Excel in Mathematics - BOOK 8 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 101 Vedanta Excel in Mathematics - Book 8 Solution: Here, M.P. of the article = Rs 3,200 Discount percent = 10% a) Discount amount = 10% of Rs 3,200 = 10 100 × Rs 3,200 = Rs 320 b) S.P. of the article = M.P. – Discount = Rs 3,200 – Rs 320 = Rs 2,880 c) Again VAT amount = 13% of S.P. = 13 100 × Rs 2,880 = Rs 374.40 d) Now, S.P. with VAT = S.P. + VAT amount = Rs 2,880 + Rs 374.40 = Rs 3,254.40 Hence, the customer should pay Rs 3,254.40. Example 4: A retailer bought a woolen jacket for Rs 4,000 and fixed its price 30% above the cost price. He/she then allows 10% discount and sold it. a) Find the marked price of the jacket. b) Calculate the discount amount. c) How much did a customer pay for it? Solution: Here, C.P. of the jacket = Rs 4,000 a) Now, M.P. of the jacket = Rs 4,000 + 30% of Rs 4,000 = Rs 4,000 + 30 100 × Rs 4,000 = Rs 4,000 + Rs 1,200 = Rs 5,200 Hence, the required M.P. is Rs 5,200. b) Again, discount amount = 10% Rs 5,200 = 10 100 × Rs 5,200 = Rs 520 Hence, the required discount amount is Rs 520. c) Then, S.P. of the jacket = M.P. – Discount = Rs. 5,200 – Rs 520 = Rs 4,680 Hence, a customer paid Rs 4,680 for it. Example 5: Bijaya marked the price of a watch as Rs 2,400. He sold it allowing a discount of 15% and made a profit of Rs 240. Find the cost price of the watch. Solution: Here, M.P. of the a watch = Rs 2,400 Discount percent = 15% Profit amount = Rs 240 C.P. of the watch = ? Direct process M. P.= (100 + 30)% of C. P. = 130% of Rs 4,000 = 130 100 × Rs 4,000 = Rs 5,200 Direct process S. P. = (100 – discount)% of M. P. = (100 – 10)% of Rs 5,200 = 90% of Rs 5,200 = 90 100 × Rs 5,200 = Rs 4,680 Profit and Loss

Vedanta Excel in Mathematics - Book 8 102 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Now, discount amount = Discount % of M. P. = 15 100 × Rs 2400 = Rs 360 Then, S.P. of the watch = M. P. – Discount amount = Rs 2,400 – Rs 360 = Rs 2,040 Again, C. P. of the watch = S. P. – Profit amount = Rs 2,040 – Rs 240 = Rs 1,800 Hence, the cost price of the watch was Rs 1,800. Example 6: After allowing 20% discount on the marked price of a computer, 13% Value Added Tax (VAT) was levied on it. Then, the computer was sold for Rs 48,816. a) Find the selling price of the computer without VAT. b) Find the marked price of the computer. Solution: Here, discount percent = 20% , VAT percent = 13% and S. P. with VAT = Rs 48,816 a) Now, to find S.P. without VAT: S.P. + 13% of S.P. = Rs 48,816 or, S.P. + 13 100 × S.P. = Rs 48,816 or, 113 S.P. 100 = Rs 48,816 or, S.P. = Rs 48,816 × 100 113 = Rs 43,200 b) Again, M. P. – D% of M. P. = S.P. without VAT or, M. P. – 20% of M. P. = Rs 43,200 or, M.P. – 20 100 of M.P. = Rs 43,200 or, 4 M.P. 5 = Rs 43,200 or, M.P. = Rs 43,200 × 5 4 = Rs 54,000 Hence, the marked price of the computer was Rs 54,000 Direct process S. P. = (100 – 15)% of M. P. = 85% of Rs 2,400 = Rs 2,040 Direct process (100 + 13)% of S.P. = S.P. with VAT or, 113% of S.P. = Rs 48,816 or, S.P. = Rs 48,816 113% = Rs 48,816 113% 100 = Rs 48,816 × 100 113 = Rs 43,200 Direct process (100 – 20)% of M.P. = S.P. without VAT or, 80% of M.P. = Rs 43,200 or, 80 100 of M.P. = Rs 43,200 M.P. = 100 × Rs 43,200 80 = Rs 54,000 Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 103 Vedanta Excel in Mathematics - Book 8 EXERCISE 7.2 General Section 1. Say and write the answers as quickly as possible. a) M.P. = Rs 200, discount = Rs 30, S.P. = ……………. b) M.P. = Rs 400, S.P = Rs 375, discount = ……………. c) S.P. = Rs 550, discount = Rs 60, M.P. = ……………. 2. a) M. P. = Rs 100, discount = Rs 25, discount percent = ……………. b) M.P. = Rs 100, S.P. = Rs 90, discount percent = ……………. c) M.P. = Rs 100, discount percent = 10% , discount amount = ……………. d) M.P. = Rs 100, discount percent = 20% , S. P. = ……………. 3. a) S. P. = Rs 100, VAT amount = Rs 13, VAT percent = ……………. b) S. P. = Rs 100, VAT percent = 15%, VAT amount = ……………. c) S.P. = Rs 100, S.P with VAT = Rs 113, VAT percent = ……………. d) S.P. = Rs 100, VAT percent = 13% , S.P. with VAT = ……………. Creative Section - A 4. a) If M.P. = Rs 720, discount = Rs 36, find the discount percent. b) If M.P. = Rs 1,530, S.P. = Rs 1,377, find the discount and discount percent. c) If M.P. = Rs 940, discount percent = 5%, find the discount and S.P. d) If S.P. = Rs 450, S.P. with VAT = Rs 495, find VAT percent. e) If S.P. = Rs 800, VAT percent = 12%, find VAT amount and S.P with VAT. 5. a) The marked price of a bag is Rs 960 and the shopkeeper allows 10% discount to the customer while selling. (i) Find the amount of discount. (ii) Find its selling price after the discount. b) The marked price of a mobile is Rs 2,500 and the shopkeeper allows 15% discount. How much should a customer pay for buying it? c) On the occasion of ‘Dashain festival’ , 15% discount is given on every item in a supermarket. If the labeled price of a pair of shoes is Rs 3,200, find its selling price. Profit and Loss

Vedanta Excel in Mathematics - Book 8 104 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) Let the marked price be Rs x, S.P. is Rs 180 and discount percent is 10%. Find the marked price. [Hints: x – 10% of x = 180] b) If a shopkeeper allows 25% discount and sells an umbrella for Rs 450, find the marked price of the umbrella. Creative Section - B 7. a) A shopkeeper marked the price of a calculator as Rs 1,600 and she/he sold it allowing a discount of 10%.Then, she made a profit of Rs 140. (i) Find the amount of discount. (ii) Find the selling price after allowing discount. (iii) Find the cost price of the calculator. b) Mr. Muhammad fixed the marked price of a cycle as Rs 6,000. He sold it after allowing a discount of 15% and made a profit Rs 500. (i) Find the amount of discount. (ii) Find the selling price after allowing discount. (iii) Find the cost price of cycle. c) Kamala marked the price of a cosmetic item as Rs 400. She offered her customers a discount of 20% and made a loss of Rs 30, what was the actual cost of the item to her? 8. a) The marked price of a fan is Rs ,1500. The shopkeeper allows 20% discount on it. (i) Find the amount of discount. (ii) Find the selling price after giving discount. (iii) If 13% VAT is charged on the selling price, find the amount of VAT. (iv) Find the selling price with VAT. b) The price of a rice cooker is marked as Rs 2,500. Then, 40% discount is allowed and 13% VAT is charged while selling the cooker. Find the selling price of the cooker with VAT. c) The marked price of a jacket is Rs 5,000. What is the price of the jacket if 10% VAT was levied after allowing 20% discount? 9. a) A retailer bought a watch for Rs 800 and fixed its price 25% above the cost price. He then allows 10% discount. (i) Find the marked price of the watch. (ii) Find the discount amount. (iii) Find the selling price of the watch. (iv) How much should a customer pay for it with 13% VAT? Profit and Loss

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 105 Vedanta Excel in Mathematics - Book 8 b) A shopkeeper bought an electric fan for Rs 2,000 and fixed its price 20% above the cost price. If he then gives 15% discount, how much should a customer pay for it with 15% VAT? c) A dealer marks the price of his mobiles 40% above the cost price and allows 20% discount. If his purchase price of a mobile is Rs 7,500, how much should a customer pay for it with 13% VAT? 10. a) A gift house allowed 20% discount on the marked price of a doll. Then, 13% VAT was levied on it and the doll was sold at Rs 1,808. (i) Find the selling price of the doll without VAT. (ii) Find the marked price of the doll. b) After allowing 16% discount on the marked price of a watch, 13% Value Added Tax (VAT) was levied on it. If the watch was sold for Rs 4,746, calculate the marked price of the watch. It’s your time - Project work and Activity section! 11. a) Let’s become a problem maker and problem solver. Write the values of the variables of your own. Then, solve each problem to find unknown variable. M.P. = ................... Discount % = .............. Find S.P. M. P. = ........................ S. P. = ........................ Find discount percent S. P. = ........................ VAT% = ........................ Find S.P. with VAT S. P. with VAT = ......... VAT% = ................ Find S.P. without VAT b) Let’s search the rate of VAT of different countries in the available website. Compare these VAT rates with the VAT rate of our country. c) Let’s make group of 5 friends and visit your local market to find the selling price of some electrical and electronic items. Calculate the S. P. of these items with the current VAT rate system. Profit and Loss

Vedanta Excel in Mathematics - Book 8 106 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment - II 1. A school distributes a total of Rs 13,800 as scholarship to its three students, each student from the classes 5, 8 and 10 in the ratio of the number of their classes. Answer the following questions. (a) Write the ratio at which the amounts is distributed to the students. (b) How much amount of scholarship does each student receive? (c) If the same amount of money were distributed to only two students each from class 5 and 10 in the ratio of the number of their classes, how much amount would each student receive? 2. In a tea corner, the shopkeeper mixes the water and milk in the ratio of 5 : 2. A group of five people come to the corner and order 5 cups of tea. He mixes 200 ml of milk to prepare 5 cups of tea. (a) How many milliliters of water does he use to fulfill the order? (b) Find the quantity of tea for each of the prepared cup. 3. Two numbers are in the ratio 4 : 5. When 4 is added to each term, the ratio of the numbers becomes 5 : 6. (a) Find the numbers. (b) What number should be subtracted from both the numbers so that their ratio becomes 3:4? 4. 20 workers can construct a house in 24 days working 9 hours a day. (a) In how many days would 1 worker construct the same house with the same working hours? (b) In how many days would 27 workers construct the same house working 10 hours a day? (c) How many workers would construct the same house in 30 days working 8 hours a day? 5. There are 25 workers in a factory and the daily wages of all the workers are same. 25 workers earn Rs 4,00,000 in 20 days. (a) How much does 1 worker earn in 1 day? (b) How much do 10 workers earn in 15 days at the same rate of daily wages? (c) In how many days, do 15 workers earn Rs 84,000 at the same rate of daily wages? 6. Rakshya fixed the marked price of a sari as Rs 11,000. She sold it after allowing a discount of 20% and made a profit Rs 1,760. (a) Find the amount of discount. (b) Find the selling price after allowing discount. (c) Find the cost price of sari. (d) Find her profit or loss percent. 