197 (d) when, x = 3 when, x = 8 tn = arn – 1 = 4 – 1 2 n – 1 tn = arn – 1 = 9 – 1 3 n – 1 7. (i) The 6 times of the fifth term of a GP is 7500. If its common ratio is 5 then find its first term. (ii) The 4 times of the fourth term is equal to the one-fourth of the sixth term. If the first term is 1 512 , find its common ratio. Solution: (i) Here, r = 6, 6t5 = 7500 Now, 6t5 = 6 × ar5 – 1 or, 7500 = 6 × a × 54 or, 7500 = 6 × a × 625 or, 7500 3750 = a a = 2 (ii) Here, a = 1 512 and 4t4 = 1 4 t6 or, 4 × ar4 – 1 = 1 4 × ar6 – 1 or, 16r3 = r5 or, 16 = r2 r = ± 4 8. (i) The third term and the seventh term of a GS are 4 and 64 respectively. (a) Find the first term of the GS. (b) Find common ratio of the GS. (ii) The fifth term and the ninth term of a GP are 8 and 2048 respectively. (a) Find the first term of the GS. (b) Find the common ratio of the GS. (c) Find the eleventh term of the GS. Solution: (i) Here, t3 = 4 and t7 = 64 (b) we have, tn = arn – 1 or, t3 = ar3 – 1 or, 4 = ar2 .......... (i) and t6 = ar7 – 1 or, 64 = ar6 ........ (ii) Dividing equation (ii) by (i), ar6 ar2 = 64 4 or, r4 = 16 r = 2 (a) putting the value of r in equation (i), a × 22 = 4 or, 4a = 4 a = 1 (ii) Here, t5 = 8 and t9 = 2048 (b) we have, tn = arn – 1 or, t5 = ar5 – 1 or, 8 = r4 .............. (i) and t9 = ar9 – 1 or, 2048 = ar8 ...... (ii)
198 Dividing equation (ii) by (i), ar8 ar4 = 2048 8 or, r4 = 256 or, r4 = 44 r = 4 (a) putting the value of r in equation (i) 8 = a × 44 or, 8 = a × 256 or, 8 256 = a a = 1 32 (c) t11 = ar11 – 1 = 1 32 × 410 = 1 32 × 1048576 = 32768 9. (i) If the nth terms of the geometric sequences 1 32 , 1 8 , 1 2 , ... and 4, 8, 16,… are equal, find the number of terms. (ii) The nth terms of the geometric sequences 1, 3, 9, ... and 1 6561 , 1 729 , 1 81 , ... are equal. Find the value of n. Solution: (i) In sequence 1 32 , 1 8 , 1 2 , ... a = 1 32 r = 1 8 1 32 = 4 and tn = arn – 1 = 1 32 × 4n – 1 In sequence, 4, 8, 16, ... a = 4 r = 8 4 = 2 and tn = arn – 1 = 4 × 2n – 1 Now, 1 32 × 4n – 1 = 4 × 2n – 1 or, 4n – 1 2n – 1 = 128 or, 2n – 1 = 128 or, 2n – 1 = 27 then, n – 1 = 7 n = 8 (ii) In sequence 1, 3, 9, ... a = 1 r = 3 1 = 3 And, tn = arn – 1 = 1 × 3n – 1 = 3n – 1
199 In sequence, 1 6561 , 1 729 , 1 81 a = 1 6561 r = 1 729 1 6561 = 9 And, tn = arn – 1 = 1 6561 × 9n – 1 Now, 3n – 1 = 1 6561 × 9n – 1 or, 6561 = 9n – 1 3n – 1 = 9 3 n – 1 or, 38 = 3n – 1 then, 8 = n – 1 n = 9 10. (i) Three successive numbers are in a GP and its product is 512. Their sum is 42. (ii) The sum of three consecutive terms in GP is 62 and their product is 1000. (a) Suppose the three consecutive numbers in AP. (b) Find these three numbers. Solution: (i) Let, a r , a and ar are three term of GS then, a r × a × ar = 512 or, a3 = 83 a = 8 Again, a r + a + ar = 42 or, a 1 r + 1 + r = 42 or, 8 1 + r + r2 r = 42 or, 4 + 4r + 4r2 = 21r or, 4r2 – 17r + 4 = 0 or, 4r2 – 16r – r = 4 = 0 or, 4r(r – 4) – 1(r – 4) = 0 or, (r – 4) (4r – 1) = 0 r = 1 4 , 4 If r = 1 4 If r = 4 a r = 8 1 4 = 32 a r = 8 4 = 2 a = 8 a = 8 ar = 8 × 1 4 = 2 ar = 8 × 4 = 32 Hence, required term of GP are 32, 8, 2 or 2, 8, 32.
200 (ii) Let, a r , a and ar in GP then, a r × a × ar = 1000 or, a3 = 103 a = 10 Also, a r + a + ar = 62 or, a 1 r + 1 + r = 62 or, 10 1 + r + r2 r = 62 or, 5 + 5r + 5r2 = 31r or, 5r2 – 26r + 5 = 0 or, (r – 5) (5r – 1) = 0 r = 1 5 , 5 If r = 1 5 If r = 5 a r = 10 1 5 = 50 a r = 10 5 = 2 a = 10 a = 10 ar = 10 × 1 5 = 2 ar = 10 × 5 = 50 Hence, terms of GP are 50, 10, 2 or 2, 10, 50. 11. (i) In a geometric sequence, the fifth term is 8 times of the second term and the fourth term is 24. (a) Find the first term of the GS. (b) Find common ratio of the GS. (c) Find the geometric series. (ii) Three numbers are in the ratio of 1:4:13. If 1 is added to all three numbers then they form a GP. Find the original numbers. Solution: (i) (b) Here, t5 = 8t2 or, ar5 – 1 = 8 × ar2 – 1 or, r4 = 8r or, r3 = 23 r = 2 (a) Again, t4 = 24 or, ar4 – 1 = 24 or, a × 23 = 24 or, a = 24 8 a = 3 (c) the terms of GS a + ar + ar2 + ar3 + ... 3 + 6 + 12 + 24 + ...
201 (ii) Let, x, 4x and 13x are numbers in 1 : 4 : 13 Now, x + 1, 4x + 1 and 13x + 1 are in GP So, 4x + 1 x + 1 = 13x + 1 4x + 1 or, 16x2 + 8x + 1 = 13x2 + 14x + 1 or, 3x2 – 6x = 0 or, 3x(x – 2) = 0 or, x – 2 = 0 x = 2 Hence, x = 2 4x = 8 13x = 26 are required numbers. 12. (i) A rich man saves Re. 1 in the first day, Rs. 2 in the second day, Rs. 4 in the third day, Rs. 8 in the fourth day and so on. How much money does he save in the 30th day? (ii) A Ping-Pong ball is dropped from a height of 20 meter on to a horizontal roof of a house and always reduces 1 8 of the distance of the previous fall in each bounce. What height does it fall at 7th time? Solution: (i) Here, 1, 2, 4, 8, ... is sequence of saving a = 1 r = 2 1 = 2 n = 30 we have, tn = arn – 1 or, t30 = 1 × 230 – 1 = 229 = 536870912 Hence, A man saves 536870912 in 30th day. (ii) Here, a = 20 r = 1 – 1 8 = 7 8 Now, tn = arn – 1 = 20 × 7 8 7 – 1 = 20 × 76 86 = 20 × 117649 262144 = 8.98 Hence, the height bounced by ball at 7th time = 8.98 m
202 CHAPTER FACTORIZATION 9 9.1 Factorization in the Form of (a ± b)3 and a3 ± b3 PRACTICE 9.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) What is factorization of algebraic expression? (b) What are the factors of a2 – b2 ? (c) Write down the formula for (a + b)2 in expanded form. (d) What are the factors of a3 – b3 ? (e) What are the factors of a4 – b4 ? Solution: (a) The process of finding factors from the given algebraic expression is called its factorization. (b) The factors of a2 – b2 are (a + b) and (a – b). i.e. a2 – b2 = (a + b) (a – b) (c) (a + b)2 = a2 + 2ab + b2 (d) a3 – b3 = (a – b) (a2 + ab + b2 ) (e) a4 – b4 = (a + b) (a – b) (a2 + b2 ) 2. Find the square of the following expressions by using formula. (a) (a + 3) (b) (2 + b) (c) (4 – x) (d) (2x – 3y) (e) (4m + 3) (f) m – 1 2m (g) xy + 1 x (h) a2 – 1 3a2 (i) p2 + 1 5pq2 Solution: (a) (a + 3)2 = a2 + 2.a.3 + 32 = a2 + 6a + 9 (b) (2 + b)2 = 22 + 2.2.b + b2 = 4 + 4b + b2 (c) (4 – x)2 = 42 – 2.4.x + x2 = 16 – 8x + x2 (d) (2x – 3y)2 = (2x)2 – 2.2x.3y + (3y)2 = 4x2 – 12xy + 9y2 (e) (4m + 3)2 = (4m)2 + 2.4m.3 + 32 = 16m2 + 24m + 9 (f) m – 1 2m 2 = m2 – 2.m. 1 2m + 1 2m 2 = m2 – 1 + 1 4m2 (g) xy + 1 x 2 = (xy)2 + 2.xy. 1 x + 1 x 2 = x2 y2 + 2y + 1 x2 (h) a 2 – 1 3a2 2 = (a2 ) 2 – 2.a2 . 1 3a2 + 1 3a2 2 = a4 – 2 3 + 1 9a4 (i) p2 + 1 5pq2 2 = (p2 ) 2 + 2.p2 . 1 5pq2 + 1 5pq2 2 = p4 + 2p q2 + 1 25p2 q4
203 3. Factorize by using formulae: (a) x2 – y2 (b) 9x2 – y2 (c) 4a2 – 9 (d) m2 – 1 m2 (e) a2 – 1 4a2 (f) 16p2 – 1 64a2 (g) p4 – q4 (h) 4a4 – 9b4 (i) 16x4 – 81y4 (j) 4x4 + y4 (k) 64a4 + 1 (l) 36a4 + 9 Solution: (a) x2 – y2 = (x)2 – (y)2 = (x + y) (x – y) (b) 9x2 – y2 = (3x)2 – (y)2 = (3x + y) (3x – y) (c) 4a2 – 9 = (2a)2 – 32 = (2a + 3) (2a – 3) (d) m2 – 1 m2 = m2 – 1 m 2 = m + 1 m m – 1 m (e) a 2 – 1 4a2 = a2 – 1 2a 2 = a + 1 2a a – 1 2a (f) 16p2 – 1 64a2 = (4p)2 – 1 8a 2 = 4p + 1 8a 4p – 1 8a (g) p4 – q4 = (p2 ) 2 – (q2 ) 2 = (p2 + q2 ) (p2 – q2 ) = (p2 + q2 ) (p + q) (p – q) (h) 4a4 – 9b4 = (2a2 ) 2 – (3b2 ) 2 = (2a2 + 3b2 ) (2a2 – 3b2 ) (i) 16x4 – 81y4 = (4x2 ) 2 – (9y2 ) 2 = (4x2 + 9y2 ) (4x2 – 9y2 ) = (4x2 + 9y2 ) {(2x)2 – (3y)2 } = (4x2 + 9y2 ) (2x + 3y) (2x – 3y) (j) 4x4 + y4 = (2x2 ) 2 + (y2 ) 2 = (2x2 + y2 ) 2 – 2.2x2 .y2 = (2x2 + y2 ) 2 – 4x2 y2 = (2x2 + y2 ) 2 – (2xy)2 = (2x2 + y2 + 2xy) (2x2 + y2 – 2xy) = (2x2 + 2xy + y2 ) (2x2 – xy + y2 ) (k) 64a4 + 1 = (8a2 ) 2 + 12 = (8a2 + 1)2 – 2.8a2 .1 = (8a2 + 1) – 16a2 = (8a2 + 1)2 – (4a)2 = (8a2 + 1 + 4a) (8a2 + 1 – 4a) = (8a2 + 4a + 1) (8a2 – 4a + 1) (l) 36a4 + 9 = 9(4a4 + 1) = 9{(2a2 ) 2 + 12 } = 9{(2a2 + 1)2 – 2.2a2 .1} = 9{(2a2 + 1)2 – 4a2 } = 9{(2a2 + 1)2 – (2a)2 } = 9(2a2 + 1 + 2a) (2a2 + 1 – 2a) = 9(2a2 + 2a + 1) (2a2 – 2a + 1) Check Your Performance 4. Find the cube of the following expressions by using formula. (a) (x + 2) (b) (3 + y) (c) (2 – y) (d) (3x – 2y) (e) (5m + 2n) (f) 2y – 1 2y Solution: (a) (x + 2)3 = x3 + 3.x2 .2 + 3.x.22 + 23 = x3 + 6x2 + 12x + 8 (b) (3 + y)3 = 33 + 3.32 .y + 3.3.y2 + y3 = 27 + 27y + 9y2 + y3 (c) (2 – y)3 = 23 – 3.22 .y + 3.2.y2 – y3 = 8 – 12y + 6y2 – y3 (d) (3x – 2y)3 = (3x)3 – 3.(3x)2 .2y + 3.3x.(2y)2 + (2y)3 = 27x3 – 54x2 y + 36xy2 + 8y3 (e) (5m + 2n)3 = (5m)3 + 3.(5m)2 .2n + 3.5m.(2n)2 + (2n)3 = 125m3 + 150m2 n + 60mn2 + 8n3 (f) 2y – 1 2y 3 = (2y)3 – 3.(2y)2 . 1 2y + 3.2y. 1 2y 2 – 1 2y 3 = 8y3 – 6y + 3 2y2 – 1 8y3
