 \t\t = 5[(a + c)2 \u2212 b2] + 5b2 \t\t = 5(a + c)2 \t\t= \u22126a2b \u2212 2b3 + a3 + 3ab2 \t\t\\\\\t Square root of the given expression is 2 \t\t 5(a + c )2 = 5(a + c ). \t\t = a3 \u2212 3a2b + 3ab2 \u2212 b3 \t\t = (a \u2212 b)3. 6\t 5.\t (3a + 2b + 3c)2 \u2212 (2a + 3b + 2c)2 + 5b2 \t\t = (\u0007 3a + 2b + 3c + 2a + 3b + 2c) (3a + 2b + 3c \u2212 2a \u2212 3b \u2212 2c) + 5b2 Hints and Explanation","142CChhaapptteerr ELKiqniuneeaamtrioatnicss and Inequations REmEmBER Before beginning this chapter, you should be able to: \u2022 Understand basic terms such as linear equation and inequations in mathematics \u2022 Solve word problems and applications of simple algebraic equations KEY IDEAS After completing this chapter, you should be able to: \u2022 Solve two simultaneous linear equations \u2022 Learn the graphical representation of linear equations and inequations \u2022 Understand the application of simultaneous linear equations \u2022 Solve word problems on linear equations and inequations \u2022 Obtain the absolute value Figure 1.1","4.2 Chapter 4 INTRODUCTION In most cases, while solving the problem, first we need to frame an equation. In this chapter, we learn how to frame and solve equations. Framing an equation is more difficult than solving it. Now, let us review the basic concepts related with equations and inequations. Algebraic Expressions Expressions of the form 2x, 3x + 5, 4x - 2y, 2x2 + 3 y , 3x3 are algebraic expressions. 3x and 2y 5 are the terms of 3x + 5, and 4x and 2y are the terms of 4x - 2y. Algebraic expressions are made up of numbers, symbols and the basic arithmetical operations. Open Sentences Sentences which are true for some values of the variable and false for the other values of the variable are called open sentences. When a certain value is substituted for the variable, the sentence becomes a statement, either true or false. Example: x + 1 = 4. This is an open sentence. When we substitute 3 for x, we get a true statement. When we substitute any other value, we get a false statement. Example: x + 1 < 3. This is another open sentence. If we substitute any value less than 2 for x we get a true statement. If we substitute 2 or any number greater than 2, we get a false statement. EQUATION An open sentence containing the equality sign is an equation. In other words, an equation is a sentence in which there is an equality sign between two algebraic expressions. Example: 2x + 5 = x + 3, 3y - 4 = 20, 5x + 6 = x + 1 are equations. Linear Equation An equation, in which the highest index of the unknowns present is one, is a linear equation. Example: 2(x + 5) = 18, 3x - 2 = 5, x + y = 20 and 3x - 2y = 5 are some linear equations. Simple Equation A linear equation which has only one unknown is a simple equation. 3x + 4 = 16 and 2x - 5 = x + 3 are examples of simple equations. The part of an equation which is to the left side of the equality sign is known as the left hand side, abbreviated as LHS. The part of an equation which is to the right side of the equality sign is known as the right hand side, abbreviated as RHS. The process of finding the value of an unknown in an equation is called solving the equation. The value (values) of the unknown found after solving an equation is (are) called the solution(s) or the root(s) of the equation. Before we learn how to solve an equation, let us review the basic properties of equality. Solving an Equation in One Variable The following steps are involved in solving an equation: Step 1:\u2002 Always ensure that the unknown quantities are on the LHS and the known quantities or constants on the RHS.","Linear Equations and Inequations 4.3 Step 2:\u2002 Add all the terms containing the unknown quantities on the LHS and all the known quantities on the RHS so that each side of the equation contains only one term. Step 3:\u2002 Divide both sides of the equation by the coefficient of the unknown. Example 4.1 If 2x + 10 = 40, find the value of x. Solution Step 1:\u2002 Group the known quantities as the RHS of the equation, i.e., 2x = 40 - 10 Step 2:\u2002 Simplify the numbers on the RHS \u21d2 2x = 30. Step 3:\u2002 Since 2 is the coefficient of x, divide both sides of the equation by 2. 2x = 30 \u21d2 x = 15. 2 2 Example 4.2 Solve for x: 5x - 8 = 3x + 22. Solution Step 1:\u20025x - 3x = 22 + 8 Step 2:\u20022x = 30 Step 3:\u2002 2x = 30 \u21d2x = 15. 2 2 Example 4.3 A teacher has 45 chocolates. After giving two chocolates to each student, she is left with 7\u00a0chocolates. How many students are there in the class? Solution Let the number of students in the class be x. Total number of chocolates distributed = 2x Total number of chocolates left with the teacher = 7 2x + 7 = 45 \u21d2 2x = 45 - 7 \u21d2 2x = 38 \u21d2 x = 19. There are 19 students in the class. SIMULTANEOUS LINEAR EQUATIONS We have learnt how to solve an equation with one unknown quantity. Very often we come across equations involving more than one unknown quantity. In such cases we require more than one condition or equation. Generally, when there are two unknowns, we require two equations to solve the problem. When there are three unknowns, we require three equations and so on.","4.4 Chapter 4 We need to find the values of the unknowns that satisfy all the given equations. Since the values satisfy all the given equations we call them simultaneous equations. In this chapter, we deal with simultaneous (linear) equations with two unknowns. Let us consider the equation, 3x + 4y = 15, which contains two unknown quantities x and y. Here, 4y = 15 \u2212 3x \u21d2 y = 15 \u2212 3x \b(1) 4 In the above equation for every value of x, there exists a corresponding value for y. When x = 1, y = 3. When=x 2=, y 9 and so on. 4 If there is another equation, of the same kind, say 5x \u2212 y = 2, from this we get, y = 5x \u2212 2\b (2) If we need the values of x and y such that both the equations are satisfied, then 15 \u2212 3x = 5x \u2212 2 \u21d2 15 \u2212 3x = 20x \u22128 4 \u21d2 23x = 23 \u21d2 x = 1. On substituting the value of x = 1 in the first equation, we get, y = 15 \u2212 3(1) \u21d2 y = 3. 4 If both the equations are to be satisfied by the same values of x and y, there is only one solution. Thus we can say that when two or more equations are satisfied by the same values of unknown quantities then those equations are called simultaneous equations. Solving Two Simultaneous Equations (1) When two equations, each in two variables, are given, they can be solved in four ways. (2) 1.\t Elimination by cancellation. 2.\t Elimination by substitution. 3.\t Adding the two equations and subtracting one equation from the other. 4.\t Graphical method. Elimination by Cancellation Example 4.4 If 4x + 3y = 25 and 5x + 2y = 26, then find the values of x and y. Solution \t4x + 3y = 25\b \t5x + 2y = 26","Linear Equations and Inequations 4.5 Using this method, the two equations are reduced to a single variable equation by eliminating one of the variables. Step 1:\u2002 Here, let us eliminate the y term, and in order to eliminate the y term, equate the coefficient of y in both the equations. \t(4x + 3y = 25)2 \u21d2 8x + 6y = 50\b (3) \t(5x + 2y = 26)3 \u21d2 15x + 6y = 78\b (4) Step 2:\u2002 Subtract equation (3) from (4), \t(15x + 6y) \u2212 (8x + 6y) = 78 \u2212 50 \u21d2 7x = 28 \t \u21d2 x = 4. Step 3:\u2002 Substitute the value of x in Eq. (1) or Eq. (2) to find the value of y. Substituting the value of x in the first equation, we have, 4(4) + 3y = 25 \u21d2 3y = 25 \u2212 16 \u21d2 3y = 9 \u21d2 y = 3. \u2234 The solution of the given pair of equations is (x, y), i.e., (4, 3). Elimination by Substitution Example 4.5 If 3x \u2212 2y = 12 and 6x + y = 9, then find the values of x and y. Solution (1) 3x - 2y = 12\b (2) 6x + y = 9\b Using this method, the two equations are reduced to a single variable equation by substituting the value of one variable, obtained from one equation, in the other equation. Step 1:\u2002 Using the first equation, find x in terms of y, i.e., 3x \u2212 2y = 12 \u21d2 3x = 12 + 2y \u21d2 x = 12 + 2y \b(3) 3 Step 2:\u2002Substitute the value of x in the second equation to find the value of y, i.e., 6x + y = 9 \u21d2 6 \uf8eb 12 + 2y \uf8f6 + y = 9 \uf8ed\uf8ec 3 \uf8f8\uf8f7 Step 3:\u2002 Simplify the equation in terms of y and find the value of y. \u20092(12 + 2y) + y = 9 \u21d2 24 + 4y + y = 9 \u2009\u21d2 5y = 9 \u2212 24 \u2009 \u21d2 y = \u22123.","4.6 Chapter 4 Step 4:\u2002 Substituting the value of y in Eq. (3) we get x = 2. \u2234 The solution for the given pair of equations is (x, y), i.e., (2, -3). Adding Two Equations and Subtracting One Equation from the Other We use this method, in case of solving the system of linear equations of the form ax + by = p and bx + ay = q. Example 4.6 (1) Solve 4x + 5y = 37 and 5x + 4y = 35. (2) Solution (3) Given,\t4x + 5y = 37\b and\t5x + 4y = 35\b (4) Step 1:\u2002 Adding both equations, we get \u20099\u2009 x + 9y = 72 \u21d2 9(x + y) = 9 \u00d7 8 \u21d2 x + y = 8\b Step 2:\u2002 Subtracting Eq. (2) from the Eq. (1), (4x + 5y) \u2212 (5x + 4y) = 37 \u2212 35 \u21d2 \u2212x + y = 2\b Step 3:\u2002 Adding the Eqs. (3) and (4), (x + y) + (\u2212x + y) = 8 + 2 \u21d2 2y = 10 \u21d2 y = 5. Substituting y = 5 in any of the Eqs. (1), (2), (3) or (4), we get x = 3. \u2234 The solution of the pair of equations is (x, y), i.e., (3, 5). \u2009\u2009Note\u2002 \u2002 Choosing a particular method to solve a pair of equations makes the simplification easier. One can learn which method is easiest to solve a pair of equations by becoming familiar with the different methods of solving the