7. Jitendra borrowed a loan of Rs 2,00,000 for the foreign employment in Saudi Arab at the rate of 12% p.a. for 1 years 6 months. (a) How much simple interest did he pay? (b) How much amount did he pay to clear the debt?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 107 Vedanta Excel in Mathematics - Book 8 8.1 Indices - Looking back Classwork - Exercise Let's discuss the answers of the following questions. 1. In a licensed network business, each member should make 2 members everyday. In this way, in the first day, 1 member makes 2 members. In the second day, 2 members make 4 members. In the third day 4 members make 8 members, and so on. a) How many members are made in the 4th day? ......................................... b) How many members are made in the 5th day? ......................................... c) How many members are made in the 6th day? ......................................... d) How many members are made in the 7th day? ......................................... 2. 2x3 is a given algebraic expression. a) What is the coefficient of this expression? ............................................ b) What is the base of this expression? ............................................ c) What is the index (or exponent) of the base? ............................................ 8.2 Laws of Indices - Review Let’s consider an algebraic term 4x3 . Here, 4 is the coefficient, x is the base, and 3 is the exponent of the base. The exponent of a base is also called its index (plural: indices). Power is an expression that represents repeated multiplication of the same number (or variable) whereas exponent refers to a quantity that represents the power to which the number (or variable) is raised. Both terms are often used interchangeably in mathematical operations. There are certain rules for the operations of indices of different algebraic terms. These rules are also called the Laws of indices. Unit 8 Laws of Indices

Vedanta Excel in Mathematics - Book 8 108 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (i) Product law of indices Study the following illustrations and investigate the idea of the product law of indices. 1 2 3 4 2 22 = 4 unit squares = 21 × 21 = 21 +1 32 1 2 3 4 5 6 7 8 9 3 3 = 9 unit squares = 31 × 31 = 31 +1 x x2 = x1 × x1 = x1 +1 Similarly, y y2 = y1 × y1 = y1 +1 x 2 y Again, 2 23 = 8 unit cubes = 21 × 21 × 21 = 21 + 1 + 1 5 6 7 8 3 4 1 2 2 2 3 33 = 27 unit cubes = 31 × 31 × 31 = 31 + 1 + 1 3 3 x x3 = x1 × x1 × x1 = x1 + 1 + 1 x x y y3 = y1 × y1 × y1 = y1 + 1 + 1 y y Also, 22 × 2 = 22+1 = 23 , 32 × 33 = 32 + 3 = 35 , and so on. Thus, if am and an are any two terms with the same base a and the powers m and n respectively, then, am × an = am + n (ii) Quotient law of indices Let’s study the following illustrations and investigate the idea of the quotient law of indices. 22 ÷ 2 = 22 2 = 2 × 2 2 = 2 = 22 – 1 24 ÷ 22 = 24 22 = 2 × 2 × 2 × 2 2 × 2 = 22 = 24 – 2 23 ÷ 25 = 23 25 = 2 × 2 × 2 2 × 2 × 2 × 2 × 2 = 1 22 = 1 25 – 3 Thus, if am and an are any two terms with the same base a and powers m and n respectively, then, am ÷ an = am – n if m > n and am ÷ an = 1 an – m if m < n It’s easier! In the case of division of the same bases, smaller power is subtracted from the greater power!! Laws of Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 109 Vedanta Excel in Mathematics - Book 8 (iii) Power law of indices Study the following illustrations and investigate the idea of the power law of indices. (22 ) 2 = 22 × 22 = 24 = 22 × 2 (23 ) 2 = 23 × 23 = 26 = 23 × 2 (24 ) 3 = 24 × 24 × 24 = 212 = 24 × 3 Thus, if am is any term with the base a and the index m, (am) n = am × n Furthermore, if a and b are any two terms, (a × b) m = am × bm and m a b = am bm (iv)Law of negative index If a– m is a term with base a and power – m, then, a– m = 1 am or, 1 am = a– m or am = 1 a–m For example, 2–2 = 1 22 , 1 24 = 2–4, 34 = 1 3–4 (v) Law of zero index Study the following illustrations and investigate the idea of the law of zero index. 2° = 21 – 1 = 21 ÷ 21 = 21 21 = 1 3° = 31 – 1 = 31 ÷ 31 = 31 31 = 1 Thus, if a° is any term with base a and power 0, then a° = 1 (vi) Root law of indices 2 (or only ) is the 2nd order of root, 3 is the 3rd order of root, and so on. Therefore, n is the nth order of root. Again, the square root of 4 = 4 = 22 or 2 22 = 2 2 2 = 2 The square root of 16 = 16 = 24 or 2 24 = 4 2 2 = 22 = 4 The cube root of 27 = 33 = 3 33 = 3 3 3 = 3 In this way, 2 22 = 2 2 2 , 2 24 = 4 2 2 , 3 33 = 3 3 3 ... and so on. In the similar way, n am = m n a . Thus, if m n a is a term with base a and power m n , then, m n a = n am I got it! (x2 ) 3 = x2×3 = x6 (y3 ) 4 = y3 × 4 = y12 Verification x3 x5 = x × x × x x × x × x × x × x = 1 x2 From the quotient law, x3 x5 = x3–5 = x–2 So, 1 x2 = x –2 Laws of Indices

Vedanta Excel in Mathematics - Book 8 110 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember (i) Product law am × an = am + n (ii) Quotient law am ÷ an = am – n when m > n am ÷ an = 1 an – m when m < n (iii) Power law (am) n = am × n, (ab) m = ambm, m a b = am bm (iv) Law of negative index a– m = 1 am or am = 1 a–m (v) Law of zero index a° = 1, b° = 1, x° = 1 and so on (vi) Root law of index m n a = n am or n am = m n a Worked-out Examples Example 1: Find the products in their exponential forms. (i) 5 × 52 × 53 × 55 (ii) 32 × 92 × 272 (iii) (2a)4 × (2a)– 12 × (2a)2 Solution: (i) 5 × 52 × 53 × 55 = 51 + 2 + 3 + 5 = 511 (ii) 32 × 92 × 272 = 32 × (32 ) 2 × (33 ) 2 = 32 × 32 × 2 × 33×2 = 32 × 34 × 36 = 312 (iii) (2a) 4 × (2a) –12 × (2a) 2 = (2a) 4 – 12 + 2 = (2a) – 6 = 1 (2a) 6 Example 2: Find the quotient in exponential forms. (i) (73 ) 5 ÷ 78 (ii) 164 ÷ 82 (iii) 272 ÷ 96 In product law, am × an = am + n In power law, (am) n = am × n Law of negative index: a–m = 1 am Solution: (i) (73 ) 5 ÷ 78 = 715 ÷ 78 = 715 – 8 = 77 (ii) 164 ÷ 82 = (24 ) 4 ÷ (23 ) 2 = 216 ÷ 26 = 216 – 6 = 210 In quotient law of indices. am ÷ an = am – n if m > n (iii) 272 ÷ 96 = (33 ) 2 ÷ (32 ) 6 = 36 ÷ 312 = 1 312 – 6 = 1 36 In quotient law of indices, am ÷ an = 1 an – m if m < n Laws of Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 111 Vedanta Excel in Mathematics - Book 8 Example 3: Evaluate(i) 8 27 2 3 (ii) 25 36 0.5 (iii) 125 64 1 3 Solution: (i) 8 27 2 3 = 23 33 2 3 = 2 3 2 3 3 × = 2 3 2 = 4 9 (ii) 25 36 0.5 = 25 36 1 2 = 52 62 1 2 = 5 6 1 2 2 × = 5 6 (iii) 125 64 1 3– = 64 125 1 3 = 43 53 1 3 = 4 5 1 3 3 × = 4 5 125 64 1 3– 125 64 1 3– 1 3– 125 64 1 3 1 1 3 1 64 125 1 3 1 3 1 125 1 3 1 64 1 3 = = = × = = 64 125 1 3 Example 4: Simplify (i) 125x6 y 3 3 , (ii) 3 64 Solution: (i) 125x6 y 3 3 = 53 x6 y3 3 = 5 3 3 x 6 3 y 3 3 = 5x2 y (ii) 3 64 = 3 8 = 3 23 = 2 3 3 = 2 Example 5: Simplify (i) 83 × 153 64 × 103 (ii) (xa + b) a – b × (xb + c) b – c × (xc – a) c + a Solution: (i) 83 × 153 64 × 103 = (23 ) 3 × (3 × 5)3 (2 × 3)4 × (2 × 5)3 = 29 × 33 × 53 24 × 34 × 23 × 53 = 29 × 33 × 53 27 × 34 × 53 = 29 – 7 × 53 – 3 34 – 3 = 22 × 50 3 = 4 3 Root law of indices: n am = m n a (ii) (xa + b) a – b × (xb + c) b – c × (xc – a) c + a = (x) (a + b) (a – b) × (x) (b + c) (b – c) × (x) (c – a) (c + a) = xa2 – b2 × xb2 –c2 × xc2 – a2 = xa2 – b2 + b2 – c2 + c2 – a2 = x0 = 1 Laws of Indices

Vedanta Excel in Mathematics - Book 8 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 8.1 General Section - Classwork Let’s say and write the answers using the laws of indices as quickly as possible. 1. a) x2 × x = ............ b) a2 × a3 = ............ c) pa × pb = ............ d) y3 ÷ y = ............ e) z5 ÷ z2 = ............ f) xa ÷ xb = ............ g) (x2 ) 3 = ............ h) (y3 ) 2 = ............ i) (pa ) b = ............ j) (x2 y2 ) 2 = ............ k) (ab3 ) 2 = ............ l) (xp yq ) m = ............ m) x° = ............ n) 5° = ............ o) (x + 2)° = ............ 2. Let’s use the law of negative index and answer with positive index. a) x – 2 = ............ b) p – 3 = ............ c) a – x = ............ d) 1 a–3 = ............ e) 1 x –4 = ............ f) 1 p–q = ............ 3. Let’s use root law of index and write the answers. a) x 1 2 = ............ b) 3 2 3 = ............ c) a m n = ............ d) 3 = e) 3 52 = ............ f) x ay = ............ 4. a) For what power of q is its value 1? .............................. b) If 3x = 1, what is the value of x? .............................. c) What is the value of 5x ×5–x ? .............................. d) What is the value of 1a + b? .............................. e) What should be the power of x so that its value will be 1 x2? .............................. f) For what power of (a + 2) the value of 3 a + 2 is 3? .............................. g) For what power of p will it be 3 p2 ? .............................. Creative Section - A 5. Let’s find the products or quotients in the exponential forms by using laws of indices. a) 22 × 25 × 2–1 b) 93 × 3– 3 × 272 c) (5x) 3 × (25x2 ) 2 × (5x) – 12 d) (xy) 7 × (xy) – 5× (xy) – 6 e) 3ax2 × xa3 × ax4 × a2 f) 1 x2 × 1 x –3 × 3 x6 g) 38 ÷ 33 h) 256 ÷ 57 i) 82 ÷ 210 j) (y2 ) – 3 ÷ 1 y – 8 k) xm – 3 ÷ xm – 4 l) pa – b ÷p3a + b Laws of Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 113 Vedanta Excel in Mathematics - Book 8 6. Let’s apply the laws of indices and simplify. a) (x2 ) 3 b) (p2 q3 ) 2 c) (xy2 ) – 4 d) 1 ab –3 e) xy x2 y2 –4 f) x2 g) 3 y3 h) 3 x3 y3 i) 3 x y 9 j) 3 x6 7. Let’s evaluate. a) (22 ) 3 b) (36 ) 1 2 c) (52 ) 1 2 d) (7–3) 1 3 e) 22 32 1 2 – f) 1 2 4 g) 1 2 9 h) 2 3 8 i) 1 4 16 j) – 1 3 8 k) – 3 4 81 l) 16 25 1 2 m) 16 81 1 4– n) 625 81 – 0.5 o) 243 32 – 0.4 8. Let’s evaluate. a) 22 b) 24 – 1 c) 3 53 d) 3 1 2–6 e) 3 29 f) 3 8 g) 4 16 h) 3 27–1 i) 4 1 81 j) 5 32–2 k) 3 64 125 l) 3 43 m) 3 64 n) 4 256 o) 3 274 Creative Section - B 9. Let’s simplify: a) 22 × 23 × 24 2 × 25 b) 33 × 35 × 37 32 × 95 c) 42 × 24 × 82 22 × 45 d) 253 × 52 × 625 54 × 1253 e) 44 × 55 252 × 162 f) 84 × 95 162 × 273 g) 42 × 93 × 64 82 × 274 h) 35 × 255 × 225 93 × 1254 10. Let’s simplify: a) xb – c × xc – a × xa – b b) (xa ) b – c × (xb ) c – a × (xc ) a – b c) (xa – b) c × (xb – c) a × (xc – a) b d) (xa + b) a – b × (xb+ c) b – c × (xc + a) c – a e) xa + b + 1 × xa – b + 2 x2a + 3 f) xp – q + 1 × xq – r + 1 × xr – p + 1 x3 g) xa + b × xb + c × xc + a x2a × x2b × x2c h) xa + b xc + b × xc + d xd + a i) xa xb a + b × xb xc b + c × xc xa c + a j) xa x –b a – b × xb xc b + c × x – a x – c c + a k) (a2 ) x + y × (a2 ) y + z × (a2 ) z + x (ax .ay .az ) 4 l) xa – b × xb – c × xc – a Laws of Indices