204 5. Expand: (a) (2x + y)3 (b) (x – 3y)3 (c) (2m – n)3 (d) (2p – 3q)3 (e) x + 1 x 3 (f) 3m + 1 3m 3 Solution: (a) (2x + y)3 = (2x)3 + 3.(2x)2 .y + 3.2x.y2 + y3 = 8x3 + 12x2 y + 6xy2 + y3 (b) (x – 3y)3 = x3 – 3.x2 .3y + 3.x.(3y)2 + (3y)3 = x3 – 9x2 y + 27xy2 + 27y3 (c) (2m – n)3 = (2m)3 – 3.(2m)2 .n + 3.2m.n2 – n3 = 8m3 – 12m2 n + 6mn2 – n3 (d) (2p – 3q)3 = (2p)3 – 3.(2p)2 .3q + 3.2p.(3q)2 – (3q)3 = 8p3 – 36p2 q + 54pq2 – 27a3 (e) x + 1 x 3 = x3 + 3.x2 . 1 x + 3.x. 1 x 2 + 1 x 3 = x2 + 3x + 3 x + 1 x3 (f) 3m + 1 3m 3 = (3m)3 + 3.(3m)2 . 1 3m + 3.3m. 1 3m 2 + 1 3m 3 = 27m3 + 9m + 1 m + 1 27m3 6. Express as the square of binomial. (a) x2 + 2x + 1 (b) y2 + 2y + 1 (c) 4x2 – 4x + 1 (d) 4m2 – 16mn + 16n2 (e) 9x2 + 12xy + 4y2 (f) 16m2 – 72mn + 81n2 (g) y2 + 2 + 1 y (h) 4m2 – 2 + 1 4m2 Solution: (a) x2 + 2x + 1 = x2 + 2.x.1 + 12 = (x + 1)2 (b) y2 + 2y + 1 = y2 + 2.y.1 + 12 = (y + 1)2 (c) 4x2 – 4x + 1 = (2x)2 – 2 . 2x . 1 + 12 = (2x – 1)2 (d) 4m2 – 16mn + 16n2 = 4(m2 – 4mn + 4n2 ) = 4{m2 – 2.m.2n + (2n)2 } = 4(m – 2n)2 = {2(m – 2n)}2 = (2m – 4n)2 (e) 9x2 + 12xy + 4y2 = (3x)2 + 2.3x.2y + (2y)2 = (3x + 2y)2 (f) 16m2 – 72mn + 81n2 = (4m)2 – 2.4m.9n + (9n)2 = (4m – 9n)2 (g) y2 + 2 + 1 y = y2 + 2.y.1 y + 1 y 2 = y + 1 y 2 (h) 4m2 – 2 + 1 4m2 = (2m)2 – 2.2m. 1 2m + 1 2m 2 = 2m – 1 2m 2 7. Express as the cube of binomial. (a) a3 + 3a2 + 3a + 1 (b) 8x3 + 12a2 + 6a + 1 (c) 27m3 – 54m2 n + 36mn2 – 8n3 (d) 8p3 – 36p2 q + 54pq2 – 27q3 Solution: (a) a3 + 3a2 + 3a + 1 = a3 + 3.a2 .1 + 3.a.1 + 13 = (a + 1)3 (b) 8x3 + 12a2 + 6a + 1 = (2x)3 + 3.(2x)2 .1 + 3.2x.12 + 13 = (2x + 1)3 (c) 27m3 – 54m2 n + 36mn2 – 8n3 = (3m)3 – 3.(3m)2 .2n + 3.3m.(2n)2 – (2n)3 = (3m – 2n)3 (d) 8p3 – 36p2 q + 54pq2 – 27q3 = (2p)3 – 3.(2p)2 .3q + 3.2p.(3q)2 – (3q)3 = (2p – 3q)3 8. Factorize: (a) (x + 2y)2 (b) (3p + 2q)3 (c) 125a3 + 75a2 b + 15ab2 + b3 (d) 343x3 – 294x2 y + 84xy2 – 8y3 Solution: (a) (x + 2y)2 = (x + 2y) (x + 2y) (b) (3p + 2q)3 = (3p + 2q) (3p + 2q) (3p + 2q)
205 (c) 125a3 + 75a2 b + 15ab2 + b3 = (5a)3 + 3.(5a)2 .b + 3.5a.(b)2 + b3 = (5a + b)3 = (5a + b) (5a + b) (5a + b) (d) 343x3 – 294x2 y + 84xy2 – 8y3 = (7x)3 – 3.(7x)2 .2y + 3.7x.(2y)2 – (2y)3 = (7x – y)3 = (7x – 2y) (7x – 2y) (7x – 2y) 9. Simplify: (a) (a + 1) (a – 2)2 (b) (3x + 1) (2x – 1)2 (c) (2a + 5) (a – 4)2 (d) –6x (2x + 9) (2x – 9) (e) (2y + 5) (3y + 2) (2y – 5) (f) (3x + 2y) (3x – 2y) (9x2 – 4y2 ) Solution: (a) (a + 1) (a – 2)2 = (a + 1) (a2 – 4a + 4) = a(a2 – 4a + 4) + 1(a2 – 4a + 4) = a3 – 4a2 + 4a + a2 – 4a + 4 = a3 – 3a2 + 4 (b) (3x + 1) (2x – 1)2 = (3x + 1) (4x2 – 4x + 1) = 3x(4x2 – 4x + 1) + 1(4x2 – 4x + 1) = 12x3 – 12x2 + 3x + 4x2 – 4x + 1 = 12x3 – 8x2 – x + 1 (c) (2a + 5) (a – 4)2 = (2a + 5) (a2 – 8a + 16) = 2a(a2 – 8a + 16) + 5(a2 – 8a + 16) = 2a3 – 16a2 + 32a + 5a2 – 40a + 80 = 2a3 – 11a2 – 8a + 80 (d) –6x (2x + 9) (2x – 9) = – 6x[(2x)2 – 92 ] = – 6x(4x2 – 81) = – 24x3 + 486x (e) (2y + 5) (3y + 2) (2y – 5) = {(2y)2 – 52 } (3y + 2) = (4y2 – 25) (3y + 2) = 4y2 (3y + 2) – 25(3y + 2) = 12y3 + 8y2 – 75y – 50 (f) (3x + 2y) (3x – 2y) (9x2 – 4y2 ) = (3x)2 – (2y)2 (9x2 – 4y2 ) = (9x2 – 4y2 ) (9x2 – 4y2 ) = (9x2 – 4y2 ) 2 = (9x2 ) 2 – 2.9x2 .4y2 + (4y2 ) 2 = 81x4 – 72x2 y2 + 16y4 10. Find the volume of the following cuboids. (a) (b) (c) Solution: (a) Here, = b = h = (x + 2y) Volume of cube = 3 = (x + 2y)3 = x3 + 3.x2 .2y + 3.x.(2y)2 + (2y)3 = x3 + 6x2 y + 12xy2 + 8y3 (b) Here, = b = h = 2x – y Volume of cube = 3 = (2x – y)3 = (2x)3 – 3.(2x)2 .y + 3.2x.y2 – y3 = 8x3 – 12x2 y + 6xy2 – y3 (c) Here, = b = h = 3x – 1 Volume of cube = 3 = (3x – 1)3 = (3x)3 – 3.(3x)2 .1 + 3.3x.12 – 13 = 9x3 – 27x2 + 9x – 1
206 11. Find the square root of : (a) 4x2 + 4x + 1 (b) 9p2 + 24pq + 16q2 (c) 4x2 – 2 + 1 4x2 (d) 1 a2 – 1 ab + 1 4b2 (e) 16 25 a2 + 2ab + 25 16 b2 (f) y2 + 2 + 1 y2 Solution: (a) 4x2 + 4x + 1 = (2x)2 + 2.2x.1 + 12 ) = (2x + 1)2 = (2x + 1) (b) 9p2 + 24pq + 16q2 = (3p)2 + 2.3p.4q + (4q)2 = (3p + 4q)2 = (3p + 4q) (c) 4x2 – 2 + 1 4x2 = (2x)2 – 2.2x. 1 2x + 1 2x 2 = 2x – 1 2x 2 = 2x – 1 2x (d) 1 a 2 – 1 ab + 1 4b2 = 1 a 2 – 2 . 1 a . 1 2b + 1 2b 2 = 1 a – 1 2b 2 = 1 a – 1 2b (e) 16a2 25 + 2ab + 25b2 16 = 4a 5 2 + 2 . 4a 5 . 5b 4 + 5b 4 2 = 4a 5 + 5b 4 2 = 4a 5 + 5b 4 (f) y2 + 2 + 1 y2 = y2 + 2.y. 1 y + 1 y 2 = y + 1 y 2 = y + 1 y 9.2 Factorization in Form of (a4 + a2 b2 + b4 ) PRACTICE 9.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Write down the factors of x4 + 4y4 . (b) What are the factors of x4 + x2 y2 + y4 ? Solution: (a) Here, x4 + 4y4 = (x2 ) 2 + (2y2 ) 2 =(x2 + 2y2 ) 2 – 2.x2 .2y2 = (x2 + 2y2 ) 2 – (2xy)2 = (x2 + 2y2 + 2xy)(x2 + 2y2 – 2xy) = (x2 + 2xy + 2y2 ) (x2 – 2xy + 2y2 ). Hence, the factors of x4 + 4y4 are (x2 + 2xy + 2y2 ) (x2 – 2xy + 2y2 ). (b) Here, x4 + x2 y2 + y4 = (x2 ) 2 + (y2 ) 2 + x2 y2 = (x2 + y2 ) 2 + x2 y2 = (x2 + y2 ) 2 – 2x2 y2 + x2 y2 = (x2 + y2 ) 2 – (xy)2 = (x2 + y2 + xy) (x2 + y2 – xy). Hence, the factors of x4 + x2 y2 + y4 are (x2 + xy + y2 ) (x2 – xy + y2 ). Check Your Performance 2. Factorize: (a) a4 – 1 (b) a4 – 16 (c) x6 – y4 (d) (x + y)4 – (x – y)4 (e) 4y5 + 81y (f) (x + y)4 + 4(x – y)4 Solution: (a) a4 – 1 = (a2 ) 2 – 12 = (a2 + 1) (a2 – 1) = (a2 + 1) (a + 1) (a – 1) (b) a4 – 16 = (a2 ) 2 – 42 = (a2 + 4) (a2 – 22 ) = (a2 + 4) (a + 2) (a – 2) (c) x6 – y4 = (x3 ) 2 – (y2 ) 2 = (x3 + y2 ) (x3 – y2 ) (d) (x + y)4 – (x – y)4 = {(x + y)2 }2 – {(x – y)2 }2 = (x2 + 2xy + y)2 – (x2 – 2xy + y2 ) 2 = (x2 + 2xy + y2 + x2 – 2xy + y2 ) (x2 + 2xy + y2 – x2 + 2xy – y2 )
207 = (2x2 + 2y2 ) 4xy = 2.4xy(x2 + y2 ) = 8xy(x2 + y2 ) (e) 4y5 + 81y = y(4y4 + 81) = y[(2y2 ) 2 + 92 ] = y[(2y2 + 9)2 – 2.2y2 .9] = y[(2y2 + 9)2 – (6y)2 ] = y(2y2 + 6y + 9) (2y2 – 6y + 9) (f) Here, (x + y)4 + 4(x – y)4 Let, x + y = a and x – y = b Then, a4 + 4b4 = (a2 ) 2 + (2b2 ) 2 = (a2 + 2b2 ) 2 – 2.a2 .2b2 = (a2 + 2b2 ) 2 – (2ab)2 = (a2 + 2ab + 2b2 ) (a2 – 2ab + 2b2 ) = (x + y)2 + 2(x + y) (x – y) + 2(x – y)2 }{(x + y)2 – 2(x + y) (x – y) + 2(x – y)2 } = (x2 + 2xy + y2 + 2x2 – 2y2 + 2x2 – 4xy + 2y2 ) (x2 + 2xy + y2 – 2x2 + 2y2 + 2x2 – 4xy + 2y2 ) = (5x2 – 2xy + y2 ) (x2 – 2xy + 5y2 ) 3. Resolve into factors: (a) 1 + b2 + b4 (b) x4 + 3x2 + 4 (c) 4x4 + 3x2 + 1 (d) a4 – 31a2 + 9 (e) x8 + x4 y4 + y8 (f) x8 + 9x4 + 81 Solution: (a) 1 + b2 + b4 = 12 + (b2 ) 2 + b2 = (1 + b2 ) 2 – 2.1.b2 + b2 = (1 + b2 ) 2 – b2 = (1 + b2 + b) (1 + b2 – b) = (1 + b + b2 ) (1 – b + b2 ) (b) x4 + 3x2 + 4 = (x2 ) 2 + 22 + 3x2 = (x2 + 2)2 – 2.x2 .2 + 3x2 = (x2 + 2)2 – x2 = (x2 + x + 2) (x2 – x + 2) (c) 4x4 + 3x2 + 1 = (2x2 ) 2 + 12 + 3x2 = (2x2 + 1)2 – 2.2x2 .1 + 3x2 = (2x2 + 1)2 – x2 = (2x2 + x + 1) (2x2 – x + 1) (d) a4 – 31a2 + 9 = (a2 ) 2 + 32 – 31a2 = (a2 – 3)2 + 2.a2 .3 – 31a2 = (a2 – 3)2 – (5a)2 = (a2 – 3 + 5a) (a2 – 3 – 5a) = (a2 + 5a – 3) (a2 – 5a – 3) (e) x8 + x4 y4 + y8 = (x4 ) 2 + (y4 ) 2 + x4 y4 = (x4 + y4 ) 2 – 2.x4 .y4 + x4 y4 = (x4 + y4 ) 2 – (x2 y2 ) 2 = (x4 + y4 + x2 y2 ) (x4 + y4 – x2 y2 ) = [(x2 ) 2 + (y2 ) 2 + x2 y2 ] (x4 – x2 y2 + y4 ) = [(x2 + y2 ) 2 – 2x2 y2 + x2 y2 ] (x4 – x2 y2 + y4 ) = [(x2 + y2 ) 2 – (xy)2 ] (x4 – x2 y2 + y4 ) = (x2 + xy + y2 ) (x2 – xy + y2 ) (x4 – x2 y2 + y4 ) (f) x8 + 9x4 + 81 = (x4 ) 2 + 92 + 9x4 = (x4 + 9)2 – 2.x4 .9 + 9x4 = (x4 + 9)2 – (3x2 ) 2 = (x4 + 3x2 + 9) (x4 – 3x2 + 9) 4. Factorize: (a) a4 + 5a2 + 4 (b) a4 – 5a2 + 4 (c) x4 + 7x2 – 18 (d) a4 – 5a2 – 6 (e) 3x4 + 7x2 + 4 (f) 4a4 – 5a2 + 1 Solution: (a) a4 + 5a2 + 4 = a4 + a2 (4 + 1) + 4 = a4 + 4a2 + a2 + 4 = a2 (a2 + 4) + 1(a2 + 4) = (a2 + 4) (a2 + 1) (b) a4 – 5a2 + 4 = a4 – a2 (4 + 1) + 4 = a4 – 4a2 – a2 + 4 = a2 (a2 – 4) – 1(a2 – 4) = (a2 – 4) (a2 – 1) = (a + 2) (a – 2) (a + 1) (a – 1) (c) x4 + 7x2 – 18 = x4 + x2 (9 – 2) – 18 = x4 + 9x2 – 2x2 – 18 = x2 (x2 + 9) – 2(x2 + 9) = (x2 + 9) (x2 – 2) (d) a4 – 5a2 – 6 = a4 – a2 (6 – 1) – 6 = a4 – 6a2 + a2 – 6 = a2 (a2 – 6) + 1(a2 – 6) = (a2 – 6) (a2 + 1)