equation. Graphical Method: Plotting the Points If we consider any point in a plane, then we can determine the location of the given point, i.e., we can determine the distance of the given point from X-axis and Y-axis. Therefore, each point in the plane represents the distance from both the axes. So, each point is represented by an ordered pair and it consists of x-coordinate and y-coordinate. The first element of an ordered","Linear Equations and Inequations 4.7 pair is called x-coordinate and the second element of an ordered pair is called y-coordinate. In the first quadrant Q1, both the x-coordinate and y-coordinate are positive real numbers. In the second quadrant Q2, x-coordinates are negative real numbers and y-coordinates are positive real numbers. In the third quadrant Q3 both the x-coordinate and y-coordinate are negative real numbers. In the fourth quadrant Q4, x-coordinates are positive real numbers and y-coordinates are negative real numbers. And the origin is represented by (0, 0). Consider the point (2, 3). Here, 2 is the x-coordinate and 3 is the y-coordinate. The point (2,\u00a03) belongs to the first quadrant. The point (2, 3) is 2 units away from the Y-axis and 3 units away from the X-axis. If we consider the point (-3, -5), -3 is x-coordinate and -5 is y-coordinate. The point (-3, -5) belongs to Q3 and is 3 units away from the Y-axis and 5 units away from the X-axis. The method of plotting a point in a co-ordinate plane was explained by a Ren\u00e9 Descartes, a\u00a0French mathematician. To plot a point say P(-3, 4), we start from the origin and proceed 3 units towards the left hand side along the X-axis (i.e., negative direction as x-coordinate is negative), and from there we move 4 units upwards along the Y-axis (i.e., positive direction as y-coordinate is positive). Example 4.7 Plot the following points on the co-ordinate plane: A(3, 5), B(2, \u22124), C(\u22122, 7), D(\u22123, \u22124), E(0, \u22125) and F(5, 0) Solution Y 8 \u20227 \u2022A(3, 5) C( \u2212 2, 7) 6 5 4 3 2 1 0 \u2022F(5, 0) X \u22127 \u2212 6 \u2212 5 \u2212 4 \u22123 \u22122 \u22121 \u22121 1 2 34 5 67 8 \u22122 \u22123 \u2022 \u2022\u22124 B(2,\u2212 4) \u2022\u22125 D(\u2212 3, \u2212 4) \u22126 E(0, \u22125) \u22127 Figure 4.1","4.8 Chapter 4 Example 4.8 Plot the following points on the coordinate plane. What do you observe? (a)\t (-2, -3), (-1, -3), (0, -3), (2, -3) (b)\t (5, 3), (5, 2), (5, 1), (5, 0), (5, -1). Y 7 6 5 4 \u20223 (5, 3) \u20222 (5, 2) \u20221 (5, 1) \u22126 \u22125 \u22124 \u22123 \u22122 \u22121 0 1 2 3 4 5 \u2022(5, 0) 6X \u22121 \u2022 (5, \u20131) \u22122 (\u22121, \u22123) (0, \u22123) \u2022(2, \u22123) \u22123)\u2022 \u2022(\u22122, \u22123 \u22124 \u22125 \u22126 Figure 4.2 Solution (a)\t (\u22122, -3), (-1, -3), (0, \u22123), (2, \u22123) \t(i)\tThe above points lie on the same straight line which is perpendicular to the Y-axis. \t(ii)\t The y-coordinates of all the given points are the same, i.e., y = \u22123. \t(iii)\t So, the straight line passing through the given points is represented by y = \u22123. \t(iv)\t Therefore, the line y = \u22123, is parallel to X-axis which intersects Y-axis at (0, \u22123). (b)\t (5, 3), (5, 2), (5, 1), (5, 0), (5, \u22121) \t(i)\tThe above points lie on the same straight line which is perpendicular to the X-axis. \t(ii)\t The x-coordinate of each of the given points is the same, i.e., x = 5. \t(iii)\t So, the straight line passing through the given points is represented by x = 5. \t(iv)\t Therefore, the line x = 5 is parallel to the Y-axis which intersects the X-axis at (5, 0).","Linear Equations and Inequations 4.9 \u2002Notes\u2002 1.\t T\u0007 he y-coordinate of every point on the X-axis is zero, i.e., y = 0. Therefore, the X-axis is denoted by y = 0. 2.\t T\u0007 he x-coordinate of every point on the Y-axis is zero, i.e., x = 0. Therefore, the Y-axis is denoted by x = 0. Example: The line x = 4 is parallel to Y-axis which intersects X-axis at (4, 0). Example: The line y = 5 is parallel to X-axis which intersects Y-axis at (0, 5). Example 4.9 Plot the following points on the coordinate plane and what do you observe? (-3, 3), (\u22122, 2), (\u22121, 1), (0, 0), (1, \u22121), (2, \u22122), (3, \u22123) Solution (a)\t All the given points lie on the same straight line. (b)\t Every point on the straight line represents y = \u2212x. (c)\t The above line with the given ordered pairs is represented by the equation y = \u2212x. y = \u2212x Y ( \u22123, 3) 5 4 3 (\u22122, 2) 2 1 (\u22121, 1) (0, 0) \u22125 \u22124 \u22123 \u22122 \u22121 0 123 4 5X \u22121 (1, \u22121) \u22122 (2, \u22122) \u22123 (3, \u22123) \u22124 \u22125 Figure 4.3 Example 4.10 Draw the graph of the equation y = 3x where R is the replacement set for both x and y. Solution x \u22122 \u22121 0 1 2 y = 3x \u22126 \u22123 0 3 6","4.10 Chapter 4 Some of the ordered pairs which satisfy the equation y = 3x are (\u22122, \u22126), (\u22121, -3), (0, 0), (1, 3), (2, 6). By plotting the above points on the graph sheet, we get the following: Y y = 3x 6 5 (2, 6) 4 \u20223 (1, 3) 2 1 \u22126 \u22125 \u22124 \u22123 \u22122 \u22121 0 \u2022 (0, 0)123 4 5 6X \u22121 \u22122 \u2022(\u22121, \u22123) \u22123 \u22124 \u22125 \u2022(\u22122 ,\u22126 ) \u22126 Figure 4.4 Example 4.11 Draw the graph of the equations x + y = 4 and x \u2212 y = 2. What do you notice? Solution (a)\t x + y = 4 x -2 -1 0 1 2 3 4 y=4-x 6 5 4 3 2 1 0 Some of the ordered pairs which satisfy the equation x + y = 4 are (-2, 6), (-1, 5), (0, 4), (1, 3), (2, 2), (3, 1), (4, 0). (b)\t x - y = 2 x \u22122 \u22121 0 1 2 3 4 5 y = x \u22122 \u22124 \u22123 \u22122 \u22121 0 1 2 3","Linear Equations and Inequations 4.11 \u2234 Some of the ordered pairs which satisfy the equation x \u2212 y = 2 are (\u22122, \u22124), (\u22121, \u22123), (0, \u22122), (1, \u22121), (2, 0), (3, 1), (4, 2), (5, 3). x+y=4 Y (\u22122, 6) 6 (\u22121, 5) 5 4 (0, 4) 3 (1, 3) x\u2212y=2 2 (2, 2) (5, 3) (4, 2) 1 (3, 1) \u22126 \u22125 \u22124 \u22123 \u22122 \u22121 (0, 0) (2, 0) (4, 0) 6X 01 23 45 (\u22121, \u22123) \u22121 (1, \u20131) \u22122 (0, \u20132) \u22123 (\u22122 , \u22124 ) \u22124 \u22125 \u22126 Figure 4.5 From the above graph, we notice that the two given equations intersect at the point (3, 1). That is, x + y = 4 and x \u2212 y = 2 have a common point, (3, 1).Therefore, (3, 1) is the solution of the equations x + y = 4 and x - y = 2. Verification: (1) \tx + y = 4\b (2) \tx \u2212 y = 2\b Solving Eqs. (1) and (2), we get, x = 3 and y = 1. \u2234 (3, 1) is the solution of x + y = 4 and x \u2212 y = 2. \u2009\u2009Note\u2002 \u2002 From the above example, we notice that we can find the solution for simultaneous equations by representing them in a graph, i.e., by using the graphical method. Example 4.12 Solve the following equations x + 4y = 2 and 4x \u2212 y = \u22129 by the graphical method and check the result.","4.12 Chapter 4 Solution (a)\t x + 4y = 2 x 2 6 \u22122 \u22126 y = 2 \u2212 x 0 \u22121 1 2 4 Some of the ordered pairs which satisfy the equation x + 4y = 2 are (2, 0), (6, \u22121), (\u22122, 1), (\u22126, 2). (b)\t 4x \u2212 y = \u22129 x \u22121 \u22122 0 y = 4x + 9 5 1 9 Some of the ordered pairs which satisfy the equation 4x \u2212 y = \u22129 are (\u22121, 5), (\u22122, 1), (0, 9). By plotting the above points on graph sheet, we get the following graph: Y 4x \u2212 y = \u22129 9\u2022 8 7 6 \u2022 5 4 3 \u2022 2 \u22128 \u22127 \u22126 \u22126 \u22125 \u22124 \u22123 \u20221 \u2022 \u22122 \u22121 0 1 23 4 5 6\u2022 7 8 9 10 X \u22121 x + 4y = 2 \u22122 \u22123 \u22124 Figure 4.6 From the graph, it is clear that, the given equations intersect at (\u22122, 1). Therefore, it is the common solution for the equations x + 4y = 2 and 4x \u2212 y = \u2212 9. Verification: \tx + 4y = 2\b (1) \t4x \u2212 y = \u22129\b(2) By solving Eqs. (1) and (2), we get x = -2 and y = 1. \u2234 The solution is (\u22122, 1) Hence, verified.","Linear Equations and Inequations 4.13 Nature of Solutions When we try to solve a pair of equations we could arrive at three possible results. They are, having 1.\t A unique solution. 2.\t An infinite number of solutions. 3.\t No solution. Let the pair of equations be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where a1, b1, a2 and b2 are the coefficients of x and y terms, while c1 and c2 are the known constant quantities. A Pair of Equations Having a Unique Solution If, a1 \u2260 b1 , then there will be a unique solution. a2 b\u03c4 2 We have solved such equations in the previous examples of this chapter. A Pair of Equations Having Infinite Solutions If aa=12 bb=12 c1 , then the pair of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will have c2 infinite number of solutions. \u2009\u2009Note\u2002 \u2002 In fact this means that there are no two equations as such and one of the two equations is simply obtained by multiplying the other with a constant. These equations are known as dependent equations. Example 4.13 Find out the number of solutions for the following equations: 3x + 4y = 8 9x + 12y = 24 Solution For these two equations a1 = 3, a2 = 9, b1 = 4, b2 = 12, c1 = \u22128, c2 = \u221224. \u2234 a1 = b1 = c1 a2 b2 c2 Since, 3 = 4 = \u22128 4 12 \u221224 \u2234 The above pair of equations will have infinite solutions. A Pair of Equations Having no Solution at All If a1 = b1 \u2260 c1 , then the pair of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, will have a2 b2 c2 no solution. Such lines are called parallel lines. \u2002Notes\u2002 1.\t I\u0007n other words, the two equations will contradict each other or be inconsistent with each other. 