Vedanta Excel in Mathematics - Book 8 114 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 11. Let’s simplify: a) 2x + 1 + 2x 3 × 2x b) 3y + 1 + 3y 2 × 3y c) 5x + 2 – 5x 6 × 5x d) 6x + 2 – 6x 7 × 6x e) 2x + 4 – 2x 5 × 2x f) 6n + 2 – 6n 6n + 1 + 6n g) 11p + 2 – 11p 11p + 1 + 11p h) 2x + 3 + 2x + 2 2x + 2 – 2x + 1 12. If a = 1, b = 2 and c = –3, find the value of: a) 4a3 b2 b) 3b2 c2 c) ab + ba d) ac + cb – bc e) ab bc ca f) ab+c bc+a ca + b g) ab–c bc–a ca–b 13. a) If x = 2, y = 3, m = 1 and n = 2, find the value of xm + n y4m – n . b) If x = 2, y = 4, m = –1 and n = 3, find the value of xm + n × yn – m xm – n × yn + m 14. a) A rapidly spreading virus is multiplying itself into 2 viruses in every one minute time interval. Write a correct base with appropriate exponent to show the number of viruses in each of the following intervals of time. (i) in 1 minute (ii) in 2 minutes (iii) in 3 minutes (iv) in 5 minutes (v) in 8 minutes (vi) in 10 minutes b) Each member of a licensed network business should make 3 members within every week. Write a correct base with appropriate index to show the number of members made in each of the following weeks. (i) in 1st week (ii) in 2nd week (iii) in 3rd week (iv) in 4th week (v) in 5th week It’s your time - Project work and Activity section 15. a) Let’s take any base number and index number greater than 1 and less than 6. Then, verify the following laws of indices. (i) product law (ii) Quotient law (iii) Power law (iv) Law of negative index (v) Root law of indices (vi) Law of zero index b) Let’s take a bundle of flash cards. At first, a student (S) should give 3 cards to his/ her friend (F1 ). Second time, F1 should give 3 cards to each of his/her three friends F2 , F3 and F4 . Third time, each friend F2 , F3 and F4 should give 3 cards to his/her three other friends, and so on. (i) Show the above mentioned process, repeating upto five times by the network diagram in a chart paper. (ii) Write the process with the correct base with appropriate indices and find how many flash cards are distributed in each time. Laws of Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 115 Vedanta Excel in Mathematics - Book 8 9.1 Algebraic terms and expressions - Looking back Classwork - Exercise Let’s say and write the answer as quickly as possible. 1. a) In 5x2 , the coefficient is ................., base is ................., exponent is ..................... b) In 4a3 , the coefficient is ................., base is ................., exponent is ...................... 2. a) Are 3x and 3x2 like or unlike terms? ....................................................................... b) Are y3 and 7y3 like or unlike terms? ....................................................................... 3. a) How many terms are there in the expression 2xy? .................................................. b) How many terms are there in the expression 2x + y? ............................................. c) How many terms are there in the expression 2 – x + y? ......................................... 2x, 3x2 , 5y – 7a2 b2 , etc. are a few examples of algebraic terms. An algebraic term is the product form of any number and variables. There are three important parts in an algebraic term. These parts are coefficient, base, and exponent. 3x2 Exponent Base Coefficient 7y3 Exponent Base Coefficient The sum or difference of two or more algebraic terms form an algebraic expression. For example 3x + 2 is a binomial expression, a2 – 4ab + b2 is a trinomial expression, and so on. However, an algebraic term itself is a monomial expression. 2x, y2 , –3p3 , etc. are monomial expressions. 9.2 Polynomials and degree of polynomials x + y, 2x2 + 4xy – 3y2 , a3 + a2 – 2a + 5, 2 x4 + 2x2 – 7, etc. are the examples of polynomials. In these polynomials, the powers of variables are whole numbers. The algebraic expressions in which powers of variables are whole numbers are known as polynomials. On the other hand, if the power of variable of an expression is not a whole number, the expression is not a polynomial. For example, in x +1 x , the power of x in 1 x is – 1 which is not a whole number. So x + 1 x is not a polynomial. Similarly, 1 x + 7, 3 x2 – 4x + 5, etc. are not polynomials. Unit 9 Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 116 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Let’s consider an algebraic term 3x2 . In 3x2 , the variable x is multiplied 2 times. So, the degree of 3x2 is 2. Similarly, in 2y3 , the variable y is multiplied 3 times. So, the degree of 4y3 is 3. Thus, in a polynomial with a monomial expression, the exponent of the variable is the degree of the polynomial. In the case of a term that contains two or more variables, the sum of the powers of each variable is the degree of the term. For example, In 3x3 y2 , the sum of the powers of x and y is (3 + 2) = 5. So, the degree of 3x3 y2 is 5. In 7abc, the sum of the powers of a, b, and c is (1 + 1 + 1) = 3. So, the degree of 7abc is 3. In the case of a polynomial with multinomial expression, the highest power of its any term is the degree of the polynomial. For example, In 2x2 + 3x + 4, the highest power of x is 2. So, its degree is 2. In 4y5 – 5y3 – y, the highest power of y is 5. So, its degree is 5. In 3p2 q2 – pq + 5, the highest power of pq is (2 + 2) = 4. So, its degree is 4. 9.3 Evaluation of algebraic expressions Let’s take an algebraic expression 2(l + b) and evaluate it when l = 5 and b = 4. Here, if l = 5 and b = 4, then 2(l + b) = 2(5 + 4) = 18. Thus, evaluation is the process of finding the value of an algebraic expression by replacing the variables with numbers. Worked-out Examples Example 1: Which of the following expressions are polynomials? Write with reason. (i) x2 + 1 2 (ii) y2 + 1 y2 (iii) 3 x – 2 (iv) 2 x + 5 Solution: (i) x2 + 1 2 is a polynomial because the power of x is 2 which is a whole number. (ii) y2 + 1 y2 is not a polynomial because the power of y in 1 y2 is –2 which is not a whole number. (iii) 3x – 2 is a polynomial because the power of x is 1 which is a whole number. (iv) 2 x + 5 is not a polynomials because the power of x is 1 2 which is not a whole number. Example 2: Write the degrees of the following polynomials. (i) 3x2 (ii) 2a2 bc (iii) 2x3 – 4x2 + 7x (iv) x3 y2 + x2 y2 – xy Solution: (i) The degree of 3x2 is 2. (ii) The degree of 2a2 bc is 2 + 1 + 1 = 4 (iii) The degree of 3y3 – 2y2 + 5x is 3. Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 117 Vedanta Excel in Mathematics - Book 8 (iv) In x3 y2 + x2 y2 – xy, the sum of the powers of x3 y2= 3 + 2 = 5 and it is the highest sum of the powers. Therefore, it is 5th degree polynomial. Example 3: If x = 3 and y = 2, evaluate x2 – 2xy + y2 . Solution: Here, when x = 3 and y = 2, then x2 – 2xy + y2 = 32 – 2 × 3 × 2 + 22 = 9 – 12 + 4 = 1 EXERCISE 9.1 General Section - Classwork 1. Let’s say and write the polynomial expressions in the table. a) 2x + 3 b) 3y2 – 2y + 4 c) a + 1 a d) 5x + 1 e) x2 2 – 3x + 7 f) x + 2 – 3 2. Let’s say and write the degrees of these polynomials. a) 3x ................. b) 2p2 ................. c) 4a2 b ................. d) 7y – 1 ................. e) 2x3 – 3x2 + 6 ................. f) 9xy – 7x + 2y ................ 3. Let’s tell and write the values of the expressions as quickly as possible. a) If l = 4 and b= 3, then (i) l × b = …….. (ii) 2 (l + b) = …….. b) If l = 5, then (i) l 2 = …….. (ii) l 3 = …….. (iii) 6l 2 = …….. Creative Section 4. Let’s answer the following questions. a) Why is x2 2 a polynomial but 2 x2 is not a polynomial? b) Why is 2x not a polynomial but 2 x is a polynomial? c) Define the degree of a polynomial with examples. 5. Find the degrees of the following polynomials. a) 2xyz b) 5a2 bc c) 3y2 – 4y + 8 d) 7x5 – 2x4 + 5x3 – 7x2 – 3x e) x2 + y2 + z2 – 3xyz f) x3 y3 – 2x2 y2 + 7xy – 5 Polynomials Expressions Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 118 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) If l = 5 and b = 3, find the values of (i) l × b (ii) 2(l + b) b) If l = 10, b = 8 and h = 5, find the values of (i) l × b × h (ii) 2h (l + b) (iii) 2 (lb + bh + lh) c) If r = 7, π = 22 7 , find the values of (i) πr2 (ii) 2πr d) If a = 3 and b = 2, find the values of (i) a2 + 2ab + b2 and (a + b) 2 (ii) a2 – 2ab + b2 and (a –b) 2 (iii) a3 + 3a2 b + 3ab2 + b3 and (a + b) 3 (iv) a3 – 3a2 b + 3ab2 – b3 and (a – b) 3 (v) (a + b) (a –b) and (a2 – b2 ) (vi) (a + b) 2 – 2ab 7. a) Find the value of P × T × R 100 when P = 200, T = 2 and R = 5 b) Find the value of P 1 + R 100 T when P = 100, R = 10, T = 2 c) Find the value of (i) 2prh (ii) 2pr(r + h) (iii) pr2 h when p = 22 7 , r = 7 and h = 5 9.4 Special products and formulae (i) The product of (a + b) and (a + b) : Square of binomials Let’s multiply (a + b) by (a + b) (a + b) × (a + b)= a (a + b) + b (a + b) or, (a + b) 2 = a2 + ab + ab + b2 or, = a2 + 2ab + b2 Thus, (a + b) 2 = (a2 + 2ab + b2 ) Also, a2 + b2 = (a + b) 2 – 2ab When the length a of the smaller square is increased by b, each side of the bigger square ABCD is (a + b) Now, area of ABCD = a2 + ab + ab + b2 (a + b) 2 = a2 + 2ab + b2 (ii) The product of (a – b) and (a – b) Let’s multiply (a – b) by (a – b) (a – b) × (a – b) = a (a – b) – b (a – b) or, (a – b) 2 = a2 – ab – ab + b2 = a2 – 2ab + b2 Thus, (a – b) 2 = a2 – 2ab + b2 Also, a2 + b2 = (a – b) 2 + 2ab When the length and breadth of the square ABCD is decreased by b, Area of ABCD = (a – b) 2 + b (a – b) + b (a – b) + b2 or, a2 = (a – b) 2 + ab – b2 + ab – b2 + b2 or, a2 = (a – b) 2 + 2ab – b2 or, (a – b2 ) = a2 – 2ab + b2 www.geogebra.org/classroom/t6rg6csu Classroom code: T6RG 6CSU Vedanta ICT Corner Please! Scan this QR code or browse the link given below: a + b a + b a a2 b2 ab b ab b b a b D C A B www.geogebra.org/classroom/nkdkq6wh Classroom code: NKDK Q6WH Vedanta ICT Corner Please! Scan this QR code or browse the link given below: a a – b (a – b) 2 a – b b b(a – b) b(a – b) b2 a – b D A C B b b Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 119 Vedanta Excel in Mathematics - Book 8 (iii) The product of (a + b) and (a – b) (a + b) (a – b) = a (a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2 Thus,(a + b) (a – b) = a2 – b2 Worked-out Examples Example 1: Find the squares of: a) 3x + 2y b) x – 1 x Solution: a) Square of (3x + 2y) = (3x + 2y) 2 = (3x) 2 + 2.3x.2y + (2y) 2 = 9x2 + 12xy + 4y2 b) Square of x – 1 x = x – 1 x 2 = (x) 2 – 2.x. 1 x + 1 x 2 = x2 – 2 + 1 x2 Example 2: Express 16x2 – 40xy + 25y2 as a perfect square. Solution: 16x2 – 40xy + 25y2 = (4x) 2 – 2.4x.5y + (5y) 2 = (4x – 5y) 2 Example 3: Find the product of a) (x – 3y) (x + 3y) b) (4a2 + 3b2 ) (4a2 – 3b2 ) Solution: a) (x – 3y) (x + 3y)= x2 – (3y) 2 = x2 – 9y2 b) (4a2 + 3b2 ) (4a2 – 3b2 )= (4a2 ) 2 – (3b2 ) 2 = 16a4 – 9b4 Example 4: If x + y = 7 and xy = 5, find the value of x2 + y2 . Solution: We have, x2 + 2xy + y2 = (x + y) 2 a a2 – b2 a – b b A Area of ABCD = a2 Area of ABCDEF = a2 – b2 Area of ABCD a2 – b2 = (a + b) (a – b) A E F A B B B C D C D C D b a a – b(a + b) (a – b) a a + b b www.geogebra.org/classroom/qb2awvhb