208 (e) 3x4 + 7x2 + 4 = 3x4 + x2 (4 + 3) + 4 = 3x4 + 4x2 + 3x2 + 4 = x2 (3x2 + 4) + 1(3x2 + 4) = (3x2 + 4) (x2 + 1) (f) 4a4 – 5a2 + 1 = 4a4 – a2 (4 + 1) + 1 = 4a4 – 4a2 – a2 + 1 = 4a2 (a2 – 1) – 1(a2 – 1) = (a2 – 1) (4a2 – 1) = (a + 1) (a – 1) (2a + 1) (2a – 1) 5. Resolve into factors: (a) x2 + 2xy + y2 – a2 + 2ab – b2 (b) x2 + 4xy + 4y2 – 1 + 2b – b2 (c) a2 – 4a + 3 – b2 + 2b (d) x2 – 10x + 24 + 6y – 9y2 (e) (a + b)4 + 4(a + b)2 (a – b)2 + 4(a – b)4 – 4(a4 + 4a2 b2 + 4b4 ) Solution: (a) x2 + 2xy + y2 – a2 + 2ab – b2 = (x2 + 2xy + y2 ) – (a2 – 2ab + b2 ) = (x + y)2 – (a – b)2 = (x + y + a – b) (x + y – a + b) (b) x2 + 4xy + 4y2 – 1 + 2b – b2 = (x2 + 4xy + 4y)2 – (1 – 2b + b2 ) = (x + 2y)2 – (1 – b)2 = (x + 2y + 1 – b) (x + 2y – 1 + b) (c) a2 – 4a + 3 – b2 + 2b = a2 – 2.a.2 + 22 – 1 – b2 + 2b = (a – 2)2 – 1 – b2 + 2b = (a – 2)2 – (1 – 2b + b2 ) = (a – 2)2 – (1 – b)2 = (a – 2 + 1 – b) (a – 2 – 1 + b) = (a – b – 1) (a + b – 3) (d) x2 – 10x + 24 + 6y – 9y2 = x2 – 2.x.5 + 52 – 52 + 24 + 6y – 9y2 = (x – 5)2 – 1 + 6y – 9y2 = (x – 5)2 – (1 – 6y + 9y2 ) = (x – 5)2 – (1 – 3y)2 = (x – 5 + 1 – 3y) (x – 5 – 1 + 3y) = (x – 3y – 4) (x + 3y – 6) (e) (a + b)4 + 4(a + b)2 (a – b)2 + 4(a – b)4 – 4(a4 + 4a2 b2 + 4b4 ) = [(a + b)2 + {2(a – b)}2 ] 2 – 4(a2 + 2b2 ) 2 = (a2 + 2ab + b2 + 2a2 – 4ab + 2b2 ) 2 – (2a2 + 4b2 ) 2 = (3a2 – 2ab + 3b2 ) 2 – (2a2 + 4b2 ) 2 = (3a2 – 2ab + 3b2 + 2a2 + 4b2 ) (3a2 – 2ab + 3b2 – 2a2 – 4b2 ) = (5a2 – 2ab + 7b2 ) (a2 – 2ab – b2 ) 6. Factorize: (a) a6 + a4 + a2 (b) 2a4 b + 2a2 b3 + 2b5 (c) x5 y – 39x3 y + 25xy (d) 2x6 – 86x4 + 18x2 (e) 98a5 – 92a3 b2 + 18ab4 (f) 12a5 b + 105a3 b + 363ab (g) x2 y2 + 1 + y2 x2 (h) x4 + 1 + 1 x4 (i) a4 b4 + 1 + a2 b2 (j) 4p4 q4 + q4 p4 (k) a4 b4 – 3 + 9b4 a4 () x4 – 57x2 + 256 Solution: (a) a6 + a4 + a2 = a2 (a4 + a2 + 1) = a2 [(a2 ) 2 + 12 + a2 ] = a2 [(a2 + 1)2 – 2.a2 .1 + a2 ] = a2 [(a2 + 1)2 – a2 ] = a2 (a2 + a + 1) (a2 – a + 1) (b) 2a4 b + 2a2 b3 + 2b5 = 2b(a4 + a2 b2 + b4 ) = 2b[(a2 ) 2 + (b2 ) 2 + a2 b2 ] = 2b(a2 + b2 ) 2 – 2.a2 .b2 + a2 b2 = 2b[(a2 + b2 ) 2 – (ab)2 ] = 2b(a2 + ab + b2 ) (a2 – ab + b2 ) (c) x5 y – 39x3 y + 25xy = xy(x4 – 39x2 + 25) = xy[(x2 ) 2 + 52 – 39x2 ] = xy[(x2 + 5)2 – 2.x2 .5 – 39x2 ] = xy[(x2 + 5)2 – (7x)2 ] = xy(x2 + 7x + 5) (x2 – 7x + 5)
209 (d) 2x6 – 86x4 + 18x2 = 2x2 (x4 – 43x2 + 9) = 2x2 [(x2 ) 2 + 32 – 43x2 ] = 2x2 [(x2 + 3)2 – 2.x2 .3 – 43x2 ] = 2x2 [(x2 + 3)2 – (7x)2 ] = 2x2 (x2 + 7x + 3) (x2 – 7x + 3) (e) 98a5 – 92a3 b2 + 18ab4 = 2a(49a4 – 46a2 b2 + 9b4 ) = 2a[(7a2 – 3b2 ) 2 + 2.7a2 .3b2 – 46a2 b2 ] = 2a[(7a2 – 3b2 ) 2 – (2ab)2 ] = 2a(7a2 + 2ab – 3b2 ) (7a2 – 2ab – 3b2 ) (f) 12a5 b + 105a3 b + 363ab = 3ab(4a4 + 35a2 + 121) = 3ab[(2a2 + 11)2 – 2.2a2 .11 + 35a2 ] = 3ab[(2a2 + 11)2 – (3a)2 ] = 3ab(2a2 + 11 + 3a) (2a2 + 11 – 3a) (g) x2 y2 + 1 + y2 x2 = x y + y x 2 – 2 . x y . y x + 1 = x y + y x 2 – 12 = x y + 1 + y x x y – 1 + y x (h) x4 + 1 + 1 x4 = x2 + 1 x2 2 – 2 . x2 . 1 x2 + 1 = x2 + 1 x2 2 – 12 = x2 + 1 x2 + 1 x2 + 1 x2 – 1 (i) a 4 b4 + 1 + a 2 b2 = a 2 b2 + 1 2 – 2 . a 2 b2 . 1 + a 2 b2 = a 2 b2 + 1 2 – a b 2 = a 2 b2 + 1 + a b a 2 b2 + 1 – a b (j) 4p4 q4 + q4 p4 = 2p2 q2 2 – q2 p2 2 = 2p2 q2 + q2 p2 2 – 2. 2p2 q2 . q2 p2 = 2p2 q2 + q2 p2 2 – 22 = 2p2 q2 + 2 + q2 p2 2p2 q2 – 2 + q2 p2 (k) a 4 b4 – 3 + 9b4 a 4 = a 2 b2 + 3b2 a 2 2 – 2 . a 2 b2 . 3b2 a 2 – 3 = a 2 b2 + 3b2 a 2 2 – 32 = a 2 b2 + 3 + 3b2 a 2 a 2 b2 – 3 + 3b2 a 2 () x4 – 57x2 + 256 = (x2 – 16)2 + 2.x2 .16 – 57x2 = (x2 – 16)2 – (5x)2 = (x2 + 5x + 16) (x2 – 5x + 16) 7. Factorize: (a) 25p4 + 36q4 – 49r4 – 64s4 + 60p2 q2 – 112r2 s 2 (b) a4 – a3 + a2 b2 – 2a2 b – 2ab2 – b3 + b4 Solution: (a) Here, 25p4 + 36q4 – 49r4 – 64s4 + 60p2 q2 – 112r2 s 2 = 25p4 + 60p2 q2 + 36q4 – 49r4 – 112r2 s 2 – 64s4 = (5p2 + 6q2 ) 2 – (7r2 – 8s2 ) 2 = (5p2 + 6q2 + 7r2 – 8s2 ) (5p2 + 6q2 – 7r2 + 8s2 ) (b) Here, a4 – a3 + a2 b2 – 2a2 b – 2ab2 – b3 + b4 = (a2 ) 2 + (b2 ) 2 + a2 b2 – (a3 + b3 + 2a2 b + 2ab2 ) = (a2 + b2 ) 2 – 2a2 b2 + a2 b2 – [(a + b) (a2 – ab + b2 ) + 2ab(a + b)] = (a2 + b2 ) 2 – (ab)2 – (a + b) (a2 – ab + b2 + 2ab) = (a2 + ab + b2 ) (a2 – ab + b2 ) – (a + b) (a2 + ab + b2 ) = (a2 + ab + b2 ) (a2 – ab + b2 – a – b) = (a2 + ab + b2 ) (a2 – a – ab – b + b2 )
210 CHAPTER HCF AND LCM 10 10.1 Highest Common Factor (HCF) PRACTICE 10.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define HCF of algebraic expressions. (b) If there is no common factor in the given algebraic expressions, what is the HCF? (c) Write the HCF of the three expressions having the same algebraic factors. (d) What is the HCF of (3x + y) (3x – 2) and (2x + y) (3x – 2)? Solution: (a) The common factor in highest degree that divides the given algebraic expressions, is called Highest Common Factor (HCF). (b) If there is no common factor in given algebraic expression, the HCF is 1. (c) HCF of the three expressions having the same algebraic factors = common factor (d) Here, 1st expression = (3x + y) (3x – 2) 2nd expression = (2x + y) (3x – 2) then, HCF = 3x – 2 Check Your Performance 2. Find the Highest Common Factor (HCF) of: (a) a2 + ab and ab + b2 (b) a2 – b2 and a2 – ab (c) a3 – 3a2 b and a2 – 9b2 (d) (x – 2)3 and x3 – 8 (e) 3ax(a – x)3 and 2a2 x(a – x)2 (f) 4x(x2 – y2 ) and 12x2 y(x2 + 3xy + 2y2 ) Solution: (a) Here, a2 + ab and ab + b2 1st expression = a2 + ab = a(a + b) 2nd expression = ab + b2 = b(a + b) HCF = a + b (b) Here, a2 – b2 and a2 – ab 1st expression = a2 – b2 = (a + b) (a – b) 2nd expression = a2 – ab = a(a – b) HCF = a – b (c) Here, a3 – 3a2 b and a2 – 9b2 1st expression = a3 – 3a2 b = a2 (a – 3b) 2nd expression = a2 – 9b2 = a2 – (3b)2 = (a + 3b) (a – 3b) HCF = a – 3b
211 (d) Here, (x – 2)3 and x3 – 8 1st expression = (a – 2)3 = (x – 2) (x – 2)2 2nd expression = x3 – 8 = x3 – 23 = (x – 2) (x2 + 2x + 4) HCF = x – 2 (e) Here, 3ax(a – x)3 and 2a2 x(a – x)2 1st expression = 3ax(a – x)3 = 3ax(a – x)2 (a – x) 2nd expression = 2a2 x(a – x)2 = 2a.ax(a – x)2 HCF = ax(a – x)2 (f) Here, 4x(x2 – y2 ) and 12x2 y(x2 + 3xy + 2y2 ) 1st expression = 4x(x2 – y2 ) = 4x(x + y) (x – y) 2nd expression = 12x2 y(x2 + 3xy + 2y2 ) = 12x2 y(x2 + 2xy + xy + 2y2 ) = 12x2 y{x(x + 2y) + y(x + 2y)} = 3xy × 4x(x + y) (x + 2y) HCF = 4x(x + y) 3. Find the Highest Common Factor (HCF) of: (a) x4 – x and x3 + x2 + x (b) a2 – 9 and a2 – 6a + 9 (c) a3 – b3 and a2 – b2 (d) 10x2 + 7x + 1 and 20x3 + 9x2 + x (e) x2 – 5x + 6 and x2 – 4 (f) a2 + 2ab + b2 – c2 and a2 – b2 + ac – bc Solution: (a) Here, x4 – x and x3 + x2 + x 1st expression = x4 – x = x(x3 – 1) = x(x3 – 13 ) = x(x – 1) (x2 + x + 1) 2nd expression = x3 + x2 + x = x(x2 + x + 1) HCF = x2 + x + 1 (b) Here, a2 – 9 and a2 – 6a + 9 1st expression = a2 – 9 = a2 – 32 = (a + 3) (a – 3) 2nd expression = a2 – 6a + 9 = a2 – 3a – 3a + 9 = a(a – 3) – 3(a – 3) = (a – 3) (a – 3) HCF = a – 3 (c) Here, a3 – b3 and a2 – b2 1st expression = a3 – b3 = (a – b) (a2 + ab + b2 ) 2nd expression = a2 – b2 = (a + b) (a – b) HCF = a – b (d) Here, 10x2 + 7x + 1 and 20x3 + 9x2 + x 1st expression = 10x2 + 7x + 1 = 10x2 + 5x + 2x + 1 = 5x(2x + 1) + 1(2x + 1) = (2x + 1) (5x + 1) 2nd expression = 20x3 + 9x2 + x = x(20x2 + 9x + 1) = x(20x2 + 5x + 4x + 1) = x{5x(4x + 1) + 1(4x + 1)} = x(4x + 1) (5x + 1) HCF = 5x + 1 (e) Here, x2 – 5x + 6 and x2 – 4 1st expression = x2 – 5x + 6 = x2 – 3x – 2x + 6 = x(3 – x) – 2(x – 3) = (x – 3) (x – 2) 2nd expression = x2 – 4 = x2 – 22 = (x + 2) (x – 2) HCF = x – 2 (f) Here, a2 + 2ab + b2 – c2 and a2 – b2 + ac – bc 1st expression = a2 + 2ab + b2 – c2 = (a + b)2 – c2 = (a + b + c) (a + b – c) 2nd expression = a2 – b2 + ac – bc = (a + b) (a – b) + c(a – b) = (a – b) (a + b + c) HCF = a + b + c