2.\t A pair of equations are said to be consistent if they have a solution (finite or infinite).","4.14 Chapter 4 Example 4.14 Find out the number of solutions for the following equations: 4x + 5y = 20 8x + 10y = 30 Solution For these two equations, a1 = 4, a2 = 8, b1 = 5, b2 = 10, c1 = \u221220, c2 = \u221230 a1 = b1 \u2260 c1 \u21d2 4 = 5 \u2260 \u221220 a2 b2 c2 8 10 \u221230 Hence, given pair of equations have no solution at all. Word Problems and Application of Simultaneous Equations In this chapter, we have discussed earlier that it is essential to have as many equations as there are unknown quantities to be determined. In word problems also, it is required that there are as many independent conditions as there are unknown quantities to be determined. Let us understand with the help of the following examples as to how word problems can be solved using simultaneous equations. Example 4.15 The sum of two numbers is 30 and the larger number exceeds the smaller by 6. Find the numbers. Solution Let the numbers be a and b. Given that a + b = 30 and a \u2212 b = 6. By adding both the equations, 2a = 36 \u21d2 a = 18 Substituting a = 18 in the first equation, 18 + b = 30 \u21d2 b = 12 \u2234 The two numbers are 18 and 12. Example 4.16 In a fraction, if unity is added to the numerator and subtracted from the denominator, it becomes 2 . Instead, if unity is subtracted from the numerator and added to the denominator, 3 1 it becomes 2 . Find the fraction. Solution x y Let the fraction be , applying the first condition, we get,","Linear Equations and Inequations 4.15 x +1 = 2 y \u22121 3 \u21d2 3x + 3 = 2y \u2212 2 \u21d2 3x \u2212 2y = \u22125 \b(1) Applying the second condition, we get, x \u22121 = 1 y +1 2 \u21d2 2x \u2212 3 = 2y + 1 \u21d2 2x \u2212 y = 3 \b(2) By solving the Eqs. (1) and (2) using any of the methods discussed earlier, we get x = 11 and y = 19. \u2234 The fraction is 11 . 19 Example 4.17 The sum of the digits of a two digit number is 8. If 18 is added to the number, then the resultant number is equal to the number obtained by reversing the digits of the original number. Find the original number. Solution Let the number be in the form of 10x + y, where x and y are the tens digit and the units digit respectively. Applying the first condition, we get x + y = 8\b (1) Applying the second condition, given in the problem, 10x + y + 18 = 10y + x \u21d2 9x \u2212 9y = \u221218 \u21d2 x \u2212 y = \u22122\b(2) By solving Eqs. (1) and (2), we get, x = 3 and y = 5 and the number is 35. Example 4.18 Four years ago, age of a person was 4 times that of his son. Six years later, the age of the person will be 10 years less than thrice the age of his son. Find the present ages of the person and his son. Solution Let the present ages of the person and his son be x years and y years respectively. Given,","4.16 Chapter 4 \u2009\u2009x \u2212 4 = 4(y \u2212 4) \u2009\u21d2 x \u2212 4 = 4y \u2212 16 \u21d2 x \u2212 4y = \u221212\b(1) And also given, \u2009x + 6 = 3 (y + 6) \u221210 (2) \u21d2 x + 6 = 3y + 8 \u21d2 x \u2212 3y = 2\b By solving the Eqs. (1) and (2), we get, x = 44 and y = 14. Example 4.19 For what values of k is the set of equations 2x \u2212 3(2k \u2212 1)y = 10 and 3x + 4(k + 1)y = 20 are consistent? Solution Consider the equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. If this pair of equations is consistent, a1 \u2260 b1 or a1 = b1 = c1 a2 b2 a2 b2 c2 As\t a1 \u2260 b1 , a2 b2 let a1 = 2, b1 = \u22123(2k \u2212 1), a2 = 3 and b2 = 4(k + 1) Now, 2 \u2260 \u22123(2k \u2212 1) \u21d2 8k + 8 \u2260 \u221218k + 9 \u21d2 26k \u2260 1\u21d2 k \u2260 1 3 4(k + 1) 26 \u2234k \u2208 R \u2212 \uf8f11 \uf8fc\uf8fd . \uf8f3\uf8f2 26 \uf8fe LINEAR INEQUATIONS Introduction If a is any real number, then a is either positive or negative or zero. When a is positive, we write a\u00a0> 0, which is read as \u2018a is greater than zero\u2019. When a is negative, we write a < 0, which is read as \u2018a is less than zero\u2019. If a is zero, we write a = 0 and in this case, a is neither positive nor negative. The two signs > and < are called the \u2018signs of inequalities\u2019. Notation 1.\t \u2018>\u2019 denotes \u2018greater than\u2019. 2.\t \u2018<\u2019 denotes \u2018less than\u2019. 3.\t \u2018\u2265\u2019 denotes \u2018greater than or equal to\u2019. 4.\t \u2018\u2264\u2019 denotes \u2018less than or equal to\u2019.","Linear Equations and Inequations 4.17 Definition For any two non-zero real numbers a and b, 1.\t a is said to be greater than b when a \u2212 b is positive, i.e., a > b when a \u2212 b > 0 and 2.\t a is said to be less than b when a \u2212 b is negative, i.e., a < b when a \u2212 b < 0. Listed below are some properties\/results which are needed to solve problems on inequalities. The letters a, b, c, d, etc. represent real numbers. \u2022\u2022 For any two real numbers a and b, either a > b or a < b or a = b. This property is called Law of Trichotomy. \u2022\u2022 If a > b, then b < a. \u2022\u2022 If a > b and b > c, then a > c. This property is called transitive property. \u2022\u2022 If a > b then a + c > b + c. \u2022\u2022 If a > b and c > 0, then ac > bc. \u2022\u2022 If a > b and c < 0, then ac < bc. \u2022\u2022 If a > b and c > d, then a + c > b + d. \u2022\u2022 The square of any real number is always greater than or equal to 0. \u2022\u2022 If a > 0, then \u2212a < 0 and if a > b, then \u2212a <\u2212b. 1 1 \u2022\u2022 If a and b are positive numbers and a > b, then a < b . \u2022\u2022 If a and b are negative numbers and a > b, then 1 < 1 . a b 1 1 \u2022\u2022 If a > 0 and b < 0, then a > b . INEQUATION An open sentence which consists of one of the symbols, viz., >, <, \u2265, \u2264 is called an inequation. Examples:\u2003 1. 3x - 8 > 8\t\t 2. 2x2 - 3x \u2264 6 Continued Inequation Two inequations of the same type (i.e., both consisting of > or \u2265 or both consisting of < or \u2264) can be combined into a continued inequation as explained below. Examples: 1.\t If a < b and b < c, we can write a < b < c. 2.\t If a \u2265 b and b > c, we can write a \u2265 b > c. Linear Inequation An inequation in which the highest degree of the variables present is one is called a linear inequation. Examples: 1.\t 3x + 4 \u2264 8 - 3x and 8x - 64 \u2265 8 + 5y are some of the examples for linear inequations. 2.\t 5x2 + 6 > 7 and 6x3 + 6y3 \u2264 8 are some of the examples for non-linear inequations. Solving Linear Inequation in One Variable We are familiar with solving linear equations. Now let us look at solving some linear inequations in one variable.","4.18 Chapter 4 Example 4.20 Solve the following inequations: (a)\u20025x \u2013 3 \u2264 12, X \u2208 N\u2003\u2003\u2003(b)\u20022x \u2212 4 < 4, x \u2208 R\u2003\u2003\u2003(c)\u20023x \u2212 1 \u2265 5, x \u2208 Z Solution (a)\t 5x - 3 \u2264 12, X \u2208 N \u21d2 5x \u2264 15 \u21d2 x \u2264 3 \u21d2 x \u2208\u2009{1, 2, 3}. (b)\t 2x \u2212 4 < 4, x \u2208 R \u21d2 2x < 8 \u21d2 x < 4 T\u0007 he set of all the numbers which are less than 4 is the solution set of the given inequation. (c)\t 3x \u2212 1 \u2265 5, x \u2208 Z \u21d2 3x \u2265 6 \u21d2 x \u2265 2 \u0007The set of all the integers which are greater than or equal to 2 is the solution set of the given in equation. Example 4.21 Represent the following inequations on the number line. (a)\t x \u2265 \u22122 Solution \u2212\u221e \u221e \u22124 \u22123 \u22122 \u22121 \u22121 0 1 2 3 Figure 4.7 In the above figure, the ray drawn above the number line represents the solution set of the inequation. The end point of the ray is a part of the solution set. This is indicated by placing a solid dot at the end point. (b)\t x \u2264 3 Solution \u2212\u221e \u221e \u22124 \u22123 \u22122 \u22121 \u22121 0 1 2 3 Figure 4.8 In the figure above, the ray drawn above the number line represents the solution set of the inequation. The end point of the ray is the part of the solution set. This is indicated by placing a solid dot at the end point. Example 4.22 Draw the graph x \u2265 1 in the cartesian plane. Solution We first draw the line x = 1 which is parallel to Y-axis and 1 unit away from the Y-axis. Then shade the region x \u2265 1. The boundary line for the graph x \u2265 1 is x = 1.","Linear Equations and Inequations 4.19 Y x>2 x=1 4 3 2 1 \u22124 \u22123 \u22122 \u22121 0 1 2 3 4 5 X \u22121 \u22122 \u22123 \u22124 Figure 4.9 \u2002Notes\u2002 1.\t F\u0007 rom the above graph, we notice that every line in a plane divides the plane into two half planes. 2.\t W\u0007 hen the boundary line is included, the region is called a Closed Half Plane. 3.\t W\u0007 hen the boundary line is not included, the region is called Open Half Plane. Example 4.23 Draw the graph of y < 2 in the Cartesian plane. Solution Draw the line y = 2 with dots since the boundary line does not belong to the graph and then shade the region y < 2. Y 4 3 2 1 \u20134 \u22123 \u22122 \u22121 0 12 34 5 X \u22121 \u22122 y < 2 \u22123 \u22124 Figure 4.10","4.20 Chapter 4 Solving Linear Inequations in Two Variables We have already learnt about linear equations, their representation as graphs and solving linear equations in two variables with the help of graphs. These concepts are very useful in drawing the graphs representing linear inequations. Finding Solution for the Inequation of Two Variables We know that a linear equation of two variables of the form ax + by + c = 0 divides the plane into three mutually disjoint sets of points. They are as follows: 1.\t Points lying on the line ax + by + c = 0. 2.\t Points lying on one side of the line ax + by + c = 0. 