Classroom code: QB2A WVHB Vedanta ICT Corner Please! Scan this QR code or browse the link given below: (3x + 2y) (3x + 2y) Let’s consider, 3x = a and 2y = b We know that (a + b) 2 = a2 + 2ab + b2 Then, (3x + 2y) 2 = (3x) 2 + 2.3x.2y + (2y) 2 Let’s consider, x = a and 1 x = b We know that, (a – b) 2 = a2 – 2ab + b2 Then, x – 1 x 2 = (x) 2 – 2.x. 1 x + 1 x 2 (x – 1 x ) (x – 1 x ) Let’s consider 4x = a and 5y = b. Then, (4x) 2 – 2.4x.5y + (5y) 2 is in the form a2 – 2ab +b2 and a2 – 2ab + b2 = (a – b) 2 . Using (a + b) (a – b) = a2 – b2 Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 120 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, x2 + y2 = (x + y) 2 – 2xy = 72 – 2 × 5 = 49 – 10 = 39 Example 5: If a + 1 a = 6, find the values of (i) a2 + 1 a2 (ii) a – 1 a 2 . Solution: Here, a + 1 a = 6 (i) We have, a2 + b2 = (a + b) 2 – 2ab ∴ a2 + 1 a2 = a + 1 a 2 – 2.a. 1 a = 62 – 2 = 36 – 2 = 34 (ii) We have (a – b) 2 = a2 – 2ab + b2 ∴ a – 1 a 2 = a2 – 2.a. 1 a + 1 a2 = a2 + 1 a2 – 2 = 34 – 2 = 32 Example 6: Simplify (x + 3y)2 – (x – 3y)2 Solution: (x + 3y) 2 – (x – 3y) 2 = x2 + 2.x.3y + (3y) 2 – (x2 – 2.x.3y + (3y) 2 ) = x2 + 6xy + 9y2 – x2 + 6xy – 9y2 = 12xy Example 7: Simplify 2.7 × 2.7 – 1.3 × 1.3 2.7 – 1.3 Solution: 2.7 × 2.7 – 1.3 × 1.3 2.7 + 1.3 = (2.7)2 – (1.3)2 2.7 + 1.3 = (2.7 + 1.3) (2.7 – 1.3) 2.7 + 1.3 = 2.7 – 1.3 = 4 EXERCISE 9.2 General Section - Classwork 1. Let’s say and write the expanded forms of these expressions. Expressions Expanded forms a) (x + y) 2 b) (a + 1)2 c) (p + 2)2 d) (y + 3)2 2. Let’s say and write the following products in the forms of difference of two squared terms. Alternative process (x + 3y) 2 – (x – 3y) 2 = (x + 3y + x – 3y) [x + 3y – (x – 3y)] = 2x(x + 3y – x + 3y) = 2x × 6y = 12xy Expressions Expanded forms e) (x – y) 2 f) (m – 1)2 g) (a – 2)2 h) (p – 3)2 Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 121 Vedanta Excel in Mathematics - Book 8 a) (x + y) (x – y) = .......................... b) (x + 1) (x – 1) = .......................... c) (p + 2) (p – 2) = .......................... d) (m + 3) (m – 3) = .......................... e) (a + 1 a) (a – 1 a) = .......................... f) (x + 2 x) (x – 2 x) = .......................... g) (3 p + p) (3 p – p) = .......................... h) ( 4 m + m) ( 4 m – m) = .......................... Creative Section - A 3. Let’s look at these diagrams and write the areas in algebraic expression forms as shown in the example. Area of ABCD Area of PQRS Area of PREM a) D x x + 2 2.x 2.x x + 2 x2 x 2 2 C A B b) S x x R P Q M x x + y x.y x.y x + y x2 x y y E P R c) Area of ABCD d) D x-1 (x-1)2 1.(x-1) x x 1.(x-1) x–1 1 C 1 A B 12 Area of EFGH e) H x x G E F (x-2)2 Area of PQRS f) S x x R P Q x – y x – y (x – y)2 y y 4. Let’s find the area of each of these rectangles in algebraic expression forms. Area of ABCD = x2 + 1.x + 1.x + 12 (x + 1)2 = x2 + 2x + 1 D x x + 1 1.x 1.x 12 x + 1 x2 x 1 1 C A B Area of ABCD a) x x x + 1 x – 1 x2 – 12 1 1 1 12 A B D C Area of MNOP b) x x x + 2 x – 2 x2 – 22 2 2 2 M N P O Area of EFGH c) x x x + y x – y x2 – y2 y y y y2 E F H G Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 122 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5. a) Find the squares of (i) 2a + 1 (ii) x – 3y (iii) a – 1 a (iv) x + 1 x b) Expand (i) (3x – 1)2 (ii) (2y + 1 2y ) 2 6. Express a) x2 +2x + 1 b) a2 – 4ab + 4b2 c) 9p2 + 12pq + 4q2 as perfect squares. 7. Let’s find the expressions to be inserted to make these expressions perfect squares. a) x2 + ……… + 16y2 (Hint. x2 + 2.x.4y + (4y) 2 , so the required terms is 8xy) b) x2 + …....................…… + 4y2 c) x2 – …....................…… + 9y2 d) 4x2 + …....................…… + 25y2 e) 9a2 – …....................…… + 49b2 8. Let’s apply the appropriate formula to find the products. a) (a + 3) (a – 3) b) (2x + 1) (2x – 1) c) (4 + 3p) (4 – 3p) d) (2x – 3y) (2x + 3y) e) (x2 – y2 ) (x2 + y2 ) f) (2x2 + 5y2 ) (2x2 – 5y2 ) g) (a + b + c) (a + b – c) h) (x – y + z) (x + y + z) i) (p – q – r) (p + q – r ) j) (x + y) (x – y) (x2 + y2 ) k) (x + 2) (x – 2) (x2 + 4) l) (2a + y) (2a – y) (4a2 + y2 ) 9. Let’s find the products using the formula (a + b) (a – b) = a2 – b2 . Hint. 99 × 101 = (100 – 1 ) × (100 + 1) = 1002 – 12 = 10000 – 1 = 9999 a) 19 × 21 b) 49 × 51 c) 78 × 82 d) 102 × 98 10. a) If (x + y) = 5 and xy = 3, find the value of x2 + y2 . b) If (a – b) = 4 and ab = 2, find the value of a2 + b2 . c) If x + 1 x = 3, find the value of x2 + 1 x2 . d) If p – 1 p = 7, find the value of p2 + 1 p2 . 11. a) If a + 1 a = 3, find the values of (i) a2 + 1 a2 (ii) a – 1 a 2 b) If x + 1 x = 5, find the values of (i) x2 + 1 x2 (ii) x – 1 x 2 c) If p – 1 p = 4, find the values of (i) p2 + 1 p2 (ii) p + 1 p 2 d) If m2 – 1 m = 6, find the values of (i) m2 + 1 m2 (ii) m + 1 m 2 Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 123 Vedanta Excel in Mathematics - Book 8 12. Simplify. a) (a + b) 2 + (a – b) 2 b) (x + y) 2 – (x – y) 2 c) (2p – 3)2 + (2p + 3)2 d) (3x + y) 2 – (3x – y) 2 e) (x + 1 x )2 + ( x – 1 x )2 f) (2 – 1 a ) 2 – (2 + 1 a ) 2 13. Simplify. a) 2.1 × 2.1 – 0.9 × 0.9 2.1 – 0.9 b) 3.6 × 3.6 – 1.4 × 1.4 3.6 + 1.4 c) (2.5)2 + 2 × 2.5 × 1.4 + (1.4)2 2.5 + 1.4 d) (6.3)2 – 2 × 6.3 × 4.5 + (4.5)2 6.3 – 4.5 It’s your time - Project work and Activity section Paper folding: 14. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then, fold them to get square sheet of papers as shown in the diagrams. b) Again, let’s fold each square sheet of paper as shown in the diagram and complete the sums. (i) x x 1 1 121. 1.x x2 = (x + 1)2 = x2 + 1.x + 1.x + 12 = x2 + 2x + 1 (ii) = ................ = .......................... = .......................... x x y y x. y x.y (iii) = ................ = .......................... x x2 –12 x x2 x x x x x2 1 12 1 1 x – 1 x + 1 x x2 –12 x2 – 12 (iv) = ................ = .......................... x x2 –y2 x x2 x x x x x2 y y2 y y x – 1 x + y x x2 –y2 Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 124 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9.5 Factors and factorisation Classwork - Discussion Let’s discuss the answers of the following questions. In 3 × 4 = 12, what are the factors of 12 and what is the product of 3 and 4? In 2 × 3 × 4 = 24, what are the factors of 24 and what is the product of 2, 3, and 4? Similarly, In 2 × x = 2x, what are the factors of 2x and what is the product of 2 and x? In x × x2 = x3 , what are the factors of x3 and what is the product of x and x2 ? In 2y(y + 1) = 2y2 + 2y, what are the factors of 2y2 + 2y and what is the product of 2y and (y + 1)? In (a + b) (a – b) = a2 – b2 , what are the factors of a2 – b2 and what is the product of (a + b) and (a – b)? Thus, when two or more algebraic expressions are multiplied, the result is called the product and each expression is called the factor of the product. The process of finding out factors of an algebraic expression is known as factorisation. It is also called resolution of the expression into its factors. Now, let’s learn the process of factorisation of different types of expressions. (i) Factorisation of expressions which have a common factor in each of its term Let’s take an expression, ax + ay. In this expression a is present in both terms. So, a is common in both terms and it is called the common factor. In such an expression, the common factor is taken out and each term of the expression is divided by the common factor to get another factor. Worked-out Examples Example 1: Factorise (i) 2ax +6bx (ii) 6x2 y – 8xy2 + 10xy Solution: (i) 2ax + 6bx = 2x (a + 3b) (ii) 6x2 y – 8xy2 + 10xy = 2xy (3x – 4y + 5) Example 2: Resolve into factors 2x (a + b) – 3y (a + b) Solution: 2x (a + b) – 3y (a + b) = (a + b) (2x – 3y) Area = 2ax + 6ab 2x (a + 3b) 2x is the common factor in each term. 6x2 y – 8xy2 + 10xy = 3x × 2xy – 4y × 2xy + 5 × 2xy So, 2xy is the common factor in each term. (a + b) is the common factor in each term. Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 125 Vedanta Excel in Mathematics - Book 8 (ii) Factorisation of expressions having common factor in the groups of terms Let’s take an expression ax + bx – ay – by. In this expression, x is common in ax + bx and y is common in –ay – by. In such an expression, the terms are to be arranged in groups in such a way that each group has a common factor. Example 3: Factorise (i) a2 – ax + ab – bx (ii) a (x2 – y2 ) + x (y2 – a2 ) Solution i) a2 – ax + ab – bx = a (a – x) + b (a – x) = (a – x) (a + b) (ii) a (x2 – y2 ) + x (y2 – a2 )= ax2 – ay2 + xy2 – a2 x = ax2 + xy2 – a2 x – ay2 = x (ax + y2 ) – a (ax + y2 ) = (ax + y2 ) (x – a) Example 4: If (3x2 + 6x) sq. unit is the area of a rectangle, find the length and breadth of the rectangle. Solution Here, area of the rectangle = 3x2 + 6x Now, factorising 3x2 + 6x = 3x(x + 2) ∴ Length and breadth of the rectangle are 3x unit and (x + 2) unit or (x + 2) unit and 3x unit. EXERCISE 9.3 General Section Let’s factorise, say and write the answers as quickly as possible. 1. a) ax + bx = ……..............…………. b) ax – ay = ……..............…………. c) y2 – y = ……..............…………. d) 2x3 + x2 = ……..............…………. e) x2 – x3 = ……..............…………. f) x2 y + xy2 = ……..............…………. Area =a2 – ax + ab – bx (a + b ) (a – x) Terms are arranged in groups. a is common in a2 – ax and b is common in ab – bx. (a – x) is common in each term. 3x2 + 6x x + 2 3x Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 126 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 2. a) x(a + b) + y(a +b) = ………....…….. b) a(p + q) – (p + q) = ………….....….. c) (y – 2) – y (y – 2) = ………....…….. d) x(a – b) + 2(a – b) = ……….....