212 4. Find the Highest Common Factor (HCF) of: (a) 1 + 4x + 4x2 – 16x4 and 1 + 2x – 8x3 – 16x4 (b) a4 + a2 b2 + b4 and a3 + 2a2 b + 2ab2 + b3 (c) a7 – 1 a5 and a4 + 1 + 1 a4 (d) x4 + x2 + 169 and x3 + x(x + 13) + 4x2 (e) x2 + (b – a)x – ab and x2 – (a – b)x – ab (f) x2 + (p – q)x – pq and x2 + x – q2 – q Solution: (a) Here, 1 + 4x + 4x2 – 16x4 and 1 + 2x – 8x3 – 16x4 1st expression = 1 + 4x + 4x2 – 16x4 = (1 + 2x)2 – (4x2 ) 2 = (1 + 2x – 4x2 ) (1 + 2x + 4x2 ) 2nd expression = 1 + 2x – 8x3 – 16x4 = 1(1 + 2x) – 8x3 (1 + 2x) = (1 + 2x) (1 – 8x3 ) = (1 + 2x) {13 – (2x)3 } = (1+2x)(1 – 2x)(1 + 2x + 4x2 ) HCF = 1 + 2x + 4x2 (b) Here, a4 + a2 b2 + b4 and a3 + 2a2 b + 2ab2 + b3 1st expression = a4 + a2 b2 + b4 = a4 + 2a2 b2 + b4 – a2 b2 = (a2 + b2 ) 2 – (ab)2 = (a2 + b2 – ab) (a2 + b2 + ab) = (a2 – ab + b2 ) (a2 + ab + b2 ) 2nd expression = a3 + 2a2 b + 2ab2 + b3 = a3 + b3 + 2a2 b + 2ab2 = (a + b) (a2 – ab + b2 ) + 2ab(a + b) = (a + b) (a2 – ab + b2 + 2ab) = (a + b) (a2 + ab + b2 ) HCF = a2 + ab + b2 (c) Here, a7 – 1 a 5 and a4 + 1 + 1 a 4 1st expression = a7 – 1 a 5 = a a 6 – 1 a 6 = a (a2 ) 3 – 1 a 2 3 = a a 2 – 1 a 2 a 4 + a2 . 1 a 2 + 1 a 4 = a a 2 – 1 a 2 a 4 + 1 + 1 a 4 2nd expression = a4 + 1 + 1 a 4 HCF = a4 + 1 + 1 a 4 (d) Here, x4 + x2 + 169 and x3 + x(x + 13) + 4x2 1st expression = x4 + x2 + 169 = (x2 ) 2 + 26x2 + (13)2 – 25x2 = (x2 + 13)2 – (5x)2 = (x2 + 13 – 5x) (x2 + 13 + 5x) = (x2 – 5x + 13) (x2 + 5x + 13) 2nd expression = x3 + x(x + 13) + 4x2 = x(x2 + x + 13 + 4x) = x(x2 + 5x + 13) HCF = x2 + 5x + 13 (e) Here, x2 + (b – a)x – ab and x2 – (a – b)x – ab 1st expression = x2 + (b – a)x – ab = x2 + bx – ax – ab = x(x + b) – a(x + b) = (x – a) (x + b) 2nd expression = x2 – (a – b)x – ab = x2 – ax + bx – ab = x(x – a) + b(x – b) = (x – a) (x + b) HCF = (x – a) (x + b) (f) Here, x2 + (p – q)x – pq and x2 + x – q2 – q 1st expression = x2 + (p – q)x – pq = x2 + px – qx – pq = x(x + p) – q(x + p) = (x + p) (x – q) 2nd expression = x2 + x – q2 – q = x2 – q2 + x – q = (x – q) (x + q) + 1(x – q) = (x – q) (x + q + 1) HCF = x – q
213 5. Find the Highest Common Factor (HCF) of: (a) 2a4 – 32, a2 – 5a + 6 and a2 + 2a – 8 (b) a2 – a – 2, a2 + a – 6 and 3a2 – 13a + 14 (c) 3x3 + 6x2 – 9x, 9x3 – 9x, 5x3 – 5x (d) a4 + a2 b2 + b4 , a3 + b3 and a3 – a2 b + ab2 (e) x3 + 1 + 2x + x2 , x3 – 1 and x4 + x2 + 1 (f) 2x2 + 5x + 2, 3x2 + 8x + 4 and 2x2 + 3x – 2 Solution: (a) Here, 2a4 – 32, a2 – 5a + 6 and a2 + 2a – 8 1st expression = 2a4 – 32 = 2(a4 – 16) = 2{(a2 ) 2 – 42 } = 2(a2 + 4) (a2 – 4) = 2(a2 + 4) (a2 – 22 ) = 2(a2 + 4) (a + 2) (a – 2) 2nd expression = a2 – 5a + 6 = a2 – 2a – 3a + 6 = a(a – 2) – 3(a – 2) = (a – 2) (a – 3) 3rd expression = a2 + 2a – 8 = a2 + 4a – 2a – 8 = a(a + 4) – 2(a + 4) = (a + 4) (a – 4) HCF = a – 2 (b) Here, a2 – a – 2, a2 + a – 6 and 3a2 – 13a + 14 1st expression = a2 – a – 2 = a2 – 2a + a – 2 = a(a – 2) + 1(a – 2) = (a – 2) (a + 1) 2nd expression = a2 + a – 6 = a2 + 3a – 2a – 6 = a(a + 3) – 2(a + 3) = (a + 3) (a – 2) 3rd expression = 3a2 – 13a + 14 = 3a2 – 6a – 7a + 14 = 3a(a – 2) – 7(a – 2) = (a – 2) (3a – 7) HCF = a – 2 (c) Here, 3x3 + 6x2 – 9x, 9x3 – 9x, 5x3 – 5x 1st expression = 3x3 + 6x2 – 9x = 3x(x2 + 2x – 3) = 3x(x2 + 3x – x – 3) = 3x{x(x + 3) – 1(x + 3)} = 3x(x + 3) (x – 1) 2nd expression = 9x3 – 9x = 9x(x2 – 1) = 9x(x – 1) (x + 1) 3rd expression = 5x3 – 5x = 5x(x2 – 1) = 5x(x – 1) (x + 1) HCF = x(x – 1) (d) Here, a4 + a2 b2 + b4 , a3 + b3 and a3 – a2 b + ab2 1st expression = a4 + a2 b2 + b4 = a4 + 2a2 b2 + b4 – a2 b2 = (a2 + b2 ) 2 – (ab)2 = (a2 + b2 – ab) (a2 + b2 + ab) = (a2 – ab + b2 ) (a2 + ab + b2 ) 2nd expression = a3 + b3 = (a + b) (a2 – ab + b2 ) 3rd expression = a3 – a2 b + ab2 = a3 – a2 b + ab2 = a(a2 – ab + b2 ) HCF = a2 – ab + b2 (e) Here, x3 + 1 + 2x + x2 , x3 – 1 and x4 + x2 + 1 1st expression = x3 + 1 + 2x + x2 = (x + 1) (x2 – x + 1) + 2x(x + 1) = (x + 1) (x2 – x + 1 + 2x) = (x + 1) (x2 + x + 1) 2nd expression = x3 – 1 = (x – 1) (x2 + x + 1) 3rd expression = x4 + x2 + 1 = (x2 ) 2 + 2x2 + 1 – x2 = (x2 + 1)2 – x2 = (x2 + 1 – x) (x2 + 1 + x) = (x2 – x + 1) (x2 + x + 1) HCF = x2 + x + 1 (f) Here, 2x2 + 5x + 2, 3x2 + 8x + 4 and 2x2 + 3x – 2 1st expression = 2x2 + 5x + 2 = 2x2 + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (x + 2) (2x + 1) 2nd expression = 3x2 + 8x + 4 = 3x2 + 6x + 2x + 4 = 3x(x + 2) + 2(x + 4) = (x + 2) (3x + 2) 3rd expression = 2x2 + 3x – 2 = 2x2 + 4x – x – 2 = 2x(x + 2) – 1(x + 2) = (x + 2) (2x – 1) HCF = x + 2
214 10.2 Lowest Common Multiple (LCM) PRACTICE 10.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) What is the LCM of algebraic expressions ? (b) If there is no common factor in the given algebraic expressions, what is the LCM ? (c) Write the relation between HCF, LCM, first expression and second expression. (d) What is the LCM of (2b + a) (3a – 1) and (2b + a) (3a – 1) ? Solution: (a) LCM is common multiple of lowest degree which is divisible by each given algebraic expression. (b) If there is no common factor in the given algebraic expression, the LCM is product of all algebraic expression. (c) HCF × LCM = first expression × second expression. (d) 1st expression = (2b + a) (3a – 1) 2nd expression = (2b + a) (3a – 1) LCM = (2b + (a) (3a – 1) Check Your Performance 2. Find the Lowest Common Multiples (LCM) of: (a) x – 4 and (x – 4) (x + 4) (b) a2 – 4 and a2 + 2a (c) x2 – 4x + 4 and x2 – 4 (d) x2 – y2 and (x – y)2 (e) (x – 2)2 and x3 – 8 (f) 3ab(a – b)3 and 2a2 b(a –b)2 Solution: (a) Here, x – 4 and (x – 4) (x + 4) 1st expression = x – 4 2nd expression = (x – 4) (x + 4) LCM = (x – 4) (x + 4) (b) Here, a2 – 4 and a2 + 2a 1st expression = a2 + 2a = a(a + 2) 2nd expression = a2 – 4 = a2 – 22 = (a + 2) (a – 2) LCM = a(a + 2) (a – 2) (c) Here, x2 – 4x + 4 and x2 – 4 1st expression = x2 – 4x + 4 = (x – 2)2 2nd expression = x2 – 4 = x2 – 22 = (x – 2) (x + 2) LCM = (x – 2) (x – 2) (x + 2) (d) Here, x2 – y2 and (x – y)2 1st expression = x2 – y2 = (x + y) (x – y) 2nd expression = (x – y)2 (x – y) (x – y) LCM = (x – y) (x + y) (x – y)
215 (e) Here, (x – 2)2 and x3 – 8 1st expression = (x – 2)2 = (x – 2) (x – 2) 2nd expression = x3 – 8 = x3 – 23 = (x – 2) (x2 + 2x + 4) LCM = (x – 2) (x – 2) (x2 + 2x + 4) (f) Here, 3ab(a – b)3 and 2a2 b(a –b)2 1st expression = 3ab(a – b)3 = 3ab(a –b)2 (a – b) 2nd expression = 2a2 b(a –b)2 = 2a.ab(a – b)2 LCM = ab(a – b)2 . 3(a – b)2a = 6a2 b(a – 3)3 3. Find the lowest common multiples (LCM) of: (a) 6x2 + 19x + 15 and 9x2 + 9x – 10 (b) a2 – a – 2 and a2 + a – 6 (c) x4 + 4 and 2x3 – 4x2 +4x (d) x4 + 2px3 + p2 x2 and x3 – p2 x Solution: (a) Here, 6x2 + 19x + 15 and 9x2 + 9x – 10 1st expression = 6x2 + 19x + 15 = 6x2 + 9x + 10x + 15 = 3x(2x + 3) + 5(2x + 3) = (2x + 3) (3x + 5) 2nd expression = 9x2 + 9x – 10 = 9x2 + 15x – 6x – 10 = 3x(3x + 5) – 2(3x + 5) = (3x + 5) (3x – 2) LCM = (3x + 5) (2x + 3) (3x – 2) (b) Here, a2 – a – 2 and a2 + a – 6 1st expression = a2 – a – 2 = a2 – 2a + a – 2 = a(a – 2) + 1(a – 2) = (a – 2) (a + 1) 2nd expression = a2 + a – 6 = a2 + 3a – 2a – 6 = a(a + 3) – 2(a + 3) = (a + 3) (a – 2) LCM = (a – 2) (a + 1) (a + 3) (c) Here, x4 + 4 and 2x3 – 4x2 +4x 1st expression = x4 + 4 = (x2 ) 2 + 4x2 + 4 – 4x2 = (x2 + 2)2 – (2x)2 = (x2 + 2 – 2x) (x2 + 2 + 2x) = (x2 – 2x + 2) (x2 + 2x + 2) 2nd expression = 2x3 – 4x2 +4x = 2x(x2 – 2x + 2) LCM = 2x(x2 – 2x + 2) (x2 + 2x + 2) (d) Here, x4 + 2px3 + p2 x2 and x3 – p2 x 1st expression = x4 + 2px3 + p2 x2 = x2 (x2 + 2px + p2 ) = x2 (x + p)2 = x2 (x + p) (x + p) 2nd expression = x3 – p2 x = x(x2 – p2 ) = x(x – p) (x + p) LCM = x(x + p) . x(x + p) (x – p) = x2 (x + p) (x + p) (x – p) 4. Find the lowest common multiples (LCM) of: (a) 24(y3 – x3 ), 32x2 (y2 – x2 ) and 48(xy3 – x3 y2 ) (b) x2 + x – 6, x2 + 4x + 3 and x2 – x – 2 (c) a3 – b3 , a2 – b2 and a4 – b4 (d) x2 + x + 1, x6 – 1 and x3 – 1 Solution: (a) Here, 24(y3 – x3 ), 32x2 (y2 – x2 ) and 48(xy3 – x3 y2 ) 1st expression = 24(y3 – x3 ) = 8 × 3(y – x) (y2 + yx + x2 ) 2nd expression = 32x2 (y2 – x2 ) = 8 × 2x × 2x(y – x) (y + x)