3.\t Points lying on the other side of the line ax + by + c = 0. \u0007The line ax + by + c = 0 divides the plane into two half planes and they are represented by ax\u00a0+\u00a0by + c < 0 and ax + by + c > 0. 1.\t T\u0007 he solution set of ax + by + c > 0 is the region that contain (0, 0) when c > 0 and the region does not contain (0, 0) when c < 0. 2.\t T\u0007 he solution set of ax + by + c < 0 is the region that contain (0, 0) when c < 0 and the region that does not contain (0, 0) when c > 0. 3.\t \u0007When c = 0 then origin lies on the line ax + by = 0. In this case, we choose any orbitary point that does not lie on ax + by = 0 and substitute in the given inequation. If it results a true statement then the region containing the point is the solution region otherwise the other region is the solution region of the inequation. Example 4.24 Draw the graph x \u2212 y \u2264 1 in the cartesian plane. Steps Y x\u2212y\u22641 (a)\t First draw the line x \u2212 y = 1, i.e., y = x \u2212 1 4 x 1 0 \u22121 3 y 0 \u22121 \u22122 2 1 As a first step, plot the ordered pairs (1, 0), (0, \u22121), (\u22121, \u22122) and then join them with a \u22124 \u22123 \u22122 \u22121 0 1 2 3 4 5 X line. \u22121 (b)\t F\u0007 or shading the region we consider any point \u22122 in one of the half planes. \u22123 Let us consider the point (0, 0) and substitute the coordinates of the point in the inequation x \u2212 y \u22124 \u2264 1 \u21d2 0 \u2212 0 \u2264 1 \u21d2 0 \u2264 1, which is true. \u2234 (0, 0) belongs to the graph of the inequality Figure 4.11 x \u2212 y \u2264 1. Now shade the region which includes the point (0, 0). Here, boundary line x \u2212 y = 1 is included.","Linear Equations and Inequations 4.21 Example 4.25 Y X Draw graph of x < \u2212y 4 Solution 3 (a)\t Draw x = \u2212y 2 1 X \u22121 0 1 Y 1 0 \u22121 \u22124 \u22123 \u2212 2 \u22121 O 1 2 3 4 5 \u22121 (b)\t Substitute (1, 1) in x < -y \u22122 \u21d2 1 < \u22121 which is not true \u22123 \u0007\u2234 (1, 1) does not belong to the graph \u22124 represented by x < \u2212y. Figure 4.12 (c)\t \u0007Now shade the region (half plane) which does not contain the point (1, 1). \u2009\u2009Note\u2002 \u2002 The dotted line indicates the non-inclusion of the line x = \u2212y in the graph. System of Inequations The graph of a linear inequation in two variables is one of the half planes separated by the boundary line. Any ordered pair which satisfies the inequation is a solution. Now, let us learn to solve the simultaneous inequations in two variables. Example 4.26 Y Construct the region represented 4 by\u00a0 the inequations x + 3y \u2265 3 and 3 3x\u00a0+ y \u2264 3. 2 \u20221 Solution (a)\t Consider x + 3y = 3 \u21d2y = 3\u2212x x + 3y \u2265 3 3 x0 3 \u22124 \u22123 \u22122 \u22121 01 2 34 5 X y1 0 3x + y \u2264 3 \u2022 \u22121 3x + y = 3 \u22122 Substitute (0, 0) in x + 3y \u2265 3 \u21d20+0\u22653 \u22123 \u21d2 0 \u2265 3 which is false \u2234 x + 3y \u2265 3 does not contain (0, 0). \u22124 (b)\t Consider 3x + y = 3 \u21d2 y = 3 - 3x \u22125 Figure 4.13 x10 y03 Substitute (0, 0) in the inequation 3x + y \u2264 3","4.22 Chapter 4 \u21d2 0 + 0 \u2264 3 which is true \u2234 3x + y \u2264 3 contains (0, 0). The region common to the two half planes is the graph of the given system of inequations. Example 4.27 Construct the region represented by the inequations 2x + y \u2264 2 and x \u2212 3y \u2264 3. Solution (a)\t Consider 2x + y = 2 \u21d2 y = 2 \u2212 2x x01 y20 Substitute (0, 0) in 2x + y \u2264 2 \u21d2 2(0) + 0 \u2264 2 \u21d2 0 \u2264 2 which is true \u2234 2x + y \u2264 2 contains (0, 0). (b)\t Consider x \u2212 3y = 3 y = x \u2212 3 3 x0 3 y \u22121 0 Substitute (0, 0) in x \u2212 3y \u2264 3 \u21d2 0 \u2212 0 \u2264 3 \u21d2 0 \u2264 3 which is true \u2234 x \u2212 3y \u2264 3 contains (0, 0). The region common to the two half planes is the graph of the system of inequations. Y 2x + y \u2264 2 4 x \u2212 3y \u2264 3 3 \u20222 \u20221 \u22124 \u2212 3 \u2212 2 \u22121 0 \u2022 \u2022 5 X 12 34 \u2022 \u22121 \u22122 \u22123 \u22124 \u22125 Figure 4.14","Linear Equations and Inequations 4.23 ABSOLUTE VALUE If a is any real number, then 1.\t |x| \u2264 a \u21d2 -a \u2264 x \u2264 a 2.\t |x| \u2265 a \u21d2 x \u2265 a or x \u2264 -a Properties of Modulus 1.\t x = 0 \u21d4 |x| = 0 2.\t For all values of x, |x| \u2265 0 and -|\u2009 x| \u2264 0 3.\t For all values of x, |x + y| \u2264 |x| + |y| 4.\t ||\u2009x| - |y||\u2009 \u2264 |x \u2212 y| 5.\t \u2212|x| \u2264 x \u2264 |x| 6.\t |xy| = |x| |y| 7.\t x = x ,y \u22600 y y 8.\t |x|2 = x2 Interval Notation We have seen above that the solution set or the range of values that satisfy inequalities are not discrete. Instead, we have a continuous range of values. Such ranges can be represented using the interval notation. The set of all real numbers between a and b (where a < b) excluding a and b is represented as (a, b) read as \u2018the open interval a, b\u2019. [a, b] read as \u2018the closed interval a, b\u2019 means all real numbers between a and b including a and b (a < b). [a, b) means all numbers between a and b, with a being included and b excluded (a < b). Example 4.28 Solve |2x \u2212 3| < 5. Solution |2x \u2212 3| < 5 \u21d2 \u22125 < 2x \u22123 < 5 \u21d2 \u22122 < 2x < 8 \u21d2 \u22121 < x < 4 \u2234 Solution set is {x\/\u22121 < x < 4} or x \u2208 (\u22121, 4). Example 4.29 Solve |3x + 2| \u2265 7. Solution |3x + 2| \u2265 7 3x + 2 \u2265 7 or 3x + 2 \u2264 \u22127","4.24 Chapter 4 3x \u2265 5 or 3x \u2264 \u22129 x \u2265 5 or x \u2264 \u22123 3 \u2234 Solution set is \uf8f2\uf8f1x\/x \u2265 5 or x \u2264 \u2212 3\uf8fd\uf8fc or (\u2212\u221e, \u22123] \u222a \uf8ee5 , \u221e\uf8f6\uf8f7\uf8f8 . \uf8f3 3 \uf8fe \uf8f0\uf8ef 3 Example 4.30 Solve |5x - 7| < -18. Solution The modulus of any number has to be 0 or positive. Thus, there are no values of x which satisfy the given inequality. The solution set is \u03d5. Example 4.31 Find the solution set of 1 < 0. 2x \u2212 4 Solution 1 < 0. 2x \u2212 4 \u21d2 2x \u2212 4 < 0 \u21d2 x < 2 \u2234 The solution set is {x\/x < 2} or (\u2212\u221e, 2). Example 4.32 Person A can assemble 10 machines per hour and Person B can assemble 15 machines per hour. Person A works for x hours per day and Person B works for y hours per day and both the persons together can assemble at the most 200 machines in a day. Frame one inequation to represent the above data. Solution Person A can assemble 10x machines in a day. Person B can assemble 15y machines in a day. Both A and B can assemble at the most 200 machines. \u21d2 10x + 15y \u2264 200 is the required inequation.","Linear Equations and Inequations 4.25 Example 4.33 Find the two whole numbers such that their sum is utmost 10 and the difference is atleast 4 and also the resultant sum is maximum. Solution Let the two whole numbers be x and y. The required inequations subject to the given conditions are x + y \u2264 10\b (1) x \u2212 y \u2265 4\b (2) x \u2265 0, y \u2265 0\b (3) Tracing the solution set of the above four inequations, clearly the solution set is the triangular region ABC. For the sum of the numbers to be maximum, the solution occurs at the vertices of the region ABC. The vertices are (4, 0), (10, 0) and (7, 3). Out of these (10, 0) and (7, 3) give the maximum value for x + y. \u2234 The solutions are (10, 0), (7, 3). Y x + y \u2264 10 12 \u2022 10 8 6 4 C\u2022 x\u2212y\u22654 2 \u221212 \u221210 \u22128 \u2212 6 \u22124 \u22122 0 A\u2022 8 B\u2022 X 2 46 10 12 \u22122 \u22124 \u22126 \u22128 \u221210 \u221212 Figure 4.15","4.26 Chapter 4 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t The system of equations a + b = 3 and 3a + 3b = 9 1\t 7.\t If x 1 y = 1 and x 1 y = 1 , then x = _____ and is _____. (consistent\/inconsistent) + 2 \u2212 3 \t2.\t The equations px + qy + r = 0 and kpx + kqy y = _____. + kr = 0 are _____. (dependent\/inconsistent) 1\t 8.\t If a + b = c and a \u2212 b = d, then the value of b is \t3.\t If the equations 4x + py = 12 and qx + 3y = 6 are _____. (In terms of c and d). dependent, then the values of p and q are _____ and respectively. \t19.\t Every system of dependent equations is consistent. Is the converse always true? \t4.\t If 2x + 3y = 10 and 3x + 2y = 5, then the value of \t20.\t For how many pairs of non-zero integers, x + x + y is _____. y = 0 and x - y = 0? \t5.\t If 2x - 3y = 0, then the value of 2x + 3y in terms \t21.\t Any line in a plane divides the plane into three of y is _____. disjoint parts. (True\/False) \t6.\t The equations x + y = 1 and x + y =1 are 2\t 2.\t If x + y \u2264 7 and x \u2212 y \u2264 3, then x \u2264 5. (True\/False). a b b a inconsistent if _____. \t23.\t Boundary line for the region y \u2264 x + 4 is _____. \t7.\t If sum of two numbers is 10 whereas their differ- ence is 4, then the greater number is _____. 2\t 4.\t Inequation that represents the following graph is _____. \t8.\t If a + 2b + 3c = 20 and 2a + 4b + c = 25, then Y c = _____. \t9.\t If 2a + 3b + 4c = 35 and 3a + 5b + 7c = 30, then a 5 + b + c = _____. 4 3 PRACTICE QUESTIONS 1\t 0.\t If a > b, then a < b for all a, b and c \u2208 R, where 2 c c 1 c < 0. (True\/False) \t11.\t If a > b, then ac > bc for all a, b and c \u2208 R, where X c > 0. (True\/False) \u22125 \u2212 4 \u22123 \u2212 2 \u22121 0 123 45 \u22121 1\t 2.\t The number of common integral solutions of the \u22122 inequations x > -5 and x < 5 is _____. \u22123 \u22124 \t13.\t The solution set of ax + by + c < 0, if c < 0 is the _____ (region that contains (0, 0)\/region that \u22125 does not contains (0, 0)). \t14.\t Solution set for the inequation x 1 1 > 0 is\u2009_____. \t25.