…….. Creative Section - A 3. Let’s factorise a) 2ax + 4ay b) 6x2 – 9x c) 8px3 + 12px2 d) 4mx2 – 6mx + 2m2 x e) 9b2 y3 – 6b3 y2 +15b2 y2 f) –9a5 – 6a3 x2 – 15a2 g) x(2x + 3) + 4(2x + 3) h) 2y(y – 4) – 3(y – 4) i) 7x (x + 2y) + y(x + 2y) 4. Let’s apply the process of factorisation and simplify. a) 15 × 6 + 15 × 4 b) 3.4 × 103 – 2.6 × 103 c) 3.5 × 10–2 + 2.3 × 10–2 5. Let’s resolve into factors. a) ax + ay + bx + by b) px + qx – py – qy c) x2 + 2x + 3x + 6 d) x2 – 4x – 3x + 12 e) 6x2 – 2y2 + 4xy – 3xy f) ax2 + ay2 + bx2 + by2 g) x3 + x2 + x + x2 y + xy + y h) x2 – x (y + z) + yz i) a2 – a (2x – y) – 2xy j) x2 – (y – 3)x – 3y k) x (a2 – b2 ) + a (b2 – x2 ) l) ab (c2 + 1) – c (a2 + b2 ) Creative Section - B 6. Let’s find the length and breadth of the following rectangles of the given area. a) Area = 3a2 b + 6ab Area = x2 + 2x + xy + 2y Area = x2 + 3x + 4ax + 12a 7. a) If (4ax + 6ay) sq. unit is the area of a rectangle, find the length and breadth of the rectangle. b) Find the length and breadth of a rectangle of area (2x2 + 2xy + xy + y2 ) sq. unit. Also, find the perimeter of the rectangle. c) The area of a rectangular field is (x2 + 3xy + 2xy + 6y2 ) sq. unit. Find the perimeter of the field. (iii) Factorisation of expressions having the difference of two square terms Let’s find the product of the expressions (a + b) and (a – b). (a + b) (a – b) = a(a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2 Here, the expression a2 – b2 is the difference of two square terms a2 and b2 . From the above illustration, its clear that (a + b) and (a – b) are the factors of a2 – b2 . Thus, to factorise an expression of the form a2 – b2 , we should use the formula, a2 – b2 = (a + b) (a – b) b) c) Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 127 Vedanta Excel in Mathematics - Book 8 Worked-out Examples Example 1: Factorise (i) x2 – 9 (ii) 4x2 – 25y2 (iii) a4 – 16 Solution: (i) x2 – 9 = x2 – 32 = (x + 3) (x – 3) (ii) 4x2 – 25y2 = (2x) 2 – (5y) 2 = (2x + 5y) (2x – 5y) (iii) a4 – 16 = (a2 ) 2 – 42 = (a2 + 4) (a2 – 4) = (a2 + 4) (a2 – 22 ) = (a2 + 4) (a + 2) (a – 2) Example 2: Factorise (i) (x2 + y2 ) 2 – x2 y2 (ii) 9 – (p – q)2 Solution: (i) (x2 + y2 ) 2 – x2 y2 = (x2 + y2 ) 2 – (xy) 2 = (x2 + y2 + xy) (x2 + y2 – xy) = (x2 + xy + y2 ) (x2 – xy + y2 ) (ii) 9 – (p – q) 2 = 32 – (p – q) 2 = (3 + p – q) [3 – (p – q)] = (3 + p – q) (3 – p + q) Example 3: Resolve into factors x4 + x2 y2 + y4 Solution: x4 + x2 y2 + y4 = (x2 ) 2 + (y2 ) 2 + x2 y2 = (x2 + y2 ) 2 – 2x2 y2 + x2 y2 = (x2 + y2 ) 2 – x2 y2 = (x2 + y2 ) 2 – (xy) 2 = (x2 + y2 + xy) (x2 + y2 – xy) = (x2 + xy + y2 ) (x2 – xy + y2 ) Example 4: Simplify by factorisation process (i) 452 – 252 (ii) 101×99 Solution: (i) 452 – 252 = (45 + 25) (45 – 25) = 70 × 20 = 1400 (ii) 101 × 99 = (100 + 1) × (100 – 1)= 1002 – 12 = 10000 – 1 = 9999 x2 – 9 x – 3 x + 3 It is in the form a2 – b2 Using a2 – b2 = (a+ b) (a – b) 4x2 – 25y2 2x – 5y 2x + 5y Square root of 4x2 = 2x and square root of 25y2 = 5y Using a2 – b2 = (a + b) (a – b) Square root of a4 = a2 and square root of 16 = 4 Using a2 – b2 = (a + b) (a – b) a2 – 4 = a2 – 22 is further factorised to (a + 2) (a – 2) Consider, x2 + y2 = a and xy = b Then, a2 – b2 = (a + b) (a – b) (x2 + y2 ) 2 – (xy) 2 = (x2 + y2 + xy) (x2 + y2 – xy) Using x2 + y2 = (x+ y)2 – 2xy Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 128 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 9.4 General Section 1. Let’s say and write the following expressions as the product of their factors. Expressions Product of factors a) m2 – n2 b) m2 – 4 c) x2 – y2 d) x2 – 9 e) p2 – 16 Expressions Product of factors f) 4x2 – 1 g) 9y2 – 4 h) p2 – 49 i) a2 – 1 4 j) b2 – 1 9 2. Let’s say and write the differences of these squared numbers as quickly as possible. a) 102 – 92 = ……………...................... b) 72 – 52 = ……………...................... c) 212 – 202 = ……………...................... d) 502 – 492 = ……………...................... e) 202 – 102 = ……………...................... f) 402 – 302 = ……………...................... Creative Section 3. Let’s resolve into factors. a) x2 – 36 b) a2 – 49 c)25 – y2 d) 16 – 81p2 e) x3 – 16x f) 2a2 – 72 g) 5p3 – 80p h) 3y3 – 27y i) 5a3 – 20ab2 j) 8x5 y – 18x3 y3 k) 25x2 – 1 49 l) 1 4x2 – 1 81 4. Let’s factorise: a) x4 – 16 b) a4 – 81 c) x4 – y4 d) 16p4 – q4 e) 81x4 – 625 f) 32y4 – 162 g) a8 – b8 h) 256x8 – y8 5. Let’s factorise: a) (a – b) 2 – 9 b) (x + y) 2 – 25 c) (a2 + b2 ) 2 – 4 d) (a2 – b2 ) 2 – a2 b2 e) 16 – (x + y) 2 f) 49 – (a – b) 2 g) (x + 3)2 – (y + 2)2 h) (a – 5)2 – (b – 4)2 6. Let’s resolve into factors. a) x4 + x2 y2 + y4 b) x4 + x2 + 1 c) x4 + 7x2 y2 + 16y4 d) a4 – 7a2 b2 + b4 e) x4 – 3x2 y2 + 9y4 f) a4 + 6a2 b2 + 25b4 g) 4x4 + 3x2 y2 + 9y4 h) 25x4 + 4x2 y2 + 4y4 i) 4a4 – 13a2 b2 + 9b4 7. Let’s simplify by factorisation process. a) 252 – 152 b) 642 – 442 c) 962 – 862 d) (101)2 – (100)2 e) 51 × 49 f) 82 × 78 g) 101 × 99 h) 103 × 97 Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 129 Vedanta Excel in Mathematics - Book 8 8. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b). a) b) c) (v) Factorisation of trinomial expressions of the form x2 + px + q Let’s take any two binomial expressions (x + a) and (x + b) and find their product. The product of (x + a) and (x + b) = (x + a) (x + b) = x (x + b) + a (x + b) = x2 + bx + ax + ab = x2 + (a + b)x + ab Now, let’s compare the product x2 + (a + b)x + ab to the trinomial x2 + px + q. Here, p = a + b and q = ab. Thus, while factorising a trinomial expression of the form x2 + px + q, we should find p = a + b and q = ab. From the known values of a + b and ab , we should find the values of a and b. Then, the expression is factorised by grouping. Worked-out Examples Example 1: If (a + b) = 5 and ab = 6, find the value of a and b. Solution: Here, ab = 6 = 2 × 3 which is a × b Also, a + b = 2 + 3 = 5 ∴ Either a = 2 and b = 3 or, a = 3 and b = 2 Example 2: If (a + b) = – 8 and ab = 12, find the value of a and b. Solution: Here, ab = 12 = 2 × 2 × 3 = 2 (2 × 3) = 2 × 6 Also, a + b = –8 = –2 – 6 ∴ Either a = –2 and b = –6, or, a = –6 and b = –2 Example 3: Factorise x2 + 7x + 12 Solution: x2 + 7x + 12 = x2 + (3 + 4)x + 12 = x2 + 3x + 4x + 12 12 cm 3 cm 3 cm 12 cm 15 m 6 m 6 m 15 m 18 cm 5 cm 18 cm 5 cm 24 m 10 m 24 m 10m 30 ft 12 ft 30 ft 12 ft 40 m 15 m 15 m 40 m d) e) f) x x b a ax bx x + a x + b ab x2 x x 3 4 x x x x x x x x + 3 x + 4 x Product ab = 12 2 12 = 2 × 2 × 3 = 4 × 3 sum a + b = 7 = 4 + 3 So, x2 + 7x + 12 = x2 + (4 + 3)x + 12 Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 130 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = x (x + 3) + 4 (x + 3) = (x + 3) (x + 4) Example 4: Resolve into factors x2 – 10x + 21 Solution: x2 – 10x + 21= x2 – (3 + 7)x + 21 = x2 – 3x – 7x + 21 = x (x – 3) – 7 (x – 3) = (x – 3) (x – 7) Example 5: Factorise x2 + 9x – 36 Solution: x2 + 9x – 36 = x2 + (12 – 3)x – 36 = x2 + 12x – 3x – 36 = x (x + 12) – 3 (x + 12) = (x + 12) (x – 3) Example 6: Factorise x2 – 7x – 18. Solution: x2 – 7x – 18 = x2 – (9 – 2)x – 18 = x2 – 9x + 2x – 18 = x (x – 9) + 2 (x – 9) = (x – 9) (x + 2) Example 7: Factorise (a + b)2 + 9(a + b) + 20 Solution: Here, (a + b) 2 + 9(a + b) + 20 Let (a + b) = x Then, (a + b) 2 + 9(a + b) + 20 = x2 + 9x + 20 = x2 + (5 + 4)x + 20 = x2 + 5x + 4x + 20 = x(x + 5) + 4(x + 5) = (x + 5) (x + 4) Replacing the value of x, we get, = (a + b + 5) (a + b + 4) ∴ (a + b) 2 + 9(a + b) + 20 = (a + b + 5) (a + b + 4) EXERCISE 9.5 General Section - Classwork 1. Let’s express the following expressions as the product of two factors. a) a(a + 1) + 2(a + 1) = ……….....…… b) x (x – 3) – 2(x – 3) = ……….....…… c) y(5 + y) – 2(5 + y) = ……….....…… d) (x – 2) + x(x – 2) = ……….....…… e) p(p – 7) – (p –7) = ……….....…… f) xy (y + z) – 4(y + z) = ……….....…… x x x x x x x x x x x x 3 x – 7 7 x2 –10x+21 In x2 – 10x + 21 ab = 21 = (–3) × (–7) a + b = (–3 – 7) = – 10 Now, I can factorise! In x2 + 9x – 36, I should express 9 as the sum or difference of any two factors of 36. 36 = 12 × 3 9 = 12 – 3 3 x + 12 x x x – 3 12 x2 +9x–36 x x x + 2 x – 9 9 x2 – 7 x – 18 2 In x2 – 7x – 18 18 = 9 × 2 and 7 = 9 – 2 20 = 1 × 20, but 1 + 20 = 21 20 = 2 × 10, but 2 + 10 = 12 20 = 4 × 5 and 4 + 5 = 9 So, x2 + 9x + 20 = x2 +(4 + 5) x + 20 Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 131 Vedanta Excel in Mathematics - Book 8 2. Let’s study carefully, the following tricky ways of factorisation. x2 + 5x + 6 = (x + 2) (x + 3) x2 – 5x + 6 = (x – 2) (x – 3) x2 + x – 6 = (x – 2) (x + 3) x2 – x – 6 = (x + 2) (x – 3) Now, let’s apply the tricky ways of factorization. Say and write the answers as quickly as possible. a) x2 + 3x + 2 = ……..........……… b) x2 – 3x + 2 = ……..........……… c) x2 + x – 2 = ……..........……… d) x2 – x – 2 = ……..........……… e) x2 + 9x + 20 = ……..........……… f) x2 – 9x + 20 = ……..........……… 3. Let’s write the trinomial expressions from the product of the given factors. a) (x + 1) (x + 2) = x2 + 3x + ................ (x + 1) (x + 2) = x2 + 3x + 2 b) (x + 2) (x + 3) = ……..........……… c) (x + 3) (x + 4) = ……..........……… d) (x + 3) (x – 1) = ……..........……… e) (x – 3) (x + 1) = ……..........……… Creative Section - A 4. Let’s add the area of squares and rectangles. Then, express the area as the product of length and breadth. 2+3 –2+3 –2×3 2×3 – 2–3 2–3 2×(–3) – 2×(–3) 1+2 2–1 -1-2 -2+1 1×2 2×(–1) -1×-2 -2×1 1+2 1×2 D 2 x+2 x+1 x 1.x 1 x2 1.x 1.x 1 1 x C A B a) Area of ABCD = x2 + x + x + x + 1 + 1 (x + 2) (x + 1) = x2 + 3x + 2 c) H x+3 (x+1) G E F x x 3 1 Area of EFGH = ....................................................... ........................... = ....................................................... b) S x+2 (x+2) x2 1.x 1.x 1.x 1.x 1 1 1 1 R P Q x 2 x 2 Area of PQRS = ....................................................... ........................... = ........................................................ Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 132 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) P O M N x 2 x 3 Area of MNOP = ....................................................... ........................... = ....................................................... 5. a) If a + b = 7 and ab = 10, find the value of a and b. b) If a + b = – 9 and ab = 18, find the value of a and b. c) If a + b = 4 and ab = – 32, find the value of a and b. d) If a + b = – 1 and ab = – 30, find the value of a and b. 6. Let's resolve into factors. a) x2 + 3x + 2 b) x2 + 7x + 10 c) x2 + 7x + 12 d) x2 + 8x + 15 e) x2 + 6x + 8 f) x2 + 9x + 20 g) x2 + 9x + 18 h) x2 + 11x + 30 7. Let's factorise. a) x2 – 3x + 2 b) x2 – 5x + 6 c) x2 – 6x + 5 d) x2 – 10x + 24 e) x2 – 10x + 21 f) x2 – 8x + 12 g) x2 – 13x + 36 h) x2 – 13x + 30 8. Let's resolve into factors. a) x2 + x – 2 b) x2 + 2x – 3 c) x2 + 2x – 8 d) x2 + 3x – 18 e) x2 + 5x – 14 f) x2 + 3x – 40 g) x2 + 5x – 84 h) x2 + 4x – 45 9. Resolve into factors. a) x2 – x – 2 b) x2 – x – 6 c) x2 – 2x – 15 d) x2 – 3x – 28 e) x2 – 4x – 12 f) x2 – 3x – 40 g) x2 – 7x – 44 h) x2 – 8x – 65 10. Let's factorise. a) x3 + 15x2 + 56x b) x3 + 17x2 + 42x c) x4 – 4x3 – 21x2 d) x4 – 14x3 + 45x2 e) x3 + 13x2 – 30x f) x4 + 2x3 – 8x2 g) x5 – 18x4 + 72x3 h) x5 + 11x4 – 102x3 11. Let's factorise. a) (x + y) 2 + 5(x + y) + 6 b) (a + b) 2 + 14(a + b) + 33 c) (p – q) 2 – 11(p – q) + 30 d) (x – y) 2 – 9(x – y) – 22 12. a) The area of a rectangle is a2 + 6a + 8 sq. units. Find its length and breadth. Also find the perimeter of the rectangle. b) The area of a rectangular ground is x2 + 5x – 36 sq. unit. If its length and breadth are reduced by 2/2 unit, find the new area of the ground. It’s your time - Project Work and Activity Section 13. Let’s identify the inappropriate term in these expressions of the form x2 + px + q. Then, write the appropriate term and factorise the expressions. a) x2 + (3 + 2)x + 7 b) x2 + (3 – 2)x + 6 c) x2 + (4 + 5) – 20 d) x2 – (5 – 4)x – 20 14. Let’s write any two trinomial expressions of your own in each of the given forms. Then, factorise