216 3rd expression = 48(xy3 – x3 y2 ) = 8 × 2 × 3xy2 (y2 – 2) = 8 × 2 × 3xy2 (y – x) (y + x) LCM = 8 × 2 × 3 × 2x × x × y2 (y – x) (y + x) (y2 + yx + x2 ) = 96x2 y2 (y – x) (y + x) (y2 + yx + x2 ) (b) Here, x2 + x – 6, x2 + 4x + 3 and x2 – x – 2 1st expression = x2 + x – 6 = x2 + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x + 3) (x – 2) 2nd expression = x2 + 4x + 3 = x2 + x + 3x + 3 = x(x + 1) + 3(x + 1) = (x + 1) (x + 3) 3rd expression = x2 – x – 2 = x2 – 2x + x – 2 = x(x – 2) + 1(x – 2) = (x – 2) (x + 1) LCM = (x – 2) (x + 3) (x + 1) (c) Here, a3 – b3 , a2 – b2 and a4 – b4 1st expression = a3 – b3 = (a – b) (a2 + ab + b2 ) 2nd expression = a2 – b2 = (a – b) (a + b) 3rd expression = a4 – b4 = (a2 – b2 ) (a2 + b2 ) = (a – b) (a + b) (a2 + b2 ) LCM = (a – b) (a + b) (a2 + b2 ) (a2 + ab + b2 ) (d) Here, x2 + x + 1, x6 – 1 and x3 – 1 1st expression = x2 + x + 1 2nd expression = x6 – 1 = (x3 ) 2 – 12 = (x3 – 1) (x3 + 1) = (x – 1) (x2 + x + 1) (x + 1) (x2 – x + 1) 3rd expression = x3 – 1 = (x – 1) (x2 + x + 1) LCM = (x2 + x + 1) (x – 1) (x + 1) (x2 – x + 1) 5. (a) Find the LCM of the expressions a2 – 12a + 35 and a2 – 8a + 7 with the help of their HCF. (b) m2 – 5m – 14 is an expression. Find out another similar expression such that their HCF is (m – 7) and LCM is m3 – 10m2 + 11m + 70. Solution: (a) Here, 1st expression = a2 – 12a + 35 = a2 – 7a – 5a + 35 = a(a – 7) – 5(a – 7) = (a – 7) (a – 5) 2nd expression = a2 – 8a + 7 = a2 – 7a – a + 7 = a(a – 7) – 1(a – 7) = (a – 7) (a – 1) HCF = (a – 7) Remaining factors = (a – 5) (a – 1) Hence, LCM = HCF × Remaining factors = (a – 7) (a – 5) (a – 1) (b) Here, 1st expression = m2 – 5m – 14 = m2 – 7m + 2m – 14 = m(m – 7) + 2(m – 7) = (m – 7) (m + 2) HCF = m – 7 LCM = m3 – 10m2 + 11m + 70 = m3 – 7m2 – 3m2 + 21m – 10m + 70 = m2 (m – 7) – 3m(m – 7) – 10(m – 7) = (m – 7) (m2 – 3m – 10) = (m – 7) (m2 – 5m + 2m – 10) = (m – 7){m(m – 5) + 2(m – 5)} = (m – 7) (m – 5) (m + 2)
217 Now, 1st expression × 2nd expression = HCF × LCM or, (m – 7) (m + 2) × 2nd expression = (m – 7) × (m – 1) (m – 5) (m + 2) or, 2nd expression = (m – 7) (m – 5) 2nd expression = m2 – 12m + 35 6. Find the HCF and LCM of: (a) 4(a2 – 4), 6(a2 – a – 2) and 12(a2 + 3a – 10) (b) 2x2 – 3xy – 2y2 , 6x2 + xy – y2 and 3x2 – 7xy + 2y2 (c) a(c + a) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a(a + b) (d) p2 + q2 + 2pq – 1, q2 – p2 + 2q + 1 and p2 – q2 + 2p + 1 (e) 6x2 – 5x – 4, 8x2 + 2x – 15 and 12x2 – 43x + 35 Solution: (a) Here, 4(a2 – 4), 6(a2 – a – 2) and 12(a2 + 3a – 10) 1st expression = 4(a2 – 4) = 2 × a(a – 2) (a + 2) 2nd expression = 6(a2 – a – 2) = 6(a2 – 2a + a – 2) = 6{a(a – 2) + 1(a – 2)} = 3 × 2(a – 2) (a + 1) 3rd expression = 12(a2 + 3a – 10) = 12(a2 + 5a – 2a – 10) = 12{a(a + 5) – 2(a + 5)} = 2 × 2 × 3(a – 2) (a + 5) HCF = 2(a – 2) and LCM = 2(a – 2) × 2 × 3(a + 2) (a + 1) (a + 5) = 12(a – 2) (a + 2) (a + 1) (a + 5) (b) Here, 2x2 – 3xy – 2y2 , 6x2 + xy – y2 and 3x2 – 7xy + 2y2 1st expression = 2x2 – 3xy – 2y2 = 2x2 – 4xy + xy – 2y2 = 2x(x – 2y) + y(x – 2y) = (x – 2y) (2x + y) 2nd expression = 6x2 + xy – y2 = 6x2 + 3xy – 2xy – y2 = 3x(2x + y) – y(2x + y) = (2x + y) (3x – y) 3rd expression = 3x2 – 7xy + 2y2 = 3x2 – 6xy – xy + 2y2 = 3x(x – 2y) – y(x – 2y) = (x – 2y) (3x – y) HCF = HCF = 1 and LCM = (x – 2y) (2x + y) (3x – y) (c) Here, a(c + a) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a(a + b) 1st expression = a(c + a) – b(b + c) = ac + a2 – b2 – bc = ac – bc + a2 – b2 = c(a – b) + (a – b) (a + b) = (a – b) (a + b + c) 2nd expression = b(a + b) – c(c + a) = ab + b2 – c2 – ac = ab – ac + b2 – c2 = a(b – c) + (b – c) (b + c) = (b – c) (a + b + c) 3rd expression = c(b + c) – a(a + b) = bc + c2 – a2 – ab = bc – ab + c2 – a2 = b(c – a) + (c – a) (c + a) = (c – a) (a + b + c) HCF = a + b + c and LCM = (a + b + c) (a – b) (b – c) (c – a)
218 (d) Here, p2 + q2 + 2pq – 1, q2 – p2 + 2q + 1 and p2 – q2 + 2p + 1 1st expression = p2 + q2 + 2pq – 1 = (p + q)2 – q2 = (p + q – 1) (p + q + 1) 2nd expression = q2 – p2 + 2q + 1 = q2 + 2q + 1 – p2 = (q + 1)2 – p2 = (q + 1 – p) (q + 1 + p) 3rd expression = p2 – q2 + 2p + 1 = p2 + 2p + 1 – q2 = (p + 1)2 – q2 = (p + 1 – 1) (p + 1 + 1) = (p – q + 1) (p + q + 1) HCF = p + q + 1 and LCM = (P + q + 1) (p – q + 1) (p + q – 1) (q – p + 1) (e) Here, 6x2 – 5x – 4, 8x2 + 2x – 15 and 12x2 – 43x + 35 1st expression = 6x2 – 5x – 4 = 6x2 – 8x + 3x – 4 = 2x(3x – 4) + 1(3x – 4) = (3x – 4) (2x + 1) 2nd expression = 8x2 + 2x – 15 = 8x2 + 12x – 10x – 15 = 4x(2x + 3) – 5(2x + 3) = (2x + 3) (4x – 5) 3rd expression = 12x2 – 43x + 35 = 12x2 – 28x – 15x – 15x + 35 = 4x(3x – 7) – 5(3x – 7) = (3x – 7) (4x – 5) HCF = 1 and LCM = (4x – 5) (3x – 4) (2x + 1) (2x + 3) (3x – 7)
219 CHAPTER LINEAR EQUATIONS 11 11.1 Solving Simultaneous Equations by Substitution Method PRACTICE 11.1 1. (a) What is the value of x if y = 0 and 2x + y = 3? (b) What is the value of x if y = 2 and x + y = 5? Solution: (a) If y = 0 and 2x + y = 3 then, 2x + 0 = 3 or, 2x = 3 x = 3 2 (b) If y = 2 and x + y = 5 then, x + 2 = 5 x = 3 2. Solve the following equations by substitution method: (a) y = x and y = 2x + 1 (b) y = 5x and y + 3x = 16 (c) y – 2x = 0 and y + x – 9 = 0 (d) 2y = x + 3 and 3x + y + 2 = 0 Solution: (a) Here, y = x .............. (i) y = 2x + 1 ....... (ii) Substituting the value of y from equation (ii) to equation (i) we get, 2x + 1 = x or, 2x – x = – 1 x = –1 Putting the value of x in equation (i), we get y = – 1 x = –1 and y = –1 (b) Here, y = 5x ................ (i) y + 3x = 16 ........ (ii) Substituting the value of y from equation (i) to equation (ii), we get, 5x + 3x = 16 or, 8x = 16 x = 2 Putting the value of x in equation (i) y = 5 × 2 Hence, x = 2 and y = 10. (c) Given equations are y – 2x = 0 ....................... (i) or, y + x – 9 = 0 ................... (ii)
220 Substituting the value of y from (i) to (ii) we get, 2x + x – 9 = 0 or, 3x = 9 x = 3 Putting the value of x in equation (i) y = 2 × 3 = 6 Hence, x = 3 and y = 6. (d) Given equations are 2y = x + 3 or, y = x + 3 2 ................. (i) and 3x + y + 2 = 0 .......... (ii) Substituting the value of y from (i) to (ii) we get, 3x + x + 3 2 + 2 = 0 or, 6x + x + 3 + y 2 = 0 or, 7x + 7 = 0 or, 7x = – 7 x = –1 Putting the value of x in equation (i) y = – 1 + 3 2 = 2 2 = 1 Hence, x = –1 and y = 1. 3. Solve the following equations by substitution method: (a) x = 1 and x + y = 1 (b) x = 5y and 7y = 2x + 3 (c) 2x = y and 2y + 5 = 3x (d) 5x = y + 6 and 2x + y = 8 Solution: (a) Given equations are x = 1 ...................... (i) and x + y = 1 ................. (ii) Putting the value of x from (i) to (ii) 1 + y = 1 or, y = 0 Hence, x = 1 and y = 0. (b) Given equations are x = 5y .................. (i) and 7y = 2x + 3 .......... (ii) Substituting the value of x from (i) to (ii) we get, 7y = 2 × 5y + 3 or, 7y = 10y + 3 or, 7y – 10y = 3 or, –3y = 3 y = –1 Putting the value of y in equation (i) x = 5 × (–1) = –5 Hence, x = –5 and y = –1.