\t The number of solutions for the simultaneous + equations 3x + 4y = 12 and 4x \u2212 3y = 18 is\u2009_____. \t15.\t If x + y \u2264 5, then either x \u2264 5 or y \u2264 5 or both. 2\t 6.\t If the cost of 2 chocolates and 3 biscuits is (True\/False) `55 and that of 4 chocolates and 6 biscuits is `110, then the costs of one chocolate and one 1\t 6.\t If the system of linear equations is inconsistent, then biscuit are necessarily `20 and `5 respectively. the solution set is infinite. (Agree\/Disagree) (True\/False).","Linear Equations and Inequations 4.27 \t27.\t An open sentence which contains the symbol <, Y >, \u2264 or \u2265 is called an _____. 6 2\t 8.\t If x > a and x > b (where a > b), then the solution 5 set of the inequations is _____. 4 3 \t29.\t Common region for the inequations x \u2264 y and y \u2264 2 x is _____. 1 \t30.\t Inequations that represents the shaded region of 0 1 234 5 6 the following graph is _____. X Short Answer Type Questions 3\t 1.\t Solve: 11 + 2(a \u2212 b) = 11, 22 + 3(a \u2212 b) = 17. \t\t(a) (3, 0) a+b a+b \t\t(b) (-4, -4) 3\t 2.\t If px + qy + r = 0 and qx + py + r = 0 (x \u2260 y), then \t\t(c) (3, -5) \t\t(d) (2, -2) show that the value of x + y is \u2212r or \u2212r . \t\t(e) (5, 1) p q 3\t 3.\t If ax + by + c = 0, bx + cy + a = 0 and cx + ay 4\t 0.\t Find the number of solutions for the inequations x + b = 0 passes through the same ordered pair, then + y \u2264 8 and 2x + y \u2264 8. (where x and y are positive show that a3 + b3 + c3 \u2212 3abc = 0. integers). \t34.\t Find the value of x, if 2x + 3y + k = 12 and x 4\t 1.\t If the cost of 2 pencils and 3 erasers is `14. Whereas PRACTICE QUESTIONS + 6y + 2k = 18. the cost of 3 pencils and 5 erasers is `22, then find the cost of one pencil and one eraser. 3\t 5.\t In a fraction, the denominator exceeds the \t42.\t The sum of the digits of a two digit number is 7. If 9 is added to the number, the digits interchange numerator by 8. If unity is deducted from both their places. Find the number. the numerator and the denominator, the fraction becomes 3 . Find the fraction. 4\t 3.\t Harry has `2 and `5 coins with him. If he has a 7 total of 33 coins worth `120 with him, how many `5 coins does he have? 3\t 6.\t Father\u2019s age is 3 years more than thrice the son\u2019s age. After 5 years, father\u2019s age will be 12 years \t44.\t If x 1 y < 1, then show that x does not lie from more than twice the son\u2019s age. Find the father\u2019s \u2212 present age. y to y + 1. \t37.\t Sum of successors of two numbers is 40, whereas 4\t 5.\t Shade the regions that show the solution set of the their difference is 6. Find the two numbers. following inequations: 3\t 8.\t If x + y < 2 and y \u2212 2x > \u22127, then find the range \t\t(a)\u2002 x > 3 of x. \t\t(b)\u2002y < 2 3\t 9.\t Which of the following points belong to the \t\t(c)\u20022y \u2264 5 region represented by the inequations 2x \u2212 3y \u2265 5 and x \u2212 2y \u2264 3? \t\t(d)\u2002x \u2265 0, y \u2265 0 \t\t(e)\u2002x \u2265 4, y \u2264 4","4.28 Chapter 4 Essay Type Questions \t46.\t There are some chocolates with Tom and Jerry. If and second number is positive and does not exceed Tom gives certain number of chocolates to Jerry, 2 and also the resultant sum is maximum. then the number of chocolates with them will be interchanged. Instead, if Jerry gives same number 4\t 9.\t A test has 150 questions. A candidate gets 2 marks of chocolates to Tom, then the number of choco- lates with jerry will be one-fourth of the number for each correct answer and loses 1 mark for each of chocolates that Tom has. If the total number of chocolates with them is 100, then find the num- wrong answer and loses 1 mark for leaving the ber of chocolates with Tom. 2 question unattempted. A student score 165 marks. 4\t 7.\t Solve the system of inequations graphically, \t\tx + 2y \u2264 6, 2x + y \u2265 6 and x \u22644. If the student left 18 questions unattempted, find Direction for question 48: Solve the questions the number of questions he marked wrong. graphically. Direction for question 50: Solve the questions \t48.\t Find the two natural numbers so that their sum graphically. cannot exceed 6 and the difference between first \t50.\t Find the pair of whole numbers so that their sum cannot exceed 10 and difference of twice the first number and thrice the second number cannot exceed 12. CONCEPT APPLICATION Level 1 \t1.\t How many pairs of x and y satisfy the equations 2x \t6.\t The total cost of 10 erasers and 5 sharpeners is at + 4y = 8 and 6x + 12y = 24? least `65. The cost of each eraser cannot exceed `4. Find the minimum possible cost of each sharpener. \t\t(a) 0\t\t (b)1 \t\t(a) `6\t\t (b) `5.50 PRACTICE QUESTIONS \t\t(c) Infinite\t\t (d) None of these \t\t(c) `5\t\t (d) `6.50 \t2.\t Find the value of \u2018k\u2019 for which the system of linear equations kx + 2y = 5 and 3x + y = 1 has zero \t7.\t If the system of linear equations px + 3y = 9 and 4x solutions. + py = 8 has unique solution, then \t\t(a) k = 6\t\t (b) k = 3 \t\t(a) p = \u00b12 3 \t(b) p \u2260 \u00b13 2 \t\t(c) k = 4\t\t (d) None of these \t3.\t Find the minimum value of |x \u2212 3| + 11. \t\t(c) p \u2260 \u00b12 3 \t(d) p = \u00b13 2 \t\t(a) 8\t\t (b) 11 \t8.\t In a group of goats and hens, the total number of legs is 12 more than twice the total number of \t\t(c) 0\t\t (d) \u22128 heads. The number of goats is \t4.\t The maximum value of 23 \u2212 |2x + 3| is \t\t(a) 8\t\t (b) 6 \t\t(a) 20\t\t (b) 26 \t\t(c) 2\t\t (d) Cannot be determined \t\t(c) 17\t\t (d) 23 \t5.\t The product of a number and 72 exceeds the \t9.\t If x +3 < 1, then which of the following cannot product of the number and 27 by 360. Find the x \u22123 number. be the value of x? \t\t(a) 12\t\t (b) 7 \t\t(a) 0\t\t (b) 1 \t\t(c) 8\t\t (d) 11 \t\t(c) 2\t\t (d) 4","Linear Equations and Inequations 4.29 \t10.\t The system of equations px + 4y = 32 and 2qy + \t\t(a) (4, \u22122)\t\t (b) \uf8eb \u2212 1 , 1 \uf8f6 15x = 96 has infinite solutions. The value of p \u2212 q \uf8ec\uf8ed 2 4 \uf8f7\uf8f8 is \t\t(a) \u22121.\t\t (b) 1. \t\t(c) (2, \u22124)\t\t (d) \uf8eb 1 , \u22121 \uf8f6 \t\t(c) 0.\t\t (d) 11. \uf8ec\uf8ed 4 2 \uf8f7\uf8f8 1\t 1.\t If x and y are two integers where x \u2265 0 and y \u2265 \t18.\t Cost of 5 pens and 7 note books is `82 and cost of 0, then the number of ordered pairs satisfying the 4 pens and 4 note books is `52. Find the cost of 2 inequation 2x + 3y \u2264 1 is _____. note books and 3 pens. \t\t(a) 1\t\t (b) 2 \t\t(a) `34.50\t\t (b) `30.50 \t\t(c) 3\t\t (d) 4 \t\t(c) `32.50\t\t (d) `36.50 \t12.\t The common solution set of the inequations 1\t 9.\t If (a + b, a \u2212 b) is the solution of the equations 3x + 2y = 20 and 4x \u2212 5y = 42, then find the value x + y \u2264 1 and x + y > 2 is _____. of b. 2 2 \t\t(a) {(x, y)\/x < 2 and y > 2} \t\t(a) 8\t\t (b) \u22122 \t\t(b) {(x, y)\/x < 1 and y > 1} \t\t(c) \u22124\t\t (d) 5 \t\t(c) an empty set 2x \u2212 5 2 \t\t(d) {(x, y)\/x < 2 and y < 1} \t20.\t If 0 < < 7 and x is an integer, then the sum 1\t 3.\t If (1, 4) is the point of intersection of the lines 2x + of the greatest and the least value of x is by = 6 and 3y = 8 + ax, then find the value of a \u2212 b. \t\t(a) 9\t\t (b) 10 \t\t(a) 2\t\t (b) 3 \t\t(c) 6\t\t (d) 12 \t\t(c) 4\t\t (d) \u22123 \t14.\t If x be a negative integer, then the solution of the \t21.\t Number of integral values of x that do not satisfy inequation 1 \u2264 2x + 8 \u2264 11 is the inequation x\u22127 >0 is ______. PRACTICE QUESTIONS \t\t(a) {-5, -3, -4, -2, -1} x \u22129 \t\t(b) {-4, -2, -1} \t\t(c) {-6, -3, -1} \t\t(a) 4\t\t (b) 3 \t\t(d) {-3, -2, -1} \t\t(c) 2\t\t (d) 0 1\t 5.\t If 5u + 3v = 13uv and u \u2212 v = uv, then \t22.\t The solution set formed by the regions x + y (u, v) = _____. > 7 and x + y < 10 in the first quadrant represents a _____. \t\t(a) (2, 1)\t\t (b) \uf8eb 1 , 1\uf8f7\uf8f6\uf8f8 \uf8ec\uf8ed 2 \t\t(a) triangle\t\t (b) rectangle \t\t(c) \uf8eb 1, 1 \uf8f6 \t\t (d) (1, 2) \t\t(c) trapezium\t (d) rhombus \uf8ec\uf8ed 2 \uf8f7\uf8f8 2\t 3.\t Solve 3 \u2212 2x \u22644. 1\t 6.\t Solve the equations: 4(2x\u22121) + 9(3y\u22121) = 17 and 5 3(2x) \u2212 2(3y) = 6. \t\t(a) (x, y) = (2, 1)\t (b) (x, y) = (\u22122, \u22121) \t\t(a) 5 \u2264 x \u2264 35 \t(b) \u22125 \u2264 x \u2264 35 2 2 2 2 \u2212 \t\t(c) (x, y) = (1, 2)\t (d) (x, y) = (2, \u22121) 1\t 7.\t The solution set of 2 + 3 = 2 and 3 + 4 = 20 is \t\t(c) \u221235 \u2264 x \u2264 \u22125 \t (d) None of these x y x y 2 2","4.30 Chapter 4 \t24.\t In a fraction, if numerator is increased by 2 and 2\t 7.\t The fair of 3 full tickets and 2 half tickets is `204 and the fair of 2 full tickets and 3 half tickets is denominator is increased by 3, it becomes 3 and `186. Find the fair of a full ticket and a half ticket. 4 if numerator is decreased by 3 and denominator is \t\t(a) `94\t\t (b) `78 decreased by 6, it becomes 4 . \u2009Find the sum of the \t\t(c) `86\t\t (d) `62 3 numerator and denominator. 2\t 8.\t If 3 x 2y x \u2212 y = 1, then x \u2212 y = 2 + = 4 2 \t\t(a) 16\t\t (b) 18 \t\t(a) 1\t\t (b) 3 \t\t(c) 20\t\t (d) 14 \t\t(c) 2\t\t (d) 0 2\t 5.\t If 100 cm is divided into two parts such that the \t29.\t If we add 1 to the numerator and subtract 1 from sum of 2 times the smaller part and 1 of the larger the denominator a fraction becomes 1. It also 3 part, is less than 100 cm, then which of the follow- becomes 1 if we add 1 to the denominator. Then 2 the sum of the numerator and denominator of the ing is correct? fraction is \t\t(a) Larger portion is always less than 60. \t\t(a) 7\t\t (b) 8 \t\t(b) S\u0007 maller portion is always less than 60 and more than 40. \t\t(c) 2\t\t (d) 11 \t\t(c) Larger portion is always greater than 60. \t30.