RS Aggarwal Solutions Class 6 Chapter-8 Algebraic Expressions (Ex 8A) Exercise 8.1 PDF

Chapter 8: - Algebraic expressions Page No. : - 130 Exercise: - 8A Question 1: (i) Solution: x increased by 12 is (1+12) (ii) Solution: y decreased by 7 is (7-y) (iii) Solution: The difference of a and b, when a>b is (a-b). (iv) Solution...

Chapter 8: - Algebraic expressions Page No. : - 130 Exercise: - 8A Question 1: (i) Solution: x increased by 12 is (1+12) (ii) Solution: y decreased by 7 is (7-y) (iii) Solution: The difference of a and b, when a>b is (a-b). (iv) Solution: The product of x and y is my. The sum of x and y is (x+y). So, product of x and y added to their sum is xy+(x+y) (v) Solution: One third of x is (vi) x x ( a  b) Solution: One-third of x multiplied by the sum of a and b    a  b   3 3 (vii) Solution: 5 times x added to 7 times y= (5  x) + (7  y), which is equal to 5x+7y. (viii) x y Solution: Sum of x and the quotient of y by 5 is 5 (ix) Solution: x taken away from 4 is (4-x) (x) Solution: 2 less than the quotient of x by y is x/y-2. (xi) Solution: x multiplied by itself is x * x  x 2. (xii) www.vedantu.com 1  Solution: Twice x increased by y is (2*x) + y = 2x+y. (xiii) Solution: Thrice x added to y squared is (3* x)  ( y * y)  3x  y 2 (xiv) Solution: x minus twice y is x-(2*y) = x-2y (xv) Solution: x cubed less than y cubed is ( y  yy )  ( x  x  x)  y 2  x 2 (xvi) Solution: The quotient of x by 8 is multiplied by y is *y= Question 2: Solution: Ranjit’s score in English = 80 marks Ranjit’s score in Hindi = x marks. Total score in the two subjects = (Ranjit’s score in English + Ranjit’s score in Hindi). Total score in the two subjects = (80 + x) marks. Question3: (i) Solution: b*b*b*…15 times = b15 (ii) Solution: y*y*y*…20 times= y 20 (iii) Solution: 14 x a x a x a x a x b x b x b = 14x (a x a x a x a) x (b x b x b) =14 (iv) Solution: 6* x * x * y * y  6 x( x * x) x( y * y)  6 x 2 y 2 (v) 3xzxzxzxyxyxX Solution: 3* z * z * z * y * y * x  3*( z * z * z ) x( y * y)* x  3z 3 y 2 x Question 4: www.vedantu.com 2  (i) Solution: =x*x*y*y*y*y (ii) Solution: 6 =6 * y * y * y * y * y (iii) Solution: 9x z= 9 * x * y * y * z (iv) Solution: 10 = 10 * a * a * a * b * b * b * c * c * c Page No.: - 132 Exercise: - 8B Question 1: (i) Solution: Substituting a = 2 and b = 3 in the given expression: 2+3 = 5 (ii) Solution: Substituting a = 2 and b = 3 in the given expression: 22  (2  3)  4  6  10 (iii) Solution: Substituting a = 2 and b = 3 in the given expression: (2  3)  22  6  4  2 (iv) Solution Substituting a = 2 and b = 3 in the given expression: (2×2)-(3×3)=4-9=-5 www.vedantu.com 3  (v) Solution Substituting a=2 and b=3 in the given expression: 5  22  2  2  3  5  4  12  20  12  18 (vi) Solution Substituting a=2 and b=3 in the given expression: 23  33  2  2  2  3  3  3  13  27  19 Question 2: (i) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression: 3 1 – 2  2   4  5  3 – 4  20  19 (ii) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression:  13  23  5 3  1 *1 *1   2* 2* 2    5*5*5   1  8  125  134 (iii) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression:  2  12  3  22  52  2  11 – 3   2  2    5  5   2  12  25  15 (iv) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression:  1 2    2  5    5 1  2  10  5 7 (v) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression: www.vedantu.com 4   2  1  2  5  2  5  1 2 2 2  4 – 50  4  – 42 (vi) Solution: Substituting x = 1, y = 2 and z = 5 in the given expression:  13  23  53  1 * 1 * 1 – 2 * 2 * 2  – 5 * 5 * 5   1 – 