your trinomial expressions. a) x2 + px + q b) x2 – px – q c) x2 + px – q d) x2 – px + q Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 133 Vedanta Excel in Mathematics - Book 8 (vi) Factorisation of trinomial expressions of the form a2 + 2ab + b2 : Let’s find the product of (a + b) and (a + b) (a + b) (a + b) = a (a + b) + b (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 Thus, (a + b) 2 are the factors of a2 + 2ab + b2 . Similarly, (a – b) 2 are the factors of a2 – 2ab + b2 . So, the expression of the form a2 + 2ab + b2 is factorised by making it a perfect square. (vii)Factorisation of trinomial expressions of the form px2 + qx + r: In the trinomial expressions of the form px2 + qx + r, p and q are the numerical coefficients of x2 and x respectively and r is any constant term. To factorise such expressions, we need to find the two factors a and b of the product of p and r such that a + b = q. Then, the expression is expanded to four terms and factorisation is performed by grouping. Worked-out Examples Example 1: Factorise a2 + 6a + 9 Solution: a2 + 6a + 9= a2 + 2.a.3 + 32 = (a + 3)2 Example 2: Resolve into factors 4x2 – 20x + 25. Solution: 4x2 – 20x + 25 = (2x) 2 – 2.2x.5+52 = (2x – 5)2 Example 3: Resolve into factors 2x2 + 5x + 2. Solution: 2x2 + 5x + 2 = 2x2 + (4 + 1)x + 2 = 2x2 + 4x + x + 2 = 2x (x + 2) + 1 (x + 2) = (x + 2) (2x + 1) Example 4: Factorise 6x2 + x – 2 Solution: 6x2 + x – 2 = 6x2 + (4 – 3)x – 2 = 6x2 + 4x – 3x – 2 = 2x (3x + 2) – 1 (3x + 2) = (3x + 2) (2x – 1) a2 b2 ab ab a a b b b b a2 + 6a + 9 is expressed in the form a2 + 2ab + b2 4x2 – 20x + 25 is expressed in the form a2 – 2ab + b2 x x + 2 2x + 1 x x x x x x x2 x2 x 1 2 In 2x2 + 5x + 2 4 + 1 p × r = 2 × 2 = 4 q = 5 = ( 4 + 1) 2 × 2 = 4 x x2 x2 x x2 x2 x x2 x2 x x x x x x x x x 2 12x – 1 3x + 2 In 6x2 + x – 2 4 – 3 p × r = 6 × 2 = 12 q = 1 = ( 4 – 3) 6 × 2 = 12 Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 134 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 9.6 General Section - Classwork 1. Let’s write the following expressions as the product of their factors. a) 2x(x + 1) + 1(x + 1) = ....................... b) 2x(x – 1) – 1(x – 1) = ....................... c) 3x(2x + 1) + 2(2x + 1) = ....................... d) 3x(2x – 1) – 2(2x – 1) = ....................... e) 2x(3x – 2) + (3x – 2) = ....................... f) 2x(3x + 2) + (3x + 2) = ....................... 2. Let’s rewrite these expressions in the forms of a2 + 2ab + b2 or a2 – 2ab + b2 as shown. x2 + 2x + 1 = x2 + 2.x.1 + 12 9x2 – 30x + 25 = (3x)2 – 2.3x.5 + 52 a) x2 + 4x + 4 = ............................................................................ b) x2 – 8x + 16 = ............................................................................ c) 4x2 + 12x + 9 = ............................................................................ d) 9x2 – 12x + 4 = ............................................................................ Creative Section - A 3. Let's resolve into factors. a) x2 + 6x + 9 b) x2 + 8x + 16 c) x2 + 10x + 25 d) a2 + 12a + 36 e) a2 + 14a + 49 f) a2 + 16a + 64 g) x2 – 4x + 4 h) x2 – 2x +1 i) p2 – 18p + 81 j) 2 – 20p + 100 k) p2 – 24p + 144 l) x2 – 30x + 225 4. Let's factorise. a) 4a2 + 4a + 1 b) 4a2 + 20a + 25 c) 9a2 + 6a + 1 d) 9x2 + 12x + 4 e) 9a2 + 24a + 16 f) 16x2 + 24x + 9 g) 9x2 – 24x + 16 h) 25x2 – 80x + 64 i) 16a2 – 40a + 25 j) 49p2 – 14p + 1 k) 36 – 60p + 25p2 l) 144 – 120x + 25x2 5. Let's resolve into factors. a) 2x2 + 5x + 3 b) 2x2 + 7x + 5 c) 3x2 + 8x + 4 d) 3x2 + 4x + 1 e) 3x2 + 5x + 2 f) 3x2 + 10x + 3 g) 4x2 + 12x + 5 h) 4x2 + 11x + 7 i) 2x2 – 3x + 1 j) 3x2 – 10x + 8 k) 4x2 – 8x + 4 l) 8 – 22x + 9x2 m) 13 – 17x + 4x2 n) 6x2 – 19x + 14 o) 3a2 – 20a + 17 6. Let's factorise. a) 2x2 + x – 3 b) 3x2 + x – 14 c) 4x2 + 5x – 6 d) 5x2 + 13x – 6 e) 6x2 + 11x – 2 f) 7x2 + 25x – 12 g) 4x2 + 4x – 3 h) 6x2 + 11x – 10 i) 2a2 – 5a – 12 j) 3a2 – 5a – 8 k) 4a2 – 16a – 9 l) 2a2 – 7a – 15 m) 8x2 – 22x – 21 n) 6x2 – 11x – 7 o) 5x2 – 13x – 6 7. Let's resolve into factors. a) 2x2 + 5xy + 2y2 b) 3x2 + 5xy + 2y2 c) 2x2 – xy – 3y2 d) 8x2 + 2xy – 15y2 e) 15x2 – 17xy – 4y2 f) 24a2 – 38ab + 15b2 g) 15a2 – 28ab + 12b2 h) 14a2 – 38ab + 20b2 i) 8a2 – 45ab – 18b2 Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 135 Vedanta Excel in Mathematics - Book 8 Creative Section B 8. Let's factorise: a) 25x2 + 2xy + y2 25 b) 49a2 – 2ab + b2 49 c) 4x2 + 2 + 1 4x2 d) 9p2 – 1 + 1 36p2 9. Let's factorise: a) a2 + b2 – c2 – d2 + 2ab – 2cd b) p2 + q2 – r2 – s2 – 2pq – 2rs c) a2 – b2 – 2bc – c2 d) x2 – y2 + 2yz – z2 10. Let's resolve into factors: a) 2(x + y) 2 – 9(x + y) + 10 b) 3(a + b) 2 – 10(a + b) + 8 c) 2(p – q) 2 – 3(p – q) – 9 d) 6(m – n) 2 – 11(m – n) – 7 11. a) The are of a square football ground is (x2 + 8x + 16) sq. units. (i) Find the length of the square ground. (ii) Find the perimeter of the ground. (iii) If x=46m, find the actual area and perimeter of the ground. b) The area of a rectangular garden is (2x2 + 7x + 3) sq. ft. (i) Find the length and breadth of the garden. (ii) If the length is increased by 5 ft. and breadth is decreased by 2 ft., find the new area of the garden. It’s your time - Project work and Activity Section 12. Let’s write two expressions of each of the following forms. Then, factorise the expressions. a) a2 + 2ab + b2 b) a2 – 2ab + b2 [Hint: Find (ax + b) (ax + b) or (cx – d) (cx – d) then factorise the expressions] 13. Let’s write an expression of each of the following forms. Then, factorise: a) px2 + qx + r b) px2 – qx – r c) px2 + qx – r d) px2 – qx + r [Hint: Find (ax + b) (cx – d) or, (ax – b) (cx + d) or, (ax – b) (cx – d), then factorise the expressions.] 9.6 Highest Common Factor (H.C.F) Let’s take any two numbers 8 and 12. Here, all possible factors of 8 are 1, 2, 4, 8 and all possible factors of 12 are 1, 2, 3, 4, 6, 12. The common factors of 8 and 12 are 1, 2, 4 and the highest common factor is 4. Thus, H. C. F. of 8 and 12 is 4. Similarly, let’s consider any two algebraic terms x3 and x4 . All possible factors of x3 are x, x2 , x3 All possible factors of x4 are x, x2 , x3 , and x4 The factors common to x3 and x4 are x, x2 , and x3 Among these common factors, the highest common factor is x3 . So, H.C.F. of x3 and x4 is x3 . Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 136 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Thus, to find the H.C.F. of monomial expressions at first, we should find the H.C.F. of the numerical coefficient. Then, the common variable with the least power is taken as the H.C.F. of the expressions. 9.7 H.C.F of polynomial expressions To find the H.C.F. Of polynomial expression, the given polynomials are to be factorised and a common factor or the product of common factors is obtained as their H.C.F. We can also find the H.C.F. Of polynomial expression by division method. Worked-out Examples Example 1: Find the H.C.F. of 8x4 y3 and 12x3 y5 . Solution: Here, the first expression = 8x4 y3 = 2 × 2 × 2 × x4 y3 The second expression = 12x3 y5 = 2 × 2 × 3 × x3 y5 \ H.C.F. = 2 × 2 × x3 y3 = 4x3 y3 Example 2: Find the H.C.F. of 2p3 + 6p2 pnd 2p3 – 18p. Solution: The first expression = 2p3 + 6p2 = 2p2 (p + 3) = 2p × p (p + 3) The second expression = 2p3 – 18p = 2p (p2 – 9) = 2p (p2 – 32 ) = 2p (p + 3) (p – 3) \ H.C.F. = 2p ( p + 3) Example 3: Find the H.C.F. of 2a2 + 11a + 12, 6a2 + 7a – 3, 4a2 –9. Solution: The first expression = 2a2 + 11a + 12 = 2a2 + (8 + 3) a + 12 = 2a2 + 8a + 3a + 12 = 2a (a + 4) + 3 (a + 4)= (a + 4) (2a + 3) The second expression = 6a2 + 7a – 3 = 6a2 + (9 – 2)a – 3 = 6a2 + 9a – 2a – 3 = 3a (2a + 3) – 1 (2a + 3) = (2a + 3) (3a – 1) The third expression = 4a2 – 9 = (2a) 2 – 32 = (2a + 3) (2a – 3) \ H.C.F. = (2a + 3) EXERCISE 9.7 General Section - Classwork Let’s say and write the H.C.F. of these expressions as quickly as possible. 1. a) H.C.F. of a and a2 is .............. b) H.C.F. of y2 and y3 is ................. c) H.C.F of a3 and a4 is .............. d) H.C.F. of 3y2 and 6y4 is ................ In x4 y3 and x3 y5 the least power of x is x3 , the least power of y is y3 . So, H.C.F. is x3 y3 . Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 137 Vedanta Excel in Mathematics - Book 8 e) H.C.F. of 4x5 and 12x3 is ............ f) H.C.F. of xy2 and x2 y is ................. g) H.C.F. of p2 q2 and pq3 is ........... h) H.C.F. of a3 b2 and a2 b3 is ............... 2. a) H.C.F. of (x + 1) (x + 3) and (x – 2) (x + 3) is ...................... b) H.C.F. of (a + 3) (a – 2) and (a – 3) (a – 2) is ...................... c) H.C.F. of (2p – q) (p + q) and (3p – q) (2p – q) is ................... Creative Section - A 3. Let’s find the H.C.F. of the following monomial expressions: a) 2x2 y, 4xy2 b) 3a2 b3 , 6a3 b2 c) 2pxy3 , 9px3 y d) 8x4 , 12x2 , 16x5 e) 9x3 y2 , 6x2 y3 , 3x4 y4 f) 5a2 bc, 10ab2 c, 15abc2 4. Let’s find the H.C.F. of the following polynomial expressions: a) 2a + 14, 3a + 21 b) a2 + ab, a2 x + abx c) 3x + 9, x2 – 9 d) 2x2 + 4x, 2x2 – 8 e) x2 + 2xy, x3 – 4xy2 f) 6x2 – 9x, 4x2 – 9 g) (a + 2)2 , a2 – 4 h) x3 – x, x2 – x – 2 i) a2 + 6a + 9, a2 – 9 j) 2x2 + x – 3, 4x2 – 9 k) x3 – 5x2 , x2 – 6x + 5 l) 6x2 – 9x, 4x2 – 12x + 9 m) a2 + 7a + 10, a2 + 6a + 8 n) a2 + 5a – 24, a2 – 5a + 6 o) 2x2 – x – 3, 2x2 – 11x + 12 p) 6x2 – 4x – 10, 3x2 – 20x + 25 q) x4 – 1, x2 + x – 2 r) x3 – 10x2 + 25 x, x3 – 3x2 – 10x Creative Section - B 5. Find H.C.F. of the following polynomials: a) 2x + 4, x2 – 4, x2 – x – 6 b) 3x2 – 9x, x3 – 9x, x2 + x – 12 c) x2 – 4, x2 – 4x + 4, x2 + 5x – 14 d) x2 – 1, x2 + 2x – 3, x2 – 3x + 2 e) x2 + 6x + 8, x2 + 9x + 20, x2 + 7x + 10 f) 6x2 + x – 2, 8x2 – 14x + 5, 10x2 – 11x + 3 6. a) Find the greatest algebraic expression that divides 4x3 and 6x4 without leaving a remainder. b) Find the greatest algebraic expression that exactly divides (x2 + 5x + 6) and (x2 + 4x + 3) It’s your time - Project work and Activity Section 7. Let’s write a pair of trinomial expressions of your own in each of the following forms. Then, find the H.C.F. of each pair. a) x2 + bx + c b) x2 – bx – c c) x2 + bx – c d) x2 – bx + c [Hint: Let’s get 1st expression as the product of (x + a) (x + b), a and b are integers.] Let’s get 2nd expression as the product of (x + a) (x + c) or (x + c) (x + b) and so on. 9.8 Lowest Common Multiple (L.C.M) Let’s consider any two algebraic terms a2 and a3 . A few multiples of a2 are a2 , a3 , a4 , a5 , a6 ... Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 138 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur A few multiples of a3 are a3 , a4 , a5 , a6 , a7 , ... The common multiples are a3 , a4 , a5 , a6 , ... Among these common multiples, a3 is the lowest one. So, the Lowest Common Multiple (L.C.M.) of a2 and a3 is a3 . Thus, the common variable with the highest power of the given monomial expression is the L.C.M. of the expressions. In the case of polynomial expressions, their L.C.M. is obtained by the process of factorisation. By this process, the product of common factors and the factors which are not common is taken as the L.C.M. of the polynomials. Worked-out Examples Example 1: Find the L.C.M. of 6x3 y2 and 8x2 y3 . Solution: Here, the first expression = 6x3 y2 = 2 × 3 × x3 y2 The second expression = 8x2 y3 = 2 × 2 × 2 × x2 y3 \ L.C.M. = 2 × 2 × 2 × 3× x3 y3 = 24x3 y3 Example 2: Find the L.C.M. of x2 + 2xy and x3 – 4xy2 . Solution: The first expression = x2 + 2xy = x (x + 2y) The second expression = x3 – 4xy2 = x (x2 – 4y2 ) = x [x2 –(2y) 2 ] = x (x + 2y) (x – 2y) \ L.C.M. = x (x + 2y) (x – 2y) = x (x2 – 4y2 ) Example 3: Find the L.C.M. of 2x2 + 6x, 2x3 – 18x, 2x3 + 16x2 + 30x. Solution: The first expression = 2x2 + 6x = 2x (x + 3) The second expression = 2x3 – 18x = 2x (x2 – 9) = 2x (x + 3) (x – 3) The third expression = 2x3 + 16x2 + 30x = 2x (x2 + 8x + 15) = 2x (x2 + (5 + 3) x + 15) = 2x (x2 + 5x + 3x + 15) = 2x [x (x + 5) + 3 (x + 5)] = 2x (x + 5) ( x+ 3) \ L.C.M. = 2x (x + 3) (x – 3) (x + 5) = 2x (x2 – 9) (x + 5) EXERCISE 9.8 General Section - Classwork 1. Let’s say and write the L.C.M. as quickly as possible. a) L.C.M. of a and a2 is ............ b) L.C.M. of b2 and b3 is ............ c) L.C.M. of x3 and x4 is is ............ d) L.C.M. of 2x2 and 4x3 is ............ e) L.C.M. of ax3 and x5 is ............ f) L.C.M. of x2 y and xy2 is ............ In x3 y2 and x2 y3 the highest power of x is x3 , the highest power of y is y3 . So, L.C.M. is x3 y3 . Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 139 Vedanta Excel in Mathematics - Book 8 g) L.C.M. of x3 y2 and x2 y3 is ............ h) L.C.M. of axy and bx3 y2 is ............ 2. a) L.C.M of (a + 1) (a – 1) and (a – 1) is ............................. b) L.C.M. of (x – 2) (x + 3) and (x – 2) (x – 3) is ............................. c) L.C.M. of a (x + y) (x – y) and b(x – y) is ............................. Creative Section - A 3. Find the L.C.M. of the following monomial expressions. a) 2xy2 , 4x2 y b) 2a2 b2 , 3a3 b3 c) 4x2 y3 , 6x3 y2 d) 2x4 , 3x2 , 6xy4 e) 6x5 y3 ,8x3 y5 ,12xy f) 14xyz3 , 21x2 y3 z, 28x3 yz2 4. Find the L.C.M. of the following polynomial expressions. a) ax2 + ax, a2 x2 + a2 x b) 2x2 + 4x, x3 + 2x2 c) 3a2 b + 6ab2 , 2a3 + 4a2 b d) a2 x + abx, abx2 + b2 x2 e) 3x2 + 6x, 2x3 + 4x2 f) 2a + 4, a2 – 4 g) 3a2 + 3a, 6a2 – 6 h) x2 y – 5xy, x2 – 25 i) a4 x2 – a2 x4 , 7a2 x – 7ax2 j) x2 – xy, x3 y – xy3 k) 4x2 – 2x, 8x3 – 2x l) x2 – 4, x2 + 5x + 6 m) x2 + x – 6, x2 – 9 n) 2x3 – 50x, 2x2 + 7x – 15 o) 4a3 – 9a, 2a2 + 3a – 9 p) x2 + 8x + 15, x2 + 7x + 12 q) a2 – 9a + 20, a2 – 2a – 15 r) a2 + 5a – 14, a2 – 8a + 12 s) 2a2 + 5a + 2, 2a2 – 3a – 2 t) 3x2 + 8x – 16, 3x2 – 16x + 16 u) 4x2 – x – 3, 3x2 – 2x – 1 v) 2x2 + 3x – 9, 4x2 – 12x + 9 Creative Section - B 5. Find the L.C.M. of the following polynomial expressions. a) m2 + 2m, m2 – 4, m2 + 3m + 2 b) 4x – 20, x2 – 25, x2 – 3x – 10 c) (a + b) 2 , a2 – b2 , 2a2 + ab – b2 d) x2 + 4x + 4, x2 – 4, x2 + 8x + 12 e) x2 – 6x + 9, x2 – 9, x2 – 10x + 21 f) x2 + 6x + 8, x2 + x – 12, x2 + 5x + 6 g) a2 + 4a – 5, a2 + 11a + 30, a2 – 5a – 6 h) 2a2 + 5a + 2, 3a2 – 7a + 2, 6a2 + 5a + 1 It’s your time - Project Work and Activity Section 6. a) Let’s write any five pairs of monomial expressions with the same base but different exponents. Then, find the H.C.F. and L.C.M. of each pair. E.g. In 2x2 and 3x3 , H.C.F. = x2 and L.C.M. = 6x3 b) Now, find the product of each pair of expressions and the product of H.C.F. and L.C.M. Verify: Product of two expressions = H.C.F. × L.C.M. E.g. 2x2 × 3x3 = 6x5 and x2 × 6x3 = 6x5 , ∴ Product of two expressions = H.C.F × L.C.M. 7. a) Let’s write any five pairs of monomial expressions with the same base but different exponents. Then, find the H.C.F. of each pair of expressions. b) Now, find the L.C.M. of each pair of expressions using the relation: H.C.F. × L.C.M. = Product of two expressions. Algebraic Expressions

Vedanta Excel in Mathematics - Book 8 140 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 10.1 Rational expressions We have already learned about the rational numbers such as 1 2 , 2 3 , 4 5 , 7 9 , … and so on. We can express a rational number in the form p q where q ≠ 0. Similarly, algebraic expressions such as x 2, 2a 3 , x y, a + b a – b , … and so on are rational expressions. We can also express a rational expression in p q form, where q ≠ 0. However, if the denominator of an expression is 0 (Zero), it is undefined and it cannot be the rational expression. For example, In 2 x , if x = 0, then it is undefined in x + y x – y , if x = y, it is undefined. 10.2 Reduction of rational expressions to their lowest terms We can reduce a monomial rational expression to its lowest terms by using the law of indices. During the process, we simply subtract the smaller index of a variable from the higher index of the same variable. For example, 4x4 y3 6x2 y5 = = 2x4 – 2 3y5 – 3 2x 2 3y2 Alternatively, we can reduce the monomial rational expression to its lowest terms, dividing the numerator and denominator by their H.C.F. For example, In 4x4 y3 6x2 y5 , the H.C.F. of numerator and denominator is 2x2 y3 Now, dividing the numerator and denominator by 2x2 y3 4x4 y3 ¸ 2x2 y3 6x2 y5 ¸ 2x2 y3 2x4 –2y3 – 3 3x2 –2y5 – 3 2x2 y0 3x0 y2 2x2 3y = = = 2 In the case of polynomial numerator and denominator at first, we should factorise them. Then the common factors from the numerator and denominator are cancelled. For example, (x + 1) (x – 1) ax (x + 1) x – 1 ax x2 –1 ax2 +ax = = Worked-out Examples Example 1: For what value of a is 2 a – 3 undefined? Solution: To be 2 a – 3 undefined, a – 3 = 0, i.e. a = 3. \ The required value of a is 3. Unit 10 Rational Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 141 Vedanta Excel in Mathematics - Book 8 EXERCISE 10.1 General Section - Classwork 1. Let’s say and write the answers as quickly as possible. a) For what value of x is 5 x undefined ? ..................... b) For what value of y is 3 y + 2 undefined ? ..................... c) For what value of p, 5 p – 1 becomes undefined? ..................... d) For what value of a, 2 – a 3 + a becomes undefined? ..................... e) For what value of y is a + b x – y undefined? ..................... f) For what value of x is p + q 3x – 6 undefined ? ..................... 2. Let’s say and write the lowest terms of these rational expressions. a) x2 x = ................ b) 2x3 x = ................ c) 6x4 2x2 = ................ d) xy x2 y2 = ................ e) a2 – b2 a + b = ................ f) a2 – b2 a – b = ................ Creative Section 3. Reduce the following rational expressions to their lowest terms. 2x3 4x2 a) 4p5 q5 16p4 q6 f) 6x4 8x5 b) 20p7 q8 15p8 q5 g) 5x2 y 10xy2 c) –14x3 y2 z 21xy2 z2 h) 8a4 b3 12a2 b5 d) 6a2 b3 c4 – 18a3 b2 c3 i) 9ab3 6a2 b5 e) – 18p6 q4 r2 – 24p4 q6 r4 j) Example 2: Reduce 15a5 b4 10a4 b5 to its lowest terms. Solution: 15a5 b4 10a4 b5 3a5 – 4 2b5 – 4 3a 2b = = Example 3: Reduce 4x2 – 25 2x2 – x – 10 to its lowest terms. Solution: am an 1 a =a n–m m–n if m > n, = if m < n am an 4x2 – 25 2x2 – x – 10 (2x) 2 – 52 2x2 – (5 – 4)x – 10 (2x + 5) (2x – 5) 2x2 – 5x + 4x – 10 (2x + 5) (2x – 5) x (2x – 5) + 2(2x – 5) (2x + 5) (2x – 5) (2x – 5) (x +2) 2x + 5 x + 2 = = = = = Rational Expressions

Vedanta Excel in Mathematics - Book 8 142 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 4. Reduce the following rational expressions to their lowest terms. 3x + 6 6x2 + 12x x2 + xy x3 – xy2 2x2 + 5x 4x2 – 25 a2 – 4 a2 – 9a + 14 a2 – b2 a4 – b4 (x – 7)2 x2 – 49 x2 + 9x + 20 x2 + x – 12 2y – 4 4y2 – 8y 3x + 9 x2 – 9 10a2 – 15ab 6ab – 9b2 a3 – 36a a2 + 3a – 18 x + 2 x2 + 5x + 6 (x + 6)2 x2 + 5x – 6 x2 – 5x – 14 2x3 – 13x2 – 7x x2 + 4x x2 – 16 a2 – b2 a – b x – 2y x2 – 4y2 2a3 – 50a a3 + 4a2 – 45a x2 – 3x x2 – 7x + 12 x2 – 8x + 16 x2 – 2x – 8 3x2 – 5x – 12 3x2 + x – 4 a) d) g) m) j) p) s) b) e) h) n) k) q) t) c) f) i) o) l) r) u) 10.3 Multiplication of rational expressions In the case of multiplication of monomial rational expressions, we should multiply the numerators and denominators separately. Then, the product is reduced to the lowest terms by using the laws of indices. For example, 2 3 4x3 y2 6a2 b3 9a3 b2 2x2 y3 3x3 y2 a3 b2 a2 b3 x2 y3 3ax by × = = 3 2x2 + 6x x3 – x2 x2 – 1 x2 – 9 2x (x + 3) x2 (x – 1) 2 (x + 1) x (x – 3) (x + 1) (x – 1) (x + 3 (x –3) × = = × In the case of multiplication of polynomial rational expressions, the numerators and denominators are factorised. Then, they are reduced to their lowest terms. For example, 10.4 Division of rational expressions In the case of division of rational expressions, we should multiply the dividend by the reciprocal of the divisor as like the process of multiplication of rational expressions. For example, = × a2 + 5a + 6 a2 – 1 a2 – 9 a2 – 2a – 3 a2 + 3a + 2a + 6 (a + 1) (a – 1) a2 – 3a + a –3 (a + 3) (a – 3) ¸ = × = = 8x5 y3 9a4 b2 6x4 y2 15a3 b 4 8x5 y3 9a4 b2 20x5 y3 a3 b 9a4 b2 x4 y2 20xy 9ab 5 15a3 b 6x4 y ¸ 2 3 3 Rational Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 143 Vedanta Excel in Mathematics - Book 8 EXERCISE 10.2 General Section - Classwork 1. Let’s simplify mentally, say and write the answers as quickly as possible. a) x y × y2 x2 = .................... b) a2 b2 × b a = .................... c) 2x3 3y2 × 9y3 4x2 = .................... d) p2 q2 ÷ p q = .................... e) x3 y3 ÷ x y = .................... f) 10x5 6y4 ÷ 5x2 3y = .................... Creative Section - A 2. Find the products or quotients. a) 4x2 y3 3x3 y2 × 9x4 8y3 b) 12a2 b3 25x2 y4 × 5x3 y5 6a3 b4 c) 28a3 x3 y2 45p2 b3 × 27p3 b2 7a4 x2 y3 d) 6a4 9b3 ÷ 8a3 3b2 e) 8a5 b3 9x4 y2 ÷ 6a4 b2 15x3 y f) 32x7 y6 z5 27p4 q3 r2 ÷ 8x5 y4 z3 9p3 q2 r 18x3 y4 15a5 b3 25x2 yz 6a2 bc2 5x4 y5 3a6 b4 c b) × ¸ 4a4 b2 6x3 y4 12a3 2 9a4 y5 z 3a2 x2 yc 2b2 y2 z d) ¸ × 7x2 y3 8a5 b4 12ab 21xy 3a2 b2 4x3 y2 c) × ¸ x3 c2 a4 y2 x2 y3 a3 b2 xy4 c2 z2 a) ¸ × a2 – b2 a2 b + ab2 a2 b – ab2 a2 b c) ¸ 2 4x2 – 81y2 1 – 4a2 a – 2a2 2x – 9y g) × a2 + 5a + 6 a2 – 1 a2 – 9 a2 – 2a – 3 h) ¸ x2 +5x + 6 a2 + 3a – 4 x2 – 9 a2 – 16 i) ÷ x2 – a2 ax + a2 2a x – a a) × 2x + 6 x2 – 9 3x2 + 9x 2x2 – 6x b) ¸ x2 – xy x2 + xy x2 (x – y) x3 + x2 y d) ¸ 4a2 – 9b2 x2 – y2 x2 y + xy2 4a – 6b e) × 25x2 – 16y2 x2 – 4 x – 2 5x – 4y f) × 3. Simplify: 4. Simplify: = = × × a (a + 3) + 2 (a + 3) (a + 1) (a – 1) (a + 3) ( a + 2) (a + 1) (a – 1) a (a – 3) + 1 (a – 3) (a + 3) (a – 3) (a – 3) ( a +1) (a + 3) (a – 3) = a + 2 a – 1 Creative Section - B 16x2 y2 – z4 a2 – 4 (a – 2)2 4xy + z2 a2 – 4 (a +2)2 j) × ¸ x2 – 4y2 2x2 – 5xy + 2y2 x2 – xy x + y y2 – xy 2x – y k) × ¸ Rational Expressions