\t If 4x \u2212 3y = 7xy and 3x + 2y = 18xy, then (x, y) = \t\t(d) Smaller portion is always greater than 40. 1 1 \t26.\t If 2a \u2212 3b = 1 and 5a + 2b = 50, then what is the \t\t(a) \uf8eb 2 , 3 \uf8f6 \t\t (b) (3, 4) value of a \u2212 b? \uf8ed\uf8ec \uf8f7\uf8f8 \t\t(a) 10\t\t (b) 6 \t\t(c) (4, 3)\t\t (d) \uf8eb 1 , 1\uf8f6 \uf8ec\uf8ed 3 4 \uf8f7\uf8f8 \t\t(c) 7\t\t (d) 3 PRACTICE QUESTIONS Level 2 \t31.\t Jeevesh had 92 currency notes in all, some of which \t33.\t If 2|x| \u2212 |y| = 3 and 4|x| + |y| = 3, then num- were of `100 denomination and the remaining of ber of possible ordered pairs of the form (x, y) is `50 denomination. The total value of amount of all these currency notes was `6350. How much \t\t(a) 0\t\t (b) 1 amount in rupees did he have in the denomination of `50? \t\t(c) 2\t\t (d) 4 \t\t(a) 3500\t\t (b) 3350 3\t 4.\t The solution set formed by the inequations x \u2265 \u22127 and y \u2265 \u22127 in the third quadrant represents a \t\t(c) 2850\t\t (d) 2600 \t\t(a) trapezium\t (b) rectangle \t32.\t The solution set of the inequation 1 \u2264 0 is \t\t(c) square\t\t (d) rhombus + 3x 5 1 |3x \u2212 \uf8eb \u22125 \uf8f6 \uf8eb 5\uf8f6 3\t 5.\t Find the solution of the inequation 5| > 2, \uf8ec\uf8ed 3 \uf8f8\uf8f7 \uf8ec\uf8ed 3 \uf8f8\uf8f7 \t\t(a) x\u2208 , \u221e \t(b) x \u2208 \u2212\u221e, where x is a positive integer. \t\t(c) x \u2208 \uf8eb 5 , \u221e \uf8f6 \t(d) x \u2208 \uf8eb \u2212\u221e, \u22125 \uf8f6 \t\t(a) {2, 3}\t\t (b) {2, 3, 4} \uf8ec\uf8ed 3 \uf8f8\uf8f7 \uf8ec\uf8ed 3 \uf8f7\uf8f8 \t\t(c) x = 2\t\t (d) Null set","Linear Equations and Inequations 4.31 \t36.\t A father wants to divide `200 into two parts 4\t 3.\t The solution set formed by the inequations between two sons such that by adding three times x + y \u2265 3, x + y \u2264 4, x \u2264 2 in the first quadrant the smaller part to half of the larger part, then this represents a will always be less than `200. How will he divide this amount? \t\t(a) triangle\t\t (b) parallelogram \t\t(a) Smaller part is always less than 50. \t\t(c) rectangle\t (d) rhombus \t\t(b) Larger part is always greater than 160. \t44.\t Shiva\u2019s age is three times that of Ram. After 10 years Shiva\u2019s age becomes less than twice the age of \t\t(c) Larger part is always less than 160. Ram. What can be the maximum present age (in complete years) of Shiva? \t\t(d) Smaller part is always greater than 40. \t37.\t Solve |7 \u2212 2x|\u2264 13. \t\t(a) 30\t\t (b) 10 \t\t(a) 3 \u2264 x \u2264 10\t (b) \u22123 < x < 10 \t\t(c) 9\t\t (d) 29 \t\t(c) \u221210 \u2264 x \u2264 3\t (d) \u22123 \u2264 x \u2264 10 4\t 5.\t In an ICC Champions trophy series, Sachin scores 68 runs and 74 runs out of three matches. A player 3\t 8.\t If an ordered pair, satisfying the equations x + y can be placed in Grade A of ICC rankings if the = 7 and 3x \u2212 2y = 11, is also satisfies the equa- average score of three matches is at least 75 and at tion 3x + py \u2212 17 = 0, then the value of p most 85. Sachin is placed in Grade A. What is the is _____. maximum runs that he should score in the third match? \t\t(a) 2\t\t (b) \u22122 \t\t(c) 1\t\t (d) 3 \t\t(a) 105\t\t (b) 83 3\t 9.\t Solve for x: |2x + 3| < 2x + 4. \t\t(c) 113\t\t (d) 97 \t\t(a) x> -2\t\t (b) x > \u2212 7 4\t 6.\t The sum of predecessors of two numbers is 36 and 4 their difference is 4. Find the numbers. \t\t(c) x < \u2212 7 \t\t (d) x < -2 \t\tThe following are the steps involved in solving the 4 above problem. Arrange them in sequential order. \t40.\t Find the values of x and y, which satisfy the simul- \t\t(A) X \u2212 1 + Y \u2212 1 = 36 and X \u2212 Y = 4 PRACTICE QUESTIONS taneous equations 1010x + 1011y = 4040 and \t\t(B) Solve for X and Y 1011x + 1010y = 4044. \t\t(C) Let X > Y \t\t(a) x = 2, y = \u22124\t(b) x = 0, y = 4 \t\t(D) Let the numbers be X and Y \t\t(c) x = 4, y = 4\t (d) x = 4, y = 0 \t\t(a) CDAB\t\t (b) CDBA 4\t 1.\t A bus conductor gets a total of 220 coins of 25 paise, 50 paise and `1 daily. One day he got `110 \t\t(c) DCAB\t\t (d) DCBA and next day he got `80 in that the number of coins of 25 paise and 50 paise coins are inter- \t47.\t The following are the steps involved in solving the changed then find the total number of 50 paise equations 2x + 3y = 17 and 2x+1 + 3y+1 = 43 for x coins and 25 paise coins. and y. Arrange them in sequential order. \t\t(a) 180\t\t (b) 190 \t\t(A) R\u0007 ewrite the given equation in terms of p and q \t\t(c) 160\t\t (d) 200 \t\t(B) Let p = 2x and q = 3y 4\t 2.\t The common solution set of the inequations 5 \u2264 2x + 7 \u2264 8 and 7 \u2264 3x + 5 \u2264 9 is _____. \t\t(C) Find x and y \t\t(D) Solve for p and q \t\t(a) 2 \u2264x\u2264 4 \u2212 \t(b) \u22121 \u2264 x \u2264 4 \t\t(a) ABCD\t\t (b) ABDC 3 3 3 \t\t(c) BACD\t\t (d) BADC \t\t(c) 2 \u2264 x \u2264 1 \u2212 \t (d) Null set 4\t 8.\t There are two numbers. The predecessor of the 3 2 larger number exceeds the successor of the smaller","4.32 Chapter 4 number by 6.The sum of the numbers is 32. Find \t51.\t X is an integer satisfying 1 \u2264 2X + 3 \u2264 7. How the numbers. many values can it take? \t\t\t The following are the steps involved in solving \t\t(a) 4\t\t (b) 3 the above problem. Arrange them in sequential order. \t\t(c) 5\t\t (d) 6 \t\t(A) M + N = 32 and M \u2212 1 \u2212 (N + 1) = 6 \t52.\t Solve for x: 5x + 4 \u2265 x + 12. \t\t(B) Let N < M \t\t(a) x \u2265 0\t\t (b) x \u2265 1 \t\t(C) Solve for M and N \t\t(c) x \u2265 2\t\t (d) x \u2265 3 \t\t(D) Let the numbers be M and N 5\t 3.\t Y is an integer satisfying \u22123 \u2264 4Y \u22127 \u2264 5. How many values can it take? \t\t(a) BDCA\t\t (b) BDAC \t\t(c) DBCA\t\t (d) DBAC \t\t(a) 2\t\t (b) 4 4\t 9.\t The following are two steps involved in finding \t\t(c) 3\t\t (d) 5 the values of p and q from 3p + 5q = 52 and 3p\u2009\u2212\u20091 + 5q\u2009\u2212\u20091 = 14. Arrange them in sequential order. 5\t 4.\t N is a three-digit number. It exceeds the number formed by reversing the digits by 792. Its hundreds \t\t(A) Let x = 3p and y = 5q digit can be \t\t(B) Solve for x and y \t\t(a) 9 \t\t(C) Find p and q \t\t(b) 8 \t\t(D) Rewrite the given equations in terms of x and y \t\t(c) Either (a) or (b) \t\t(a) ABCD\t\t (b) ADCB \t\t(d) Neither (a) nor (b) \t\t(c) ACBD\t\t (d) ADBC \t55.\t X is a three-digit number. The number formed by reversing the digits of X is 891 less than X. Find its 5\t 0.\t Solve for x: 2x - 3 \u2264 5x + 9. units digit. \t\t(a) x \u2265 -4\t\t (b) x \u2265 -3 PRACTICE QUESTIONS \t\t(c) x \u2265 -2\t\t (d) x \u2265 -1 \t\t(a) 0\t\t (b) 1 \t\t(c) 2\t\t (d) Cannot be determined Level 3 5\t 6.\t An examination consists of 160 questions. One mark Sachin faced 10 balls less than the balls faced by is given for every correct option. If one-fourth mark Lara. How many runs were scored by Sachin? is deducted for every wrong option and half mark is deducted for every question left, then one person \t\t(a) 45\t\t (b) 60 scores 79. And if half mark is deducted for every wrong option and one-fourth mark is deducted for \t\t(c) 50\t\t (d) 55 every left question, the person scores 76, then find the number of questions he attempted correctly. \t58.\t The number of ordered pairs of different prime numbers whose sum is not exceeding 26 and dif- \t\t(a) 80\t\t (b) 100 ference between second number and first number cannot be less than 10. \t\t(c) 120\t\t (d) 140 \t\t(a) 8\t\t (b) 9 5\t 7.\t Runs scored by Sachin in a charity match is 10 \t\t(c) 10\t\t (d) 11 more than the balls faced by Lara. The number of balls faced by Sachin is 5 less than the runs scored \t59.\t The number of possible pairs of successive prime by Lara. Together they have scored 105 runs and numbers, such that each of them is greater than 40 and their sum is utmost 100, is","Linear Equations and Inequations 4.33 \t\t(a) 3\t\t (b) 2 will have 5 times as many marbles as Sudheer. Find \t\t(c) 4\t\t (d) 1 the number of marbles with Sudheer. \t60.\t In an election the supporters of two candidates A \t\t(a) 65\t\t (b) 55 and B were taken to polling booth in two differ- ent vehicles, capable of carrying 10 and 15 voters \t\t(c) 70\t\t (d) 50 respectively. If at least 90 vehicles were required to carry a total of 1200 voters, then find the maxi- 6\t 3.\t In a test, for each correct answer 1 mark is awarded mum number of votes by which the elections and each wrong answer half a mark is deducted. could be won by the Candidate B. The test has 70 questions. A candidate attempted all the questions in the test and scored 40 marks. How many questions did he attempt wrongly? \t\t(a) 900\t\t (b) 600 \t\t(a) 15\t\t (b) 20 \t\t(c) 300\t\t (d) 500 \t\t(c) 25\t\t (d) 10 \t61.\t A test has 60 questions. For each correct answer \t64.\t Amar and Bhavan have a certain amount with 2 marks are awarded and each wrong answer 1 them. If Bhavan gives `20 to Amar, he will have mark is deducted. A candidate attempted all the half the amount with Amar. If Amar gives `40 to questions in the test and scored 90 marks. Find Bhavan, he will have half the amount with Bhavan. the number of questions he attempted correctly. Find the amount with Bhavan. \t\t(a) 54\t\t (b) 48 \t\t(a) `70\t\t (b) `90 \t\t(c) `60\t\t (4) `80 \t\t(c) 49\t\t (d) 50 \t65.\t Solve for z: 4x + 5y + 9z = 36, 6x + 15 y 2 \t62.\t Krishna and Sudheer have some marbles with + 11z = 49. them. If Sudheer gives 10 marbles to Krishna, Krishna will have 40 more marbles than Sudheer. \t\t(a) 2\t\t (b) 1 If Sudheer gives 40 marbles to Krishna, Krishna \t\t(c) 3\t\t (d) Cannot be determined PRACTICE QUESTIONS","4.34 Chapter 4 TEST YOUR CONCEPTS \t16.