221 (c) Given equations are 2x = y ................. (i) and 2y + 5 = 3x ........ (ii) Substituting the value of y from (i) to (ii) we get, 2 × 2x + 5 = 3x or, 2x – 3x = – 5 x = – 5 Putting the value of x in equation (i) y = 2 × (–5) = – 10 Hence, x = –5 and y = – 10 (d) Given equation are 5x = y + 6 ....................... (i) and 2x + y = 8 or, y = 8 – 2x ....................... (ii) Substituting the value of y from (ii) to (i) we get, 5x = 8 – 2x + 6 or, 5x + 2x = 14 or, 7x = 14 x = 2 Putting the value of x in equation (ii), y = 8 × 2 × 2 = 8 – 4 = 4 Hence, x = 2 and y = 4. 4. Use substitution method to solve the following equations: (a) 2x + y = 8 and 3x – y = – 5 (b) x + 2y = 8 and 3x + y = 15 (c) 2x + 3y = 13 and x + 2y = 7 2 3 (d) 2x + 3y = 6 and y = – 1 3 x – 2 (e) x + y = 3 and 2x – 2y = 10 (f) x + y = 1 and x – 4y = – 5 Solution: (a) Given equations are 2x + y = 8 .................. (i) and 3x – y = – 5 or, 3x + 5 = y .................. (ii) Substituting the value of y from (ii) to (i) we get, 2x + 3x + 5 = 8 or, 5x = 8 – 5 x = 3 5 Putting the value of x in equation (ii) y = 3 × 3 5 + 5 = 9 + 25 5 = 34 5 Hence, x = 3 5 and y = 34 5 . (b) Given equations are x + 2y = 8 or, x = 8 – 2y ............... (i) and 3x + y = 15 ............. (ii)
222 Substituting the value of x from (i) to (ii) we get, 3(8 – 2y) + y = 15 or, 24 – 6y + y = 15 or, – 5y = 15 – 24 or, – 5y = – 9 y = 9 5 Putting the value of y in equation (i) x = 8 – 2 × 9 5 = 40 – 18 5 = 22 5 Hence, x = 22 5 and y = 9 5 (c) Give equations are 2x + 3y = 13 ................ (i) and x + 2y = 7 2 3 or, x = 23 3 – 2y .................. (ii) Substituting the value of x from (ii) to (i) we get, 2 23 3 – 2y + 3y = 13 or, 46 3 – 4y + 3y = 13 or, – y = 13 – 46 3 or, – y = 39 – 46 3 or, – y = – 7 3 y = 7 3 Putting the value of y in equation (ii) x = 23 3 – 2 × 7 3 = 23 3 – 14 3 = 9 3 = 3 Hence, x = 3 and y = 7 3 (d) Given equations are 2x + 3y = 6 ................... (i) and y = – 1 3 x – 2 or, y = – x – 6 3 .................. (ii) Substituting the value of y from equation (ii) to (i) we get, 2x + 3 – x – 6 3 = 6 or, 2x – x – 6 = 6 or, x = 12 Putting the value of x in equation (ii) y = – 12 – 6 3 = – 6 Hence, x = 12 and y = –6.
223 (e) Given equation are x + y = 3 or, y = 3 – x ..................... (i) and 2x – 2y = 10 or, x – y = 5 ..................... (ii) Substituting the value of y from equation (i) to (ii) we get, x – (3 – x) = 5 or, x – 3 + x = 5 or, 2x = 5 + 3 or, x = 8 2 x = 4 Putting the value of x in equation (i) y = 3 – 4 = – 1 Hence, x = 4 and y = – 1. (f) Given equations are x + y = 1 or, x = 1 – y ..................... (i) and x – 4y = –5 ................. (ii) Substituting the value of x from equation (i) to (ii) we get, 1 – y – 4y = – 5 or, – 5y = – 5 – 1 or, – 5y = – 6 y = 6 5 Putting the value of y in equation (i) x = 1 – 6 5 = – 1 5 Hence, x = – 1 5 and y = 6 5 . 5. Use substitution method to solve the following equations and test: (a) x + y = 9 (x – y) and 2x + y = 3 (2x – y) – 2 (b) 3x – 2y 5 = 2x – y 3 = 1 (c) x 2 + y 3 = 11 3 and x 3 – y 2 = 1 (d) 7x 3 + y 3 + 1 = and x + y 2 + 3 2 = 0 (e) x – 1 y + 1 = 4 3 and y – 2 x + 2 = 3 4 Solution: (a) Given equations are x + y = 9(x – y) or, x + y = 9x – 9y or, 10y = 8x or, y = 4x 5 ...................... (i) and 2x + y = 3(x – y) – 2 or, 2x + y = 6x – 3y – 2 or, 2 = 6x – 3y – 2x – y or, 2 = 4x – 4y or, 1 = 2x – 2y ............... (ii)
224 Substituting the value of y from equation (i) to (ii) we get, 2x – 2 4x 5 = 1 or, 10x – 8x 5 = 1 or, 2x = 5 x = 5 2 Putting value of x in equation (i) y = 4 × 5 2 5 = 2 Hence, x = 5 2 and y = 2. (b) Given equations are 3x – 2y 5 = 1 or, 3x – 2y = 5 ............... (i) And 2x – y 3 = 1 or, 2x – y = 3 or, 2x – 3 = y ................. (ii) Substituting the value of y from equation (ii) to (i) we get, 3x – 2(2x – 3) = 5 or, 3x – 4x + 6 = 5 or, – x = – 1 x = 1 Putting value of x in equation (ii) y = 2 × 1 – 3 = – 1 Hence, x = 1 and y = – 1. (c) Given equations are x 2 + y 3 = 11 3 ............... (i) And x 3 – y 2 = 1 or, x 3 = 1 + y 2 or, x = 3 + 3y 2 ............... (ii) Substituting the value of x from equation (ii) to (i) we get, 1 2 3 + 3y 2 + y 3 = 11 3 or, 3 2 + 3y 4 + y 3 = 11 3 or, 18 + 9y + 4y 12 = 11 3 or, 13y + 18 = 44 or, 13y = 26 y = 2
225 Putting value of y in equation (ii) x = 3 + 3 × 2 2 = 6 Hence, x = 6 and y = 2. (d) Given equations are 7x 3 + y 3 + 1 = 0 or, 7x + y + 3 3 = 0 or, 7x + y + 3 = 0 or, y = – 7x – 3 ..................... (i) And x + y 2 + 3 2 + 0 ................... (ii) Substituting the value of y from equation (i) to (ii) we get, x + – 7x – 3 2 + 3 2 = 0 or, 2x – 7x – 3 + 3 2 = 0 or, – 5x = 0 x = 0 Putting value of x is equation (i) y = – 7 × 0 – 3 = – 3 Hence, x = 0 and y = – 3. (e) Given equations are x – 1 y + 1 = 3 4 or, x – 1 = 3 4 (y + 1) or, x = 3 4 (y + 1) + 1 or, x = 3y + 3 + 4 4 or, x = 3y + 7 4 ...................... (i) And y – 2 x + 2 = 3 4 ........................ (ii) Substituting the value of x from equation (i) to (ii) we get, y – 2 3y + 7 4 + 2 = 3 4 or, y – 2 = 3 4 3y + 7 4 + 2 or, y – 2 = 3 4 3y + 7 + 8 4 or, y – 2 = 3 16 (3y + 15) or, 16y – 32 = 9y + 45 or, 7y = 77 y = 11 Putting value of y in equation (i) x = 3 × 11 + 7 4 = 40 4 = 10 Hence, x = 10 and y = 11.
226 6. Solve by using substituting method and test: (a) x 3 – 2 y = 14 15 and x 4 + 3 y = 8 5 (b) 11 x – 7y = 18 and 2 x – 3y = 5 (c) 8 x – 9 y = 1 and 10 x + 6 y = 7 (d) 4 x – 3 y = 3 and 8 x – 7 y = 1 (e) x + y xy = 5 6 and y – x xy = 1 6 (f) xy x + y = 1 2 and xy x – y = 1 6 (g) 9 x + y – 4 x – y = 1 and 3 x + y + 2 x – y = 2 (h) 1 x + y + 1 x – y = 7 and 2 x + y + 4 x – y = 17 Solution: (a) Given equations are x 3 – 2 y = 14 15 or, x 3 = 2 y + 14 15 or, x = 6 y + 14 5 ............ (i) And x 4 + 3 y = 8 5 ............. (ii) Substituting the value of x from equation (i) to (ii) we get, 1 4 6 y + 14 5 + 3 y = 8 5 or, 3 2y + 7 10 + 3 y = 8 5 or, 3 + 6 2y = 8 5 – 7 10 or, 9 2y = 16 – 7 10 or, 9 y = 9 5 y = 5 Putting value of y in equation (i) x = 6 5 + 14 5 = 20 5 = 4 Hence, x = 4 and y = 5. (b) Given equations are 11 x – 7y = 18 or, 11 x – 18 = 7y or, 11 7x – 18 7 = y ................... (i) And 2 x – 3y = 5..................... (ii) Substituting the value of y from equation (i) to (ii) we get, 2 x – 3 11 7x – 18 7 = 5
227 or, 2 x – 33 7x + 54 7 = 5 or, 14 – 33 7x = 5 – 54 7 or, – 19 7x = – 19 7 or, 1 x = 1 x = 1 Putting value of x in equation (i) 11 7 × 1 – 18 7 = y or, y = 11 – 18 7 = – 7 7 = – 1 Hence, x = 1 and y = – 1. (c) Given equations are 8 x – 9 y = 1 or, 8 x = 9 y + 1 or, 1 x = 9 8y + 1 8 .................... (i) And 10 x + 6 y = 7 .................... (ii) Substituting the value of 1 x from equation (i) to (ii) we get, 10 9 3y + 1 8 + 6 y = 7 or, 90 + 10y + 48 8y = 7 or, 10y + 138 = 56y or, 138 = 46y y = 3 Putting value of y in equation (i) 1 x = 9 8 × 3 + 1 8 = 4 8 or, x = 2 Hence, x = 2 and y = 3. (d) Given equations are 4 x – 3 y = 3 or, 4 x = 3 y + 3 ....................... (i) And 8 x – 7 y = 1 or, 2 × 4 x – 7 y = 1 ................. (ii)
228 Substituting the value of 4 x from equation (i) to (ii) we get, 2 3 y + 3 – 7 y = 1 or, 6 y + 6 – 7 y = 1 or, – 1 y = – 5 y = 1 5 Putting value of y in equation (i) 4 x = 3 1 5 + 3 = 15 + 3 or, x = 4 18 = 2 9 Hence, x = 2 9 and y = 1 5 . (e) Given equations are x + y xy = 5 6 or, 1 y + 1 x = 5 6 or, 1 y = 5 6 – 1 x ................. (i) And y – x xy = 1 6 or, 1 x – 1 y = 1 6 .............. (ii) Substituting the value of 1 y from equation (i) to (ii) we get, 1 x – 5 6 – 1 x = 1 6 or, 1 x – 5 6 + 1 x = 1 6 or, 2 x = 1 6 + 5 6 or, 2 x = 6 6 x = 2 Putting value of x in equation (i) 1 y = 5 6 – 1 2 = 5 – 3 6 = 2 6 y = 3 Hence, x = 2 and y = 3.