Vedanta Excel in Mathematics - Book 8 144 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur a2 – a – 20 a2 – 25 a2 – a – 2 a2 – 2a – 8 a + 1 a2 + 5a l) × ¸ a2 – 8a – 9 a2 – 17a + 72 a2 – 25 a2 – 1 a2 +4a – 5 a2 – 9a + 8 m) × ¸ 10.5 Addition and subtraction of rational expressions Addition and subtraction of rational expressions with common denominator: In the case of addition and subtraction of rational expressions with a common denominator, we should simply add or subtract the numerators as like the addition and subtraction of like fractions in arithmetic. Then, the sum is reduced to its lowest terms. For example, Addition and subtraction of rational expressions with unlike denominators: In the case of addition or subtraction of the rational expression with different denominators, follow the process given below. (i) Find the L.C.M. of the denominators. (ii) Divide the L.C.M. by each denominator and multiply the quotient so obtained by the corresponding numerator. (iii) Carry out the operation of addition or subtraction in the numerator. (iv) Reduce the sum to its lowest terms. x2 x – y y2 x – y x2 – y2 x – y (x + y) (x – y) x – y – = = = x + y x 9a a2 a2 – b2 b2 a2 – b2 2ab a2 – b2 a2 + b2 – 2ab a2 – b2 (a – b) 2 (a + b) (a – b) (a – a) (a – b) (a + b) (a – b) a – b a + b (i) (ii) 2x 9a 3x 9a x 3a x+2x 9a + + – = = = = = = = Example 2: Simplify (i) 1 x + 1 + 1 x – 1 (ii) x + 2 x – 2 – x – 2 x + 2 Solution: (i) 1 x + 1 + 1 x – 1 = x – 1 + x + 1 (x + 1) (x – 1) = 2x x2 – 1 (ii) x + 2 x – 2 – x – 2 x + 2 = (x + 2)2 – (x – 2)2 (x – 2) (x + 2) = x2 + 2.x.2 + 22 – (x2 – 2.x.2 + 22 ) (x – 2) (x + 2) = x2 + 4x + 4 – x2 + 4x – 4 x2 – 4 = 8x x2 – 4 L.C.M. of (x+1) and (x–1) is (x+1) (x–1) L.C.M. of (x–2) and (x+2) is (x–2) (x+2) Worked-out Examples Example 1: Simplify (i) x 9a 2x 9a + (ii) a2 a2 – b2 b2 a2 – b2 2ab a2 – b + 2 – Solution: Rational Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 145 Vedanta Excel in Mathematics - Book 8 Example 3: Simplify (i) p – 2 p + 2 – p – 2 p2 – 4 (ii) 2 a + 1 + 2a a – 1 – a2 + 3 a2 – 1 Solution: (i) p – 2 p + 2 – p – 2 p2 – 4 = p – 2 p + 2 – p – 2 p2 – 22 = p – 2 p + 2 – (p – 2) (p + 2) (p – 2) = p – 2 p + 2 – 1 p + 2 = p – 2 – 1 (p + 2) = p – 3 p + 2 (ii) 2 a + 1 + 2a a – 1 – a2 + 3 a2 – 1 = 2 a + 1 + 2a a – 1 – a2 + 3 (a + 1) (a – 1) = 2(a – 1) + 2a (a + 1) – (a2 + 3) (a + 1) (a – 1) = 2a – 2+ 2a2 + 2a – a2 – 3 (a + 1) (a – 1) = a2 + 4a – 5 (a + 1) (a – 1) = a2 + 5a – a – 5 (a + 1) (a – 1) = a(a + 5) –1 (a + 5) (a + 1) (a – 1) = (a+ 5) (a – 1) (a + 1) (a – 1) = a +5 a + 1 L.C.M. of (a+1) (a–1) and (a+1) (a–1) is (a+1) (a–1) Example 4: Simplify 2 x2 + 3x + 2 + 5x x2 – x – 6 – x + 2 x2 – 2x – 3 Solution: 2 x2 + 3x + 2 + 5x x2 – x – 6 – x + 2 x2 – 2x – 3 = 2 x + 2x + x+ 2 + 5x x2 – 3x + 2x – 6 – x + 2 x2 – 3x + x – 3 = 2 x(x + 2)+1( x+ 2) + 5x x(x – 3)+2( x – 3) – x + 2 x(x – 3)+1( x – 3) = 2 (x + 2) ( x+ 1) + 5x (x – 3) ( x+ 2) – x + 2 (x – 3) ( x+ 1) = 2(x – 3) + 5x (x + 1) – (x + 2)2 (x + 1) ( x+ 2) (x – 3) = 2x – 6 + 5x2 + 5x – (x2 + 4x + 4) (x + 1) (x + 2) ( x – 3) L.C.M. of (x+2) (x+1), (x–3) (x+2) and (x–3) (x+1) is (x + 1) (x + 2) (x – 3) Rational Expressions

Vedanta Excel in Mathematics - Book 8 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = 5x2 + 7x – 6 – x2 – 4x – 4 (x + 1) (x + 2) ( x – 3) = 4x2 + 3x – 10 (x + 1) (x + 2) ( x – 3) = 4x2 + 8x – 5x – 10 (x + 1) (x + 2) ( x – 3) = 4x(x + 2) – 5 (x + 2) (x + 1) (x + 2) ( x – 3) = (x + 2) (4x – 5) (x + 1) (x + 2) ( x – 3) = 4x – 5 (x + 1) ( x – 3) EXERCISE 10.3 General Section - Classwork 1. Let’s simplify, say and write the answers as quickly as possible. a) a x + 3a x = .................... b) 4x 3a + 2x 3a = .................... c) 12y 5x – 2y 5x = .................... d) 7xy 3ab – xy 3ab = .................... e) x a + b + 3x a + b = .................... f) 2x2 a – b + x2 a – b = .................... g) x x – y – y x – y = .................... h) a2 a + b – b2 a + b = .................... Creative Section - A 2. Let’s simplify: a) x x + y + x x + y b) 3p p – q – 3q p – q c) a + b 2ab + a – b 2ab d) p2 p – q – q2 p – q e) x x2 – y2 – y x2 – y2 f) a a2 –b2 – b a2 –b2 g) x2 x2 – 4 + 4 x2 – 4 – 4x x2 – 4 h) a a2 + 8a + 15 + 3 a2 + 8a + 15 i) x2 x2 + 6x + 8 + 4x x2 + 6x + 8 j) a2 a2 – 5a + 6 – 4a a2 – 5a + 6 + 3 a2 – 5a + 6 3. Let’s simplify: a) 1 x + 1 + 1 x – 1 b) 1 p – 2 – 1 p – 1 c) 1 2a + 1 + 1 2a – 1 d) a2 b a – b – ab2 a + b e) 1 y2 (x – y) – 1 x2 (x – y) f) 1 x – 2 + 4 4 – x2 Rational Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 8 g) 1 x2 – 1 – 1 x – 1 h) 2x2 x2 – 25 – 2x x – 5 i) 1 x (x + a) + 1 x (x – a) j) 2 x – 2 – 8 x2 – 4 k) x (x + y)2 + y x2 – y2 l) 9x2 + y2 9x2 – y2 – 3x – y 3x + y m) 1 2a + 3b – 4a – 6b 4a2 – 9b2 n) 1 (a – b) (c – a) – 1 (a – c) (b – c) o) x + 2 x2 + x – 2 + 3 x2 – 1 p) a – 2 a2 + 4a + 4 – a + 1 a2 – 4 q) a – 2 a2 – 1 – a + 1 a2 – 2a + 1 r) 2 x2 + x – 1 x2 + 3x + 2 Creative Section - B 4. Let’s simplify: a) 1 x – 2 x + 1 + 1 x + 2 b) 1 x + 1 – 1 x – 1 + 3 x2 – 1 c) x x + 2 + x x – 2 – 4x x2 – 4 d) 2 x + 1 + 2x x – 1 – x2 + 3 x2 – 1 e) 3 a – 1 + 4 a + 1 + a + 2 a2 – 1 f) x 2x + 3 + x 2x – 3 – 9 4x2 – 9 g) 2x 2x + 3y + 3y 2x – 3y – 18y2 4x2 – 9y2 h) x2 + y2 xy – x2 y(x + y) – y2 x(x + y) i) 2xy x2 – y2 – x – y x + y + x + y x – y j) x + y x – y – x – y x + y – 4xy x2 + y2 5. a) 1 (x – 3) (x – 4) + 1 (x – 4) (x – 5) + 1 (x – 5) (x – 3) b) 1 (x – 3) (x + 2) – 3 (x + 2) (x – 4) + 2 (x – 4) (x – 3) c) 2 (x – 2) (x – 3) – 2 (x – 1) (x – 3) + 1 (x – 1) (x – 2) d) a (a – b) (a – c) + b (b – c) (b – a) + c (c – a) (c – b) e) 2 x2 – 5x + 6 – 2 x2 – 4x + 3 + 1 x2 – 3x + 2 f) x – 1 x2 – 3x + 2 + x – 2 x2 – 5x + 6 + x – 5 x2 – 8x + 15 g) y + 3 y2 + 6y + 9 + 6 y2 – 9 – 16y 8y2 – 24y Rational Expressions

Vedanta Excel in Mathematics - Book 8 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 11.1 Equation - Looking back Classwork - Exercise 1. Let’s say and write the equations and find the values of the variables. a) The sum of x and 3 is 7, equation is ......................... and x = .................. b) The difference of x and 2 is 5, equation is ......................... and x = .................. c) The product of 4 and y is 12, equation is ......................... and y = .................. d) The quotient of p divided by 2 is 3, equation is ......................... and p = .................. 2. Let’s say and write the values of variables as quickly as possible. a) x + 3 = 9, x = ............... b) y – 1 = 5, y = .................. c) a 5 = 2, a = .................. d) 3p = 15, p = ................... 11.2 Linear equations with one variable Let’s take an equation x + 5 = 9 It is an equation with only one variable which is x and the power of the variable x is 1. Such an equation is called a linear equation with one variable. Therefore, a first degree equation in one variable is always a linear equation. 11.3 Solution of linear equation Let’s consider an equation, x + 3 = 8. This equation can be true only for a fixed value of x which is 5. So, 5 is called the solution (or root) to the equation. The process of getting a solution to an equation is called solving equation. Unit 11 Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 8 Worked-out Examples Example 1: Solve a) 3 (x – 4) = 2 (x – 1) b) 1 y – 1 + 2 y + 1 = 3 y Solution: a) 3 (x – 4) = 2 (x – 1) or, 3x – 12 = 2x – 2 or, 3x – 2x = – 2 + 12 or, x = 10 Example 2 : If the sum of three consecutive odd numbers is 27. Find the numbers. Solution : Let the first odd number be x. Then, the second consecutive odd number = x + 2 And, the third consecutive odd number = x + 4 Now, x + (x + 2) + (x + 4) = 27 or, 3x + 6 = 27 or, 3x = 27 – 6 or, x = 21 3 = 7 The first odd number, x = 7 The second odd number = x + 2 = 7 + 2 = 9 The third odd number = x + 4 = 7 + 4 = 11 So, the required consecutive odd numbers are 7, 9, and 11. Example 3: A sum of Rs 80 is divided into two parts. If two times the greater part is 10 less three times the smaller part, find the parts of the sum. Solution : Let the greater part of the sum be Rs x. Then, the smaller part of the sum = Rs (80 – x) Now, 2x = 3 (80 – x) – 10 or, 2x = 240 – 3x – 10 or, 5x = 230 or, x = 46 The greater part of the sum = Rs x = Rs 46 The smaller part of the sum = Rs (80 – x) = Rs (80 – 46) = Rs 34. So, the required parts of the sum are Rs 46 and Rs 34. b) 1 y – 1 + 2 y+1 = 3 y or, y + 1 + 2y – 2 (y – 1) (y + 1) = 3 y or, 3y – 1 y2 – 1 = 3 y or, 3y2 – y = 3y2 – 3 or, y = 3 Checking the answers: Here, the required odd numbers are 7, 9 and 11. Now, 7 + 9 + 11= 27 which is given in the question. Checking the answer: Here, the required parts are Rs 46 and Rs 34. 2 × Rs 46= Rs 92 3 × Rs 34= Rs 102 Rs 102 – Rs 92 = Rs 10 which is given in the question. Also, Rs 46 + Rs 34 = Rs 80 Equation and Graph

Vedanta Excel in Mathematics - Book 8 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 11.1 General Section 1. Let’s say and write the value of x as quickly as possible. a) x + 5 = 9, x = .......................... b) x – 5 = 3, x = .......................... c) 3x = 12, x = .......................... d) x 4 = 5, x = .......................... e) 2 x = 1 3, x = .......................... f) x – 1 3 = 2, x = .......................... 2. Let’s say and write the answers as quickly as possible. a) The sum of x and 5 is 12, then x = ...................... b) The difference of y and 4 is 5, then y = ...................... c) The difference of 15 and x is 5, then x = ...................... d) The product of 7 and x is 21, then x = ...................... e) The quotient of x divided by 4 is 6, then x = ...................... f) The quotient of 18 divided by x is 3, then x = ...................... 3. Let x be the unknown number. Let’s say and write the required equation as quickly as possible. Then, find the value of x. a) The sum of two numbers is 12 and one of them is 4. The equation is ............................................................. then x = ......................... b) The difference of two numbers is 3 and the greater one is 9. The equation is ............................................................. then x = ......................... c) The difference of two numbers is 6 and the smaller one is 10. The equation is ............................................................. then x = ......................... d) The product of two number is 30 and one of them is 5. The equation is ............................................................. then x = ......................... e) The quotient of dividing a number by 3 is 7. The equation is ............................................................. then x = ......................... f) The quotient of dividing 24 by a number is 6. The equation is ............................................................. then x = ......................... Equation and Graph