\t Disagree Very Short Answer Type Questions 1\t 7.\t 5 , \u22121 2 2 \t1.\t consistent \t2.\t dependent \t18.\t c \u2212d \t3.\t 6, 2 2 \t4.\t 3 \t19.\t No \t5.\t 6y \t6.\t a + b = 0 \t20.\t No such pairs exist \t7.\t 7 \t8.\t 3 2\t 1.\t True \t9.\t 40 1\t 0.\t True \t22.\t True \t11.\t True \t12.\t 9 \t23.\t y = x + 4 \t13.\t region that contains (0, 0) 1\t 4.\t {x\/x > \u22121} 2\t 4.\t 4x + 3y \u2212 12 \u2265 12 1\t 5.\t True 2\t 5.\t one 2\t 6.\t False 2\t 7.\t inequation \t28.\t x > a \t29.\t x = y 3\t 0.\t x \u2264 2 and y \u2265 2 Shot Answer Type Questions \t31.\t a = 8 and b = 3 3\t 8.\t (\u2212\u221e, 3) 3\t 9.\t (3, 0), (5, 1) \t34.\t 2 4\t 0.\t 12 \t41.\t `6 \t35.\t 7 \t42.\t 34 15 4\t 3.\t 18 ANSWER KEYS \t36.\t 45 years 3\t 7.\t 22, 16 Essay Type Questions 4\t 9.\t 30 \t46.\t 60 \t48. \t(4, 2)","Linear Equations and Inequations 4.35 CONCEPT APPLICATION Level 1 \t 1.\u2002(c)\t 2.\u2002(a)\t 3.\u2002(b)\t 4.\u2002(d)\t 5.\u2002(c)\t 6.\u2002(c)\t 7.\u2002(c)\t 8.\u2002(b)\t 9.\u2002(d)\t 10.\u2002(a) \t11.\u2002(a)\t 12.\u2002(c)\t 13.\u2002(b)\t 14.\u2002(d)\t 15.\u2002(c)\t 16.\u2002(a)\t 17.\u2002(d)\t 18.\u2002(b)\t 19.\u2002(d)\t 20.\u2002(d) \t21.\u2002(b)\t 22.\u2002(c)\t 23.\u2002(b)\t 24.\u2002(a)\t 25.\u2002(c)\t 26.\u2002(d)\t 27.\u2002(b)\t 28.\u2002(b)\t 29.\u2002(b)\t 30.\u2002(d) Level 2 \t31.\u2002(c)\t 32.\u2002(d)\t 33.\u2002(a)\t 34.\u2002(c)\t 35.\u2002(d)\t 36.\u2002(b)\t 37.\u2002(d)\t 38.\u2002(c)\t 39.\u2002(b)\t 40.\u2002(d) \t41.\u2002 (d) \t 42.\u2002(d)\t 43.\u2002 (b) \t 44.\u2002(d)\t 45.\u2002 (c) \t 46.\u2002(c)\t 47.\u2002(d)\t 48.\u2002(d)\t 49.\u2002(d)\t 50.\u2002(a) \t51.\u2002 (a) \t 52.\u2002(c)\t 53.\u2002 (c) \t 54.\u2002(c)\t 55.\u2002(a) Level 3 58.\u2002(d)\t 59.\u2002 (a) \t 60.\u2002(b)\t 61.\u2002(d)\t 62.\u2002(a)\t 63.\u2002 (b) \t 64.\u2002(d)\t 65.\u2002(a) \t56.\u2002(b)\t 57.\u2002 (b) \t ANSWER KEYS","4.36 Chapter 4 CONCEPT APPLICATION Level 1 \t1.\t Find a1 , b1 and c1 and proceed. 1\t 5.\t Divide both the equations by uv on both the sides. a2 b2 c2 1\t 6.\t Assume 2x = a and 3y = b then solve. \t2.\t Apply the condition a1 b1 c1 . a2 = b2 = c2 \t17.\t Put=x1 a=, 1y b and proceed. \t18.\t Make algebraic equations and proceed. \t3.\t Minimum value of modulus of a number is zero. \t4.\t Since |x| is always positive. \t\t\u2234 a + |x| \u2264 a. \t19.\t x = a + b and y = a - b. \t5.\t Assume the number to be x and frame the equation. \t21.\t If a > 0 then a > 0, b > 0 (or) a < 0, b < 0. b \t6.\t Frame the inequations and proceed. \t22.\t Draw the regions corresponding to given \t7.\t If ax + by = e and cx + dy = f have unique solution then ad - bc \u2260 0. inequations. \t8.\t Make algebraic equation and proceed. 2\t 3.\t |x| \u2264 a \u21d2 -a \u2264 x \u2264 a. 2\t 4.\t (i) \u0007Frame equations according to the data. \t9.\t (i) \u0007Subtract one on both the sides of the given equation. \t\t(ii) A\u0007 ssume the numerator and denominator as x and y respectively. \t\t(ii) \u0007On back substitution we can easily verify the value of x. \t\t(iii) F\u0007 rame the equations in terms of x and y and solve them. Hints and Explanation 1\t 0.\t (i) \u0007System of equations has infinite solutions, if \t25.\t Let the two parts be x and (100 - x) and proceed. a1 = b1 = c1 . \t26.\t Solve the equations for a and b. a2 b2 c2 \t27.\t Let full ticket is x and half ticket be y, then frame \t\t(ii) I\u0007 f a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has the equations for x and y, then solve them. infinite solutions, then a1 = b1 = c1 . \t28. \tSolve the equations for x and y. a2 b2 c2 x \t11.\t Apply trial and error method. \t29.\t Assume the fraction as y . Form equations in \t12.\t Apply the concept of half planes in the coordinate terms of x and y according to the given data and system. find the values of x and y. \t13.\t Put x = 1; y = 4 in the given lines and obtained the 3\t 0.\t Divide the two equations by xy. Assume a = a; 1 \t14.\t values of a and b by solving them. x y If a < x + b < c then a - b < x < c - b. a 1 x = a; y = b then solve for a and b. Level 2 \t31.\t Assume the denomination as x and y. Form the 3\t 3.\t (i) We cannot find a common solution. equations in terms of x and y according to the data and solve them. \t\t(ii) \u0007Solve the two equations for |x| and|y| and write the possible values of x and y. 3\t 2.\t Multiply the numerator and denominator with 5 + 3x and solve the inequation obtained. \t34.\t Represent the inequations in the coordinate plane.","Linear Equations and Inequations 4.37 \t35.\t If |x| < a \u21d2 -a < x < a. \t47.\t It can easily be seen that the correct order is BADC. 3\t 6.\t Find x and y values (x < y) such that x + y = 200 \t48.\t It can easily be seen that the correct order is and 3x + y < 200. DBAC. 2 \t49.\t It can be easily seen that the correct order is \t37.\t |x| \u2264 a \u21d2 -a \u2264 x \u2264 a. ADBC. \t38.\t (i) S\u0007olve first two equations and then substitute \t50.\t Given 2x - 3 \u2264 5x + 9 the values in the third equation. \t\t2x - 5x \u2264 9 + 3 \t\t-3x \u2264 12 \t\t(ii) S\u0007 olve the equations x + y = 7 and 3x \u2212 2y = 11. \t\tx \u2265 -4. \t\t(iii) N\u0007 ow substitute the value of x and y in the \t51.\t 1 \u2264 2X + 3 \u2264 7 equation 13x + py - 17 = 0. \t\t \u21d2 1 \u2264 2X + 3 and 2X + 3 \u2264 7 \t\t \u21d2 -2 \u2264 2X and 2X \u2264 4 \t39.\t (i) |x| < a \u21d2 -a < x < a. \t\t \u21d2 -1 \u2264 X and X \u2264 2. \t\tX is an integer. \t\t(ii) \u0007|x + a| = x + a when x > \u2212a and \u2212(x + a) \t\t\\\\ X = -1 or 0 or 1 or 2. when x < \u2212a. \t\t\\\\ X has four possibilities. \t40.\t (i) \u0007First of all add given equations and then sub- 5\t 2.\t Given 5x + 4 \u2265 x + 12 tract from one another. \t\t5x - x \u2265 12 - 4 \t\t4x \u2265 8 \t\t(ii) S\u0007 ubtract and add the given equations and then \t\tx \u2265 2. solve them for x and y. \t53.\t \u22123 \u2264 4Y \u2212 7 \u2264 5 \u21d2 \u22123 \u2264 4Y \u2212 7 and 4Y \u2212 7 \u2264 5 4\t 1.\t (i) \u0007Frame equations according to the data. \t\t \u21d2 \u22123 + 7 \u2264 4Y and 4Y \u2264 7 + 5 \t\t \u21d2 1 \u2264 Y and Y \u2264 3. \t\t(ii) A\u0007 ssume the number of coins of 25 paise, 50 \t\tY is an integer. paise and `1 as x, y and z respectively. \t\t \\\\ Y = 1 or 2 or 3. \t\t(iii) \u0007Frame the equations in terms of x, y and z \t\t\\\\ Y has three possibilities. Hints and Explanation according to the data and then solve. 5\t 4.\t Let N be abc. 4\t 2.\t (i) S\u0007olve the given inequations individually and then check. \t\t\\\\ \u0007The number formed by reversing the digits = cba. \t\t(ii) S\u0007olve the two inequations individually, then write the common solution for x. \t\t\tabc \u2212 cba = 792 \t\t\t100a + 10b + c \u2212 (100c + 10b + a) = 792 \t43.\t (i) Draw the graph. \t\t\t99a \u2212 99c = 792. \t\t\ta \u2212 c = 8, i.e., a = c + 8 \t\t(ii) \u0007Draw the region according to the inequa- \t\t\tc \u2265 0 tions given, then identify the common region \t\t\\\\ a \u2265 8 formed by them. \t\t\\\\ a = 8 or 9. 4\t 4.\t (i) Frame inequation according to the data. \t\t(ii) A\u0007 ssume the ages of Shiva and Ram as x and y years respectively. \t\t(iii) F\u0007 rame the equation and inequation according to the data. 4\t 5.\t (i) A\u0007 s per the conditions given frame in equations and solve. \t\t(ii) 75 \u2264 x+y+z \u2264 85 3 \t\t(iii) S\u0007 ubstitute x and y and obtain the maximum value of z. \t46.\t It can be easily seen that the correct order is DCAB.","4.38 Chapter 4 \t55.\t Let x be abc. \t\ti.e., a = c + 9 \t\tThe number formed by reversing the digits = cba \t\t \u2009c \u2265 0 \t\tcba = abc \u2212 891, i.e., 891 = abc \u2212 cba \t\t891 = 100a + 10b + c \u2212 (100c + 10b + a) \t\t\\\\ a \u2265 9. \t\t = 99(a \u2212 c) \t\tBut a \u2264 9. \t\t 9 = a \u2212 c \t\t \\\\ a = 9. \t\t \\\\ c = 0. Level 3 \t56.\t (i) F\u0007 rame equations according to the data. \t\tM + 40 = 5(S \u2212 40) \u21d2 S + 20 + 40 = 5(S \u2212 40) \t\t(ii) F\u0007 rame the equations according to the data and \t\t260 = 4S, i.e., S = 65. then solve. 6\t 3.\t Let the number of questions attempted wrongly by \t57.\t (i) \u0007Frame equations according to the data. the candidate be w. \t\t(ii) A\u0007 ssume runs scored by Sachin as x and balls \t\t\\\\ Number of questions be attempted correctly faced by Sachin as y. \u2009= 70 \u2212 w. \t\t(iii) \u0007Frame the equation in terms of x and y \t\t1(70 \u2212 w) \u2212 1 w = 40 according to the data and solve the equation. 2 \t58.\t Write the prime numbers up to 23, find the order \t\t \u200970 \u2212 3 w = 40 pairs such that x + y \u2264 26 and y - x > 10. 2 \t\t w = 20. Hints and Explanation \t59.\t (i) W\u0007 rite primes as per the given conditions. 6\t 4.\t Let the amounts with Amar and Bhavan be `A and `B respectively. \t\t(ii) Assume the prime numbers as x, y. 1 \t\t(iii) \u0007Find the possible values of x and y such that \t\t(B \u2212 20) = 2 (A + 20) x\u00a0+ y < 100, where x > y > 40. 1 6\t 0.\t Find the values of x and y such that x + y > 90 and and A \u2212 40 = 2 (B + 40) 10x + 15y = 1200. 1 6\t 1.\t Let the number of questions attempted correctly \t\t2B \u2212 40 \u2212 20 = A and A = 2 B + 20 + 40 by the candidate = C. \t\tHe attempted all the questions. \t\t\\\\\u2003 A = 2B \u2212 60 = 1 B + 60 2 \t\t\\\\ Number of questions attempted wrongly by \t\t32B = 120 \u21d2 B = 80. him = 60 \u2212 C \t\t2C \u2212 1(60 \u2212 C) = 90 \t65.