229 (f) Given equations are xy x + y = 1 2 or, x + y xy = 2 or, 1 y + 1 x = 2 or, 1 y = 2 – 1 x ..................... (i) And xy x – y = 1 6 or, x – y xy = 6 or, 1 y – 1 x = 6 .................... (ii) Substituting the value of 1 y from equation (i) to (ii) we get, 2 – 1 x – 1 x = 6 or, – 2 x = 6 – 2 or, – 2 x = 4 x = – 1 2 Putting value of x in equation (i) 1 y = 2 – 1 – 1 2 or, 1 y = 2 + 2 or, 1 y = 4 y = 1 4 Hence, x = – 1 2 and y = 1 4 . (g) Given equations are 9 x + y – 4 x – y = 1 .................... (i) And 3 x + y + 2 x – y = 2 or, 3 x + y = 2 – 2 x – y ................... (ii) Substituting the value of 3 x + y from equation (ii) to (i) we get, 3 2 – 2 x – y – 4 x – y = 1 or, 6 – 6 x – y – 4 x – y = 1 or, – 10 x – y = – 5
230 or, x – y = 2 or, x = y + 2 .............................. (iii) Substituting the value of x – y and x in equation (ii) 3 y + 2 + y = 2 – 2 2 or, 3 2y + 2 = 1 or, 3 = 2y + 2 or, 1 = 2y y = 1 2 Putting value of y in equation (iii) x = 1 2 + 2 = 5 2 Hence, x = 1 2 and y = 5 2 . (h) Given equations are 1 x + y + 1 x – y = 7 or, 1 x + y = 7 – 1 x – y ............... (i) And 2 x + y + 4 x – y = 17 .............. (ii) Substituting the value of 1 x + y from equation (i) to (ii) we get, 2 7 – 1 x – y + 4 x – y = 17 or, 14 – 2 x – y + 4 x – y = 17 or, 2 x – y = 3 or, 2 3 = x – y or, 2 3 + y = x .................... (iii) Substituting the value of x – y and x in equation (i) 1 2 3 + y + y = 7 – 1 2 3 or, 1 2 3 + 2y = 7 – 3 2 or, 1 2 3 + 2y = 11 2 or, 2 3 + 2y = 2 11 or, y = 1 11 – 1 3 y = – 8 33 Putting value of y in equation (iii) x = 2 3 – 8 33 = 22 – 8 33 = 14 33 Hence, x = 14 33 and y = – 8 33 .
231 11.2 Solving Simultaneous Equations by Elimination Method PRACTICE 11.2 1. (a) What is the value of y in the equations x = 2 and x + y = 3? (b) Write the value of x in the equations y = 1 and x – y = 2. Solution: (a) If x = 2 and x + y = 3 then, 2 + y = 3 y = 1 Hence, x = 2 and y = 1 2. Solve the following equations by elimination method: (a) x + y = 3 and x – y = 1 (b) x + 2y = 3 and – x + y = 3 Solution: (a) Given equations are x + y = 3 .................. (i) and x – y = 1 ................. (ii) Adding equation (i) and (ii) x + y = 3 x – y = 1 2x x = = 4 2 Putting the value of x in equation (i) 2 + y = 3 y = 1 Hence, x = 2 and y = 1. (b) Given equations are x – 2y = 3 ................... (i) and –x + y = 3 .................. (ii) Adding equation (i) and (ii) x + 2y = 3 – x + y = 3 3y y = = 6 2 Putting the value of y in equation (i) x + 2 × 2 = 3 x = 3 – 4 = – 1 Hence, x = – 1 and y = 2. 3. Solve the following equations by elimination method: (a) x + 2y = 8 and 2x + 3y = 13 (b) 3x – y = 4 and x + 2y = 13 (c) 3x – 2y = 7 and 5x + y = 3 (d) 3x + 4y = 1 and 3x + 2y = 1 Solution: (a) Given equations are x + 2y = 8 ..................... (i) and 2x + 3y = 13 ................ (2) Subtracting equation (i) × 2 from equation (ii) (b) If y = 1 and x – y = 2 then, x – 1 = 2 x = 3 Hence, x = 3 and y = 1
232 2x + 3y = 13 2x + 4y = 16 – – – – y y = = – 3 3 Putting value of y in equation (i) x + 2 × 3 = 8 or, x = 8 – 6 x = 2 Hence, x = 2 and y = 3. (b) Given equations are 3x – y = 4 ................. (i) and x + 2y = 13 .............. (ii) Adding equation (i) × 2 and (ii) 6x – 2y = 8 x + 2y = 13 7x x = = 21 3 Putting the value of x in equation (ii) 3 × 3 – y = 4 or, 9 – 4 = y y = 5 Hence, x = 3 and y = 5 (c) Given equations are 3x – 2y = 7 ................ (i) and 5x + y = 3 ................. (ii) Adding equation (i) and (ii) × 2 3x – 2y = 7 10x + 2y = 6 13x x = = 13 1 Putting the value of x in equation (ii) 5 × 1 + y = 3 or, y = 3 – 5 y = – 2 hence, x = 1 and y – 2. (d) Given equations are 3x + 4y = 1 ................... (i) and 3x + 2y = 1 .................. (ii) Subtracting equation (ii) from (i) 3x + 4y = 1 3x + 2y = 1 – – – 2y y = = 0 0 Putting the value of y in equation (ii) 3x + 2 × 0 = 1 or, 3x = 1 x = 1 3 Hence, x = 1 3 and y = 0.
233 4. Solve the following equations by elimination method and test: (a) 4x – 3y = 3 and 8x – 7y = 1 (b) 4 = 2x - y and y = – 1 2 x + 1 (c) 2x – 3y = 0 and 6x + 6y = 5 (d) 3x – y = 1 and x – 2y = – 8 Solution: (a) Given equations are 4x – 3y = 3 ............... (i) 8x – 7y = 1 ............... (ii) Subtracting equation (i) × 2 from (i) 8x – 7y = 1 8x – 6y = 6 – + – – y y = = – 5 5 Putting the value of y in equation (i) 4x – 3 × 5 = 3 or, 4x = 3 + 15 or, x = 18 4 x = 9 2 Checking in equation (ii) 8 × 9 2 – 7 × 5 = 1 or, 36 – 35 = 1 1 = 1, true Hence, x = 9 2 and y = 5. (b) Given equations are 4 = 2x – y or, 2x – y = 4 ............. (i) and y = – 1 2 x + 1 or, 2y = – x + 2 or, x + 2y = 2 .............. (ii) Adding equation (i) and (ii) 4x – 2y = 8 x + 2y = 2 5x x = = 10 2 Putting the value of x in equation (i) 2 × 2 – y = 4 y = 0 Testing in equation (ii) 2 + 2 × 0 = 2 or, 2 = 2, true Hence, x = 2 and y = 0.
234 (c) Given equations are 2x – 3y = 0 ................. (i) and 6x + 6y = 5 or, 3x + 3y = 5 2 ............... (ii) Adding equation (i) and (ii), we get, 2x – 3y = 0 3x + 3y = 5 2 5x = 5 2 x = 1 2 Putting the value of x in equation (i) 2 × 1 2 – 3y = 0 or, 1 – 3y = 0 or, – 3y = – 1 y = 1 3 Testing the value in equation (ii) 3 × 1 2 + 3 × 1 3 = 5 2 or, 3 2 + 1 = 5 2 5 2 = 5 2 (True) Hence, x = 1 2 and y = 1 3 (d) Given equations are 3x – y = 1 ............... (i) and x – 2y = – 8 ............ (ii) Subtracting equation (ii) from (i) × 2, 6x – 2y = 2 x – 24 = – 8 – + + 5x x = = 10 2 Putting the value of x in equation (i) 3 × 2 – y = 1 or, –y = 1 – 6 = – 5 y = 5 Testing the value in equation (ii) 2 – 2 × 5 = – 8 or, 2 – 10 = – 8 – 8 = – 8 (True) Hence, x = 2 and y = 5.
235 5. Solve the following equations by elimination method: (a) 1 x + 1 y = 2 and 2 x – 1 y = 1 (b) 1 x + 1 y = 3 and 2 x + 3 y = 8 (c) 2x + 1 y = 4 and x – 1 y = 5 (d) 4x + 9y – 5xy = 0 and x + 2y xy = 2 (e) x + y = xy and x + 9y + 5xy (f) 2 x + y + 1 x – y = 5 6 and 1 x + y + 4 x – y = 1 Solution: (a) Given equations are 1 x + 1 y = 2 ........... (i) 2 x – 1 y = 1 ........... (ii) Adding equation (i) and (ii) 1 x + 1 y = 2 2 x – 1 y = 1 3 x = 3 x = 1 Putting the value of x in equation (i) 1 1 + 1 y = 2 or, 1 y = 2 – 1 or, 1 y = 1 y = 1 Hence, x = 1 and y = 1. (b) Given equations are 1 x + 1 y = 3 ................ (i) 2 x + 3 y = 8 ................ (ii) Subtracting equation (i) × 2 from equation (ii) 2 x + 3 y = 8 2 x + 2 y = 6 – – – 1 y = 2 y = 1 2 Putting the value of y in equation (i) 1 x + 1 1 2 = 3
236 or, 1 x + 2 = 3 or, 1 x = 1 x = 1 Hence, x = 1 and y = 1 2 . (c) Given equations are 2x + 1 y = 4 ................. (i) and x – 1 y = 5 ................... (ii) Adding equation (i) and (ii) 2x + 1 y = 4 x – 1 y = 5 3x x = = 9 3 Putting the value of x in equation (ii) 3 – 1 y = 5 or, – 1 y = 2 y = – 1 2 Hence, x = 3 and y = – 1 2 (d) Given equations are 4x + 9y – 5xy = 0 or, 4x + 9y = 5xy 4 y + 9 x = 5 ............... (i) and x + 2y xy = 2 or, 1 y + 2 x = 2 ............... (ii) Subtracting equation (ii) × 4 from equation (i) 4 y + 9 x = 5 4 y + 8 x = 8 – – – 1 x = – 3 x = – 1 3
237 Putting the value of x in equation (ii) 1 y + 2 – 1 3 = 2 or, 1 y – 6 = 2 or, 1 y = 8 y = 1 8 Hence, x = – 1 3 and y = 1 8 (e) Given equations are x + y = xy or, 1 y + 1 x = 1 .............. (i) and x + 9y = 5xy or, 1 y + 9 x = 5 ............... (ii) Subtracting equation (i) from (ii) d – 7y = 1 8x – 6y = 6 – + – – y y = = – 5 5 Putting the value of x in equation (i) 1 y + 1 2 = 1 or, 1 y = 1 – 1 2 or, 1 y = 1 2 y = 2 Hence, x = 2 and y = 2. (f) Given equations are 2 x + y + 1 x – y = 5 6 ................ (i) and 1 x + y + 4 x – y = 1 ................ (2) Subtracting equation (ii) × 2 from equation (i) 2 x + y + 1 x – y = 5 6 2 x + y + 8 x – y = 2 – – – – 7 x – y = – 5 6 x – y = 6 or, x = y = 6 ....................... (iii)
238 Putting the value of x and x – y in equation (i) 2 y + 6 + y + 1 6 = 5 6 or, 2 2y + 6 = 5 6 – 1 6 or, 2 2y + 6 = 4 6 or, 2y + 6 = 3 or, 2y = – 3 y = – 3 2 Putting the value of y in equation (iii) x = – 3 2 + 6 x = 9 2 Hence, x = 9 2 and y = – 3 2 11.3 Verbal Problems on Simultaneous Equations in Two Variables PRACTICE 11.3 Read / Understand / Think / Do Keeping Skill Sharp 1. Define the following terms: (a) Simultaneous equation of two variables (b) Solution of simultaneous equation of two variables Solution: (a) If the two variables linear equations can be solved to get exact value and they are valid to each equations, then the equations are called simultaneous equations. (b) Simultaneous equation of two variable are solved by either substitution or elemintion or graphical or cross multiplication or matrix method to get unique value of variable. 2. Solve the following simultaneous equations: (a) x = 2 + y and x + y = 8 (b) y = 3 – x and x + 2y = 5 (c) 2x + y = 10 and x – y = 5 (d) x – 2y = 7 an x + 2y = 9 Solution: (a) Given equation are x = 2 + y ............ (i) x + y = 8 ............ (ii) Substituting the value of x from equation (i) to equation (ii) we get, 2 + y + y = 8 or, 2y = 6 y = 3
239 Putting the value of y in equation (i) x = 2 + 3 x = 5 Hence, x = 5 and y = 3. (b) Given equations are y = 3 – x ................ (i) x + 2y = 5 .............. (ii) Substituting the value of y from (i) to (ii) we get, x + 2(3 – x) = 5 or, x + 6 – 2x = 5 or, – x = – 1 x = 1 Putting the value of x in equation (i) y = 3 – 1 y = 2 Hence, x = 1 and y = 2. (c) Given equations are 2x + y = 10 ............. (i) x – y = 5 ................ (2) Adding equation (i) and (ii) 2x + y = 10 x – y = 5 3x x = = 15 5 Putting the value of x in equation (ii) 5 – y = 5 or, – y = 0 y = 0 Hence, x = 5 and y = 0 (d) Given equations are x – 2y = 7 ............... (i) x + 2y = 9 ............... (ii) Adding equation (i) and (ii) x – 2y = 7 x + 2y = 9 2x x = = 16 8 Putting the value of x in equation (ii) 8 + 2y = 9 or, 2y = 1 y = 1 2 Hence, x = 8 and y = 1 2 .
240 Check Your Performance Answer the given questions for each situation. 3. (i) A number is twice the other, if their sum is 45. (ii) Among two numbers, one is 6 more than other and their sum is 38. (a) Find the numbers. (b) How much more percent is the bigger number than the smaller number. Solution: (i) Let x and y are two numbers such that x = 2y ............... (i) x + y = 45 ............ (ii) Substituting the value of x from (i) to (ii) 2y + y = 45 or, 3y = 45 y = 15 Putting the value of y in equation (i) x = 2 × 15 x = 30 (a) Hence, x = 30 and y = 15 (b) Now, more% = x – y y × 100% = 30 – 15 15 × 100% = 100% (ii) Let x and y are two numbers such that x = y + 6 ............ (i) x + y = 38 .......... (ii) Substituting the value of x from (i) to (ii) y + 6 + y = 38 or, 2y = 32 y = 16 Putting the value of y in equation (i) x = 16 + 6 x = 22 (a) Hence, x = 22 and y = 16 (b) More% = x – y y × 100% = 22 – 16 16 × 100% = 6 16 × 100% = 37.50% 4. (i) Seven times of a number increased by 8 is equal to 57. Find the number. (ii) A number is multiplied by 5 and then subtracted by 7 becomes 58. Find the number. Solution: (i) Let x be required number such that 7x + 8 = 57 or, 7x = 49 x = 7 Hence, required number is 7. (ii) Let x be required number such that 5x – 7 = 58 or, 5x = 65 x = 13 5. (i) The length of a rectangle is twice as long as its breadth and its perimeter is 30 m. (ii) The length of equal sides of an isosceles triangle is thrice as long as its base and its perimeter is 56 m. (a) Find the length of its sides. (b) Find its area.
241 Solution: (i) Let x and y are length and breadth of rectangle respectively such that x = 2y ................... (i) and 2(x + y) = 30 or, x + y = 15 ............ (ii) Substituting the value of x from (i) to (ii) 2y + y = 15 or, 3y = 15 y = 5 Putting the value of y in equation (i) x = 2 × 5 x = 10 (a) Hence, length (x) = 10m and breadth (y) = 5 m (b) Area (A) = xy = 10 × 5 = 50 m2 (ii) Let x and y are length of equal sides and base of isosceles triangle respectively such that x = 3y ...................... (i) 2x + y = 56 ............. (ii) Substituting the value of x from (i) to (ii) 2(3y) + y = 56 or, 7y = 56 y = 8 Putting the value of y in equation (i) x = 3 × 8 x = 24 (a) Hence, equal sides (x) = 24 and base (y) = 8 (b) Area of , (A) = y 4 4x2 – y2 = 8 4 4 × 242 – 82 = 2 2304 – 64 = 2 2240 = 2 × 47.33 = 94.66 m2 6. (i) Find two consecutive numbers whose sum is 25. (ii) Find two consecutive even numbers whose sum is 6. Solution: (i) Let, x and x + 1 are two consecutive numbers such that, x + (x + 1) = 25 or, 2x = 25 – 1 or, x = 24 2 x = 12 Hence, required numbers are x = 12 x + 1 = 13 (ii) Let, x and x + 2 are two consecutive even numbers such that x + (x + 2) = 6 or, 2x = 6 – 2 or, x = 4 2 x = 2 Hence, required numbers are x = 2 x + 2 = 2 + 2 = 4
242 7. (i) One-sixth age of a mother is the age of her daughter and the sum of their ages is 49. (ii) The ratio of ages of mother and his son is 7:3 now and the difference of their ages is 40. (a) Find their ages. (b) What was the age of the mother when his child was born? Solution: (i) Let x and y are ages of mother and daughter respectively such that 1 6 x = y or, x = 6y ................... (i) and x + y = 49 ............ (ii) Substituting the value of x from (i) to (ii) 6y + y = 49 or, 7y = 49 y = 7 Putting the value of y in equation (i) x = 6 × 7 x = 42 (a) Hence, age of mother = 42 age of daughter = 7 (b) When the daughter was born, age of mother = 42 – 7 = 35 yrs. (ii) Let x and y are ages of mother and son respectively such that x y = 7 3 or, x = 7y 3 .................... (i) and x – y = 40 .............. (ii) Substituting the value of x from (i) to (ii) 7y 3 – y = 40 or, 4y 3 = 40 y = 30 Putting the value of y in equation (i) x = 7 × 30 3 x = 70 (a) Hence, age of mother (x) = 70 age of son (y) = 30 (b) When the son was born, age of mother = 70 – 30 = 40 yrs. 8. (i) If twice the son’s age added to the father’s age, the sum is 70 and twice the father’s age added to son’s age, the sum is 95. (ii) 3 years ago, a father’s age was 4 times the age of his daughter. Now, the sum of their ages is 46. What are their present ages? (a) Find their present ages. (b) How much older is the father than his child? Solution: (i) Let x and y are ages of father and son respectively such that x + 2y = 70 .............. (i)
243 and 2x + y = 95 ............... (ii) Subtracting equation (ii) from (i) × 2 2x + 4y = 140 2x + y = 95 – – – 3y y = = 45 15 Putting the value of y in equation (i) x + 2 × 15 = 70 x = 40 (a) Hence, age of father (x) = 40 yrs age of son (y) = 15 yrs (b) Father is older than son by 40 – 15 = 25 yrs. (ii) Let x and y are present ages of father and daughter respectively such that x – 3 = 4(y – 3) or, x = 4y – 12 + 3 x = 4y – 9 .................... (i) and x + y = 46 .................... (ii) Substituting the value of x from equation (i) to (ii) 4y – 9 + y = 46 or, 5y = 55 y = 11 Putting the value of y in equation (i) x = 4 × 11 – 9 = 35 (a) Hence, present age of father (x) = 35 present age of son (y) = 11 (b) Father is older than son by = 35 – 11 = 24 yrs. 9. (i) The present age of the father is three times the age of her daughter. After 10 years, the age of the daughter is equal to the age of the father before 20 years. (ii) A year hence, a father will be 4 times as old as his son. Three years ago, the father was 3 times as old as his son will be 4 years hence. (a) Find their ages. (b) How much years younger is his child than father? Solution: (i) Let x and y are present age of father and daughter respectively such that x = 3y ................ (i) and x – 20 = y + 10 or, x – y = 30 .............. (ii) Substituting the value of x from equation (i) to (ii) 3y – y = 30 or, 2y = 30 y = 15 Putting the value of y in equation (i) x = 3 × 15 = 45 (a) Hence, present age of father (x) = 45 yrs. Present age of son (y) = 15 (b) Daughter is younger to father by = 45 – 15 = 30 yrs. (ii) Let x and y are present age of father and son respectively such that x + 1 = 4(y + 1)
244 or, x = 4y + 3 ................ (i) and x – 3 = 3(y + 4) or, x – 3 = 3y + 12 or, x – 3y = 15 ................ (ii) Substituting the value of x from equation (i) to (ii) 4y + 3 – 3y = 15 y = 12 Putting the value of y from equation (i) to (ii) x = 4 × 12 + 3 = 51 (a) Hence, present age of father (x) = 51 yrs. Present age of son (y) = 12 yrs. (b) The son is younger to father by = 51 – 12 = 39 yrs. 10. (i) The sum of two numbers is 5 and their difference is 3. Find the numbers. (ii) The sum of ages of two brothers is 45 and the difference of their ages is 3. Find their ages. Solution: (i) Let x and y are two required numbers such that x + y = 5 ................... (i) x – y = 3 ................... (ii) Adding equation (i) and (ii) x + y = 5 x – y = 3 2x x = = 8 4 Putting the value of x in equation (i) 4 + y = 5 y = 1 Hence, required numbers are x = 4 and y = 1. (ii) Let x and y are ages of two brothers such that x + y = 45 ............... (i) and x – y = 3 ................. (ii) Adding equation (i) and (ii) x + y = 45 x – y = 3 2x x = = 48 24 Putting the value of x in equation (i) 24 + y = 45 y = 21 Hence, ages of two brothers are x = 24 and y = 21. 11. (i) The sum of digits in a two-digit number is 11. The number formed by interchanging the digits of that number will be 45 more than the original number. (ii) A number consists of two digits. The quotient of these two digits being 3 and if 18 is subtracted to the number, then the digits of the number are reversed. (a) Find the original number. (b) Find the more or less percent of the original number than its reverse.
245 Solution: (i) Let 10x + y be the two digits number with digits x and y at 10th place and unit place respectively. Then, y + x = 11 ................... (i) and 10y + x = 10x + y + 45 or, 9y – 9x = 45 or, y – x = 5 ..................... (ii) Adding equation (i) and (ii) y + x = 11 y – x = 5 2y y = = 16 8 Putting the value of y in equation (i) 8 + x = 11 x = 3 (a) Hence, original two digit number is 10x + y = 10 × 3 + 8 = 38 (b) Reversed number is 83 Hence, original number is less than reversed number by = 83 – 38 83 × 100% = 4500 83 % = 54.22% (ii) let 10x + y be the two digits number with digits x and y at tenth place and unit place respectively. Then, x y = 3 or, x = 3y ...................... (i) and 10x + y – 18 = 10y + x or, 9x – 9y = 18 x – y = 2 ................... (ii) Substituting the value of x from equation (i) to (ii), 3y – y = 2 or, 2y = 2 y = 1 Putting the value of y in equation (i) x = 3 × 1 = 3 (a) Hence, original two digit number = 10x + y = 10 × 3 + 1 = 31 (b) Also, reversed number = 13 Original number is more than reversed number by = 31 – 13 13 × 100% = 1800 13 % = 138.46% 12. (i) A two-digit number is such that the sum of its digits is 3. Seven times the number is equal to four times the number obtained by reversing the digits. What is that number? (ii) The sum of the digits of a two-digit number is 17 and the number is 9 more than the other with these digits in the reverse order. Find the number. (a) Find the number. (b) Find the difference between the number and its reverse.
246 Solution: (i) Let 10x + y be required two digits number with digits x and y at tenth and unit place respectively. Then, x + y = 3 ..................... (i) and 7(10x + y) = 4(10y + x) or, 70x + 7y = 40y + 4x or, 66x – 33y = 0 or, 2x – y = 0 ..................... (ii) Adding equation (i) and (ii) x + y = 3 2x – y = 0 3x x = = 3 1 Putting the value of x in equation (i) 1 + y = 3 y = 2 (a) Hence, the two digit number = 10x + y = 10 × 1 + 2 = 12 (b) Also, reversed number = 21 and, difference in number and reverse = 21 – 12 = 9 (ii) Let, 10x + y be required two digits number with digits x and y at tenth and unit place respectively. then, x + y = 17 .................... (i) and, 10x + y = 10y + x + 9 or, 9x – 9y = 9 x – y = 1 ........................ (ii) Adding equation (i) and (ii) x + y = 17 x – y = 1 2x x = = 18 9 Putting the value of x in equation (i) 9 + y = 17 y = 8 (a) Hence, the two digit number = 10x + y = 10 × 9 + 8 = 98 (b) The reversed number = 89 And, difference in number and reverse = 98 – 89 = 9 13. (i) The sum of two digits of a two-digit number and the number obtained by reversing the order of its digits is 84 and they differ by 3. (ii) In a two-digit number, the digit of the unit’s place is greater than that of ten’s place by 2. If 3 times the sum of digits is added to the number, the digits are reversed. (a) Find the number. (b) Find the increased or decreased percent of the original number than its reverse. Solution: (i) Let 10x + y be required two digits number with digits x and y at tenth and unit place respectively. then, (10y + x) + (x + y) = 84 or, 11y + 2x = 84 ..................... (i)