\t 3(4x + 5y + 9z) = (36)3 \t\t 2C \u2212 60 + C = 90 \t\t \u21d2 12x + 15y + 27z = 108\b (1) (2) \t\t C = 50. \uf8eb 15 \uf8f6 \uf8ed\uf8ec 2 \uf8f7\uf8f8 6\t 2.\t Let the number of marbles with Krishna and \t\t\u20092 6x + y + 11z = (49)2 Sudheer be M and S respectively. \u21d2\u2009 12x + 15y + 22z = 98\b \t\tM + 10 = S \u2212 10 + 40 \u21d2 M = S + 20. \t\t(1) \u2212 (2) \u21d2 5z = 10 \u21d2 z = 2.","152CChhaapptteerr EQKxuipnaerdmersaasttiioiccsns and Equations REmEmBER Before beginning this chapter, you should be able to: \u2022 Understand simple quadratic equations \u2022 Know natural numbers, integers and fractions \u2022 Aware of basic algebra and simple indices KEY IDEAS After completing this chapter, you should be able to: \u2022 Study about quadratic expression, its zeroes and quadratic equations \u2022 Find solutions\/roots of a quadratic equation \u2022 Understand the nature and sign of the roots of a quadratic equation \u2022 Learn reciprocal equation and maximum or minimum value of a quadratic equation Figure 1.1","5.2 Chapter 5 INTRODUCTION In previous topic, we have learnt linear expression, linear equation and linear inequation. In this topic, we shall learn quadratic expression and quadratic equation. QUADRATIC EXPRESSION A polynomial of second degree in one variable is termed as a quadratic polynomial or quadratic expression. The general form of a quadratic expression in x is ax2 + bx + c where a, b, c are real numbers and a \u2260 0. Example: 2x2 + 3x, x2 \u2212 2, 3x2 + 11x \u2212 108 are some quadratic expressions. \u2009\u2009Note\u2002 \u2002 The expressions x2 + 1 \u2212 3, x2 + 3 x + 3, 2x 2 \u2212 1 + 4 are not quadratic expressions. x2 x Zeroes of a Quadratic Expression If a quadratic expression, ax2 + bx + c becomes zero for x = \u03b1, where \u03b1 is a real number, then \u03b1 is called a zero of the expression ax2 + bx + c. A quadratic expression can have at the most two zeroes. QUADRATIC EQUATION The equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a \u2260 0 is known as a quadratic equation, or an equation of the second degree. Example: 2x2 + 3x + 5 = 0, x2 \u2212 5 = 0 and 3x2 \u2212 4x + 5 = 0 are some quadratic equations. Solutions or Roots of a Quadratic Equation The values of x for which the equation ax2 + bx + c = 0 is satisfied are called the roots of the quadratic equation. A quadratic equation cannot have more than two roots. Example 5.1 Verify whether x = 2, is a solution of 2x2 + x \u2212 10 = 0. Solution On substituting x = 2 in 2x2 + x \u2212 10, we get 2(2)2 + 2 \u2212 10 = 10 \u2212 10 = 0. \u2234 2 is a solution (or) root of 2x2 + x \u2212 10 = 0. Finding the Solutions or Roots of a Quadratic Equation There are two ways of finding the roots of a quadratic equation. 1.\t Factorization method 2.\t Application of formula","Quadratic Expressions and Equations 5.3 Factorization Method to Obtain the Roots of a Quadratic Equation The steps involved in obtaining the roots of ax2 + bx + c = 0 are as follows: 1.\t Resolve ax2 + bx + c into factors and express ax2 + bx + c as a product of its factors. 2.\t \u0007For this product to be zero, one of the factors should be zero. The zeroes of the factors give the roots of the equation, ax2 + bx + c = 0. Example 5.2 (a)\t Solve x2 \u2212 15x + 26 = 0. (b)\t Solve x + 1 = 5 . x 2 Solution (a)\t First, let us resolve x2 \u2212 15x + 26 into factors. \u21d2 x2 \u2212 15x + 26 = x2 \u2212 13x \u2212 2x + 26 = x(x \u2212 13) \u22122(x \u2212 13) = (x \u2212 13)(x \u2212 2). The given equation, x2 \u2212 15x + 26 = 0 is reduced to (x \u2212 13)(x \u2212 2) = 0 \u21d2 x \u2212 13 = 0 (or) x \u2212 2 = 0 \u21d2 x = 13 (or) x = 2. \u2234 x = 2, 13 are the roots of the given equation. (b)\t x2 +1 = 5 x 2 \u21d2 2x2 + 2 = 5x \u21d2 2x2 \u2212 5x + 2 = 0 \u21d2 2x2 \u2212 4x \u2212 x + 2 = 0 \u21d2 2x(x \u2212 2) \u22121(x \u2212 2) = 0 \u21d2 (2x \u2212 1)(x \u2212 2) = 0 \u21d2 2x \u2212 1 = 0 or x \u2212 2 = 0 \u21d2 x = 1 or x = 2. 2 \u2234 x = 1 , 2 are the roots of the given equation. 2 Finding the Solutions\/Roots of a Quadratic Equation by the Application of Formula Factorization of ax2 + bx + c might not be always possible using the above method. So here, we derive a formula to find the roots of the equation, ax2 + bx + c = 0, where a \u2260 0, b, c \u2208 R.","5.4 Chapter 5 ax2 + bx + c = 0 \u21d2 ax2 + bx = \u2212c \u21d2 a \uf8eb x 2 + b x\uf8f6\uf8f7\uf8f8 = \u2212c \uf8ed\uf8ec a \u21d2 x2 + b x = \u2212c a a \u21d2 x2 + 2\u22c5 b \u22c5x = \u2212c 2a a Adding \uf8eb b \uf8f62 on both sides to make LHS a perfect square, we get, \uf8ed\uf8ec 2a \uf8f7\uf8f8 \u21d2 (x )2 + 2 \uf8eb b\uf8f6 (x) + \uf8eb b \uf8f62 = \uf8eb b \uf8f62 \u2212 c \uf8ec\uf8ed 2a \uf8f7\uf8f8 \uf8ec\uf8ed 2a \uf8f8\uf8f7 \uf8ed\uf8ec 2a \uf8f7\uf8f8 a \u21d2 \uf8eb x + b \uf8f62 = b2 \u2212 4ac \uf8ec\uf8ed 2a \uf8f7\uf8f8 4a2 \u21d2 x + b = \u00b1 b2 \u2212 4ac 2a 4a2 \u21d2 x = \u2212b \u00b1 b2 \u2212 4ac 2a ( 2a )2 \u21d2 x = \u2212b \u00b1 b2 \u2212 4ac 2a \u2234 The roots of the equation ax2 + bx + c = 0 are \u2212b \u00b1 b2 \u2212 4ac . 2a Relation Between the Roots and the Coefficients of a Quadratic Equation Let us assume that a, b are the roots of ax2 + bx + c = 0. Then, \u03b1 = \u2212b + b2 \u2212 4ac and \u03b2 = \u2212b \u2212 b2 \u2212 4ac . 2a 2a Now, the sum of the roots, \u03b1 + \u03b2 = \u2212b + b2 \u2212 4ac + \u2212b \u2212 b2 \u2212 4ac = \u22122b = \u2212b . 2a 2a 2a a \u2234 The sum of the roots, a + b = \u2212b = \u2212(coefficient of x) . a (coefficient of x 2) The product of the roots, ab = \uf8eb \u2212b + b2 \u2212 4ac \uf8f6 \uf8eb \u2212b \u2212 b2 \u2212 4ac \uf8f6 \uf8ec\uf8ed\uf8ec \uf8f7 \uf8ec \uf8f7 2a \uf8f8\uf8f7 \uf8ec\uf8ed 2a \uf8f7\uf8f8","Quadratic Expressions and Equations 5.5 ( )(\u2212b)2 \u2212 b2 \u2212 4ac 2 = 4a2 = b2 \u2212 b2 + 4ac = 4ac = c . 4a2 4a2 a \u2234 The product of the roots, \u03b1\u03b2 = c = (constant term) ) . a (coefficient of x2 Nature of the Roots of a Quadratic Equation We know that the roots of the equation ax2 + bx + c = 0 are \u03b1 = \u2212b + b2 \u2212 4ac and \u03b2 = \u2212b \u2212 b2 \u2212 4ac . 2a 2a We call, D = b2 \u2212 4ac, the discriminant of ax2 + bx + c = 0. The value of the discriminant determines the nature of the roots of the equation. In this context, we study the following cases. We shall limit ourselves to cases where the coefficients a, b and c are real. Case 1:\u2002 If b2 \u2212 4ac = 0, i.e., D = 0, then \u03b1 = \u2212b and \u03b2 = \u2212b . 2a 2a So, a = b = \u2212b . 2a Hence, when D = b2 \u2212 4ac = 0, the two roots of the equation are real and equal. Case 2:\u2002 If b2 \u2212 4ac > 0, then the roots are real and distinct. We shall consider, the nature of the roots when a, b and c are rational numbers. 1.\t If b2 \u2212 4ac > 0, i.e., D > 0 and a perfect square, then the roots are rational and distinct. 2.\t I\u0007 f b2 \u2212 4ac > 0, i.e., D > 0 and not a perfect square, then the roots are irrational and distinct. In this case one root is a surd conjugate of the other. If one root is of the form a + b, then the other root will be in the form of a \u2212 b and vice-versa. Case 3:\u2002 If b2 \u2212 4ac < 0, i.e., D < 0, then the roots of the equation are imaginary. Signs of the Roots of a Quadratic Equation When the signs of the sum of the roots and the product of the roots of a quadratic equation are known, the signs of the roots of the quadratic equation can be determined. Let us consider the following four cases."]
Search
Read the Text Version
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346
- 347
- 348
- 349
- 350
- 351
- 352
- 353
- 354
- 355
- 356
- 357
- 358
- 359
- 360
- 361
- 362
- 363
- 364
- 365
- 366
- 367
- 368
- 369
- 370
- 371
- 372
- 373
- 374
- 375
- 376
- 377
- 378
- 379
- 380
- 381
- 382
- 383
- 384
- 385
- 386
- 387
- 388
- 389
- 390
- 391
- 392
- 393
- 394
- 395
- 396
- 397
- 398
- 399
- 400
- 401
- 402
- 403
- 404
- 405
- 406
- 407
- 408
- 409
- 410
- 411
- 412
- 413
- 414
- 415
- 416
- 417
- 418
- 419
- 420
- 421
- 422
- 423
- 424
- 425
- 426
- 427
- 428
- 429
- 430
- 431
- 432
- 433
- 434
- 435
- 436
- 437
- 438
- 439
- 440
- 441
- 442
- 443
- 444
- 445
- 446
- 447
- 448
- 449
- 450
- 451
- 452
- 453
- 454
- 455
- 456
- 457
- 458
- 459
- 460
- 461
- 462
- 463
- 464
- 465
- 466
- 467
- 468
- 469
- 470
- 471
- 472
- 473
- 474
- 475
- 476
- 477
- 478
- 479
- 480
- 481
- 482
- 483
- 484
- 485
- 486
- 487
- 488
- 489
- 490
- 491
- 492
- 493
- 494
- 495
- 496
- 497
- 498
- 499
- 500
- 501
- 502
- 503
- 504
- 505
- 506
- 507
- 508
- 509
- 510
- 511
- 512
- 513
- 514
- 515
- 516
- 517
- 518
- 519
- 520
- 521
- 522
- 523
- 524
- 525
- 526
- 527
- 528
- 529
- 530
- 531
- 532
- 533
- 534
- 535
- 536
- 537
- 538
- 539
- 540
- 541
- 542
- 543
- 544
- 545
- 546
- 547
- 548
- 549
- 550
- 551
- 552
- 553
- 554
- 555
- 556
- 557
- 558
- 559
- 560
- 561
- 562
- 563
- 564
- 565
- 566
- 567
- 568
- 569
- 570
- 571
- 572
- 573
- 574
- 575
- 576
- 577
- 578
- 579
- 580
- 581
- 582
- 583
- 584
- 585
- 586
- 587
- 588
- 589
- 590
- 591
- 592
- 593
- 594
- 595
- 596
- 597
- 598
- 599
- 600
- 601
- 602
- 603
- 604
- 605
- 606
- 607
- 608
- 609
- 610
- 611
- 612
- 613
- 614
- 615
- 616
- 617
- 618
- 619
- 620
- 621
- 622
- 623
- 624
- 625
- 626
- 627
- 628
- 629
- 630
- 631
- 632
- 633
- 634
- 635
- 636
- 637
- 638
- 639
- 640
- 641
- 642
- 643
- 644
- 645
- 646
- 647
- 648
- 649
- 650
- 651
- 652
- 653
- 654
- 655
- 656
- 657
- 658
- 659
- 660
- 661
- 662
- 663
- 664
- 665
- 666
- 667
- 668
- 669
- 670
- 671
- 672
- 673
- 674
- 675
- 676
- 677
- 678
- 679
- 680
- 681
- 682