8 – 125   132 Question 3: (i) Solution: Substituting p = -2, q = -1 and r = 3 in the given expression:  (2) 2  (1) 2  (3) 2   2  2    1 1 –  3  3   4 1 – 9  4. (ii) Solution: Substituting p = -2, q = -1 and r = 3 in the given expression:  2  (2) 2  (1) 2  3  (3) 2  2   2  2    1 1  3   3  3   8  1  27  34. (iii) Solution: Substituting p = -2, q = -1 and r = 3 in the given expression:   2  –  1 –  3   2 1 3 4 www.vedantu.com 5  (iv) Solution: Substituting p = -2, q = -1 and r = 3 in the given expression:  (2)3  (1)3  33  3  (2  1 3)   2   2   2    1   1   1   3  3  3   3   6    8   1   27   18  36 (v) Solution: Substituting p = -2, q = -1 and r = 3 in the given expression:  3  (2) 2  (1)  5  (2)  (1) 2  2  (2  1 3)  3   2 x  2    1  5   2    1 1  2   2    1  3  12  10  12  10. (vi) Solution Substituting p = -2, q = -1 and r = 3 in the given expression:  (2)4  (1) 4  34   2   2   2   2   1   1   1   1  3  3  3  3   16  1 – 81  64 Question 4: (i) Solution Coefficient of x in 13x is 13. (ii) Solution Coefficient of y in -5y is -5. (iii) Answer: Coefficient of a in 6ab is 6b. (iv) Solution Coefficient of z in -7xzis -7x. (v) Solution Coefficient of p in -2pqr is -2qr. www.vedantu.com 6  (vi) Solution Coefficient of in 8xy2z is 8xz. (vii) Solution Coefficient of x3 in x3 is 1. (viii) Solution: Coefficient of x2in x 2 is -1. Question 5: - Write the numerical coefficient of (i) Solution: Numerical coefficient of ab is 1. (ii) Solution: Numerical coefficient of -6bc is -6. (iii) Solution: Numerical coefficient of 7xyz is 7 (iv) Solution: Numerical coefficient of 2x 3 y 2 z is -2. Question 6: (i) Solution: In the expression, 3 x 2  5 x  8 the constant term is 8. (ii) Solution: In the expression 2 x 2  9 , the constant term is -9. (iii) 3 3 Solution: In the expression 4 y 2  5 y  the constant term is 5 5 (iv) 8 8 Solution: In the expression z 3  2 z 2  z  , the constant term is  3 3 Question 7: www.vedantu.com 7  (i) Solution: monomial (ii) Solution: binomial (iii) Solution: monomial (iv) Solution: trinomial (v) Solution: trinomial (vi) Solution: monomial. (vii) Solution: it contains four terms. So, it does not represent any of the given types. (viii) Solution: monomial. (ix) Solution: binomial. Question 8: (i) Solution: Expression 4 -6 +7 y – 9 has four terms, namely 4 -6 +7 y and -9. (ii) Solution: Expression 9 -5 +7 y – xyz has four terms, namely 9 , -5 , 7 y and – xyz. Question 9: (i) Solution: a 2 and 2a 2 are like terms. (ii) www.vedantu.com 8  1 Solution: -yz and, zy are like terms. 2 (iii) Solution: 2xy 2 xy 2 and 5 y 2 x are like terms. (iv) Solution: ab 2 c, acb 2 , b 2 ac and ca are like terms. Page Number: - 134 Exercise: - 8C Question 1: (i) Solution: Required sum = 3x + 7x = (3+7) x = 10x (ii) Solution: 7y + (-9y) = (7-9) y = -2y (iii) Solution: 2xy+5xy -xy = (2+5-1) xy = 6xy (iv) Solution: Required sum 3x + 2y (v) Solution: 2 x 2  3x 2  7 x 2  (2  3  7) x 2  6 x 2 (vi) Solution: 7xyz -5xyz+ 9xyz -8xyz = (7 – 5 + 9 – 8) xyz = 3xyz (vii) Solution: 6a 3  4a 3  10a 3  8a 3  (6  4  10  8)a3  4a3 www.vedantu.com 9  (viii) Solution: - 5 - 4 ) + (2 + + 4 ) = (1 – 5 – 4) + (2 - 1 + 4) = -8 + 5 Question 2: (i) Solution: x – 3y – 2z 5x +7y – z -7x – 2y + 4z - x + 2y + z (ii) Solution: – 4m + 5 -2 + 6m – 6 - – 2m - 7 -2 +0xm–8  2m 2  0  8  2m 2  8 (iii) Solution: 2 – 3xy + -7 – 5xy - 2 4 + xy - 6 - - 7xy - 7 www.vedantu.com 10  (iv) Solution: 4xy – 5yz – 7zx -5xy + 2yz + zx -2xy – 3yz + 3zx - 3xy - 6yz - 3zx Question 3: (i) Solution: Sum of the given expressions = (3a -2b + 5c) + (2a + 5b – 7c) – (-a –b +c) = Rearranging and collecting the like terms = 3a + 2a –a – 2b + 5b – b +5c – 7c + c = (3 + 2 + 1) a + (-2 + 5 – 1) b + (5 – 7 + 1) c = 4a + 2b - c (ii) Solution: Sum of the given expressions = (8a – 6ab + 5b) + (-6a – ab -8b) + (-4a + 2ab + 3b) = Rearranging and collecting the like terms = (8 – 6 – 5) a + (-6 – 1 + 2) ab + (5 – 8 + 3) b = -2a – 5ab + 0 = -2a – 5ab (iii) Solution: Sum of the given expressions  (2 x3  3x 2  7 x  8)  (5 x5  2 x 2  4 x  1)  (3  6 x  5 x 2  x3 ) = Rearranging and collecting the like terms  2 x3  5 x3  x3  3x 2  2 x 2  5 x 2  7 x  4 x  6 x  8  1  3  (2  5  1) x3  (3  2  5) x 2  (7  4  6) x  4  4 x3  4 x 2  3x  4 (iv) Solution: Sum of the given expressions  (2 x 2  8 xy  7 y 2  8 xy 2 )  (2 xy 2  6 xy  y 2  3 x 2 )  (4 y 2  xy  x 2  xy 2 ) = Rearranging and collecting the like terms www.vedantu.com 11   2 x 2  3x 2  x 2  7 y 2  y 2  4 y 2  8 xy  6 xy  xy  8 xy 2  2 xy 2  xy 2  (2  3  1) x 2  (7  1  4) y 2  (8  6  1) xy  (8  2  1) xy 2  4 x 2  10 y 2 ?3xy  5 xy 2 (v) Solution: Sum of the given expressions  ( x3  y 3  z 3  3xyz )  ( x3  y 3  z 3  6 xyz )  ( x3  y 3  z 3  8 xyz ) = Rearranging and collecting the like terms  x  x3  x3  y 3  y 3  y 3  z 3  z 3  z 3  3xyz  6 xyz  8 xyz 3  (1  1  1) x3  (1  1  1) y 3  (1  1  1) z 3  (3  6  8) xyz  x3  y 3  z 3  11xyz (vi) Solution: Sum of the given expressions  (2  x  x 2  6 x3 )  (6  2 x  4 x 2  3x3 )  (2  x 2 )  (3  x3  4 x  2 x 2 ) = Rearranging and collecting the like terms  6 x  3x3  x3  x 2  4 x 2  x 2  2 x 2  x  2 x  4 x  2  6  2  3 3  (6  3  1) x3  (1  4  1  2) x3  (1  2  4) x  1  2 x3  2 x 2  3x  1 Question 4: (i) Solution: Term to be subtracted = 5x = Changing the sign of each term of the expression gives -5x. = On adding: =2x+(-5x) = 2x-5x = (2-5)x = -3x (ii) Solution: Term to be subtracted = -xy = Changing the sign of each term of the expression gives xy. = On adding: = 6xy+xy = (6+1) xy = 7xy www.vedantu.com 12  (iii) Solution: Term to be subtracted = 3a =Changing the sign of each term of the expression gives -3a. = On adding = 5b+(-3a) = 5b-3a (iv) Solution: Term to be subtracted = -7x =Changing the sign of each term of the expression gives 7x. =On adding: = 9y+7x (v) Solution: Changing the sign of each term of the expression gives 10x 2. = On adding:  7 x  (10 x 2 )  7 x 2  10 x 2 2  (7  10) x 2  17 x 2 (vi) Solution: Term to be subtracted = a2 − b2 =Changing the sign of each term of the expression gives a 2  b 2. =On adding:  b  a 2  (a 2  b 2 )  b 2  a 2  a 2  b 2 2  (1  1)b 2  (1  1)a 2  2b 2  2a 2 Question 5: (i) Solution: Term to be subtracted = 5a + 7b − 2c = Changing the sign of each term of the expression gives -5a -7b + 2c. = On adding: = (3a − 7b + 4c)+(-5a -7b + 2c) = 3a − 7b + 4c-5a -7b + 2c = (3-5)a+ (− 7-7) b + (4+2) c = -2a − 14b + 6c www.vedantu.com 13 (ii) Solution: Term to be subtracted = a − 2b − 3c = Changing the sign of each term of the expression gives -a +2b + 3c. = On adding: = (−2a + 5b − 4c)+(-a +2b + 3c) = −2a + 5b − 4c-a +2b + 3c = (−2-1)a + (5+2)b +(−4+3)c = −3a + 7b − c (iii) Solution: Term to be subtracted = 5 x 2  3xy  y 2 =Changing the sign of each term of the expression gives 5 x 2  3xy  y 2 = On adding:  (7 x  2 xy  4 y 2 )  (5 x 2  3 xy  y 2 )  7 x 2  2 xy  4 y 2  5 x 2  3 xy  y 2 2  (7  5) x 2  (2  3) xy  (4  1) y 2  2 x 2  xy  5 y 2 (iv) Solution: Term to be subtracted = 6 x3  7 x 2  5 x  3 = Changing the sign of each term of the expression gives 6 x 3  7 x 2  5 x  3 = On adding:  (4  5 x  6 x 2  8 x3 )  (6 x3  7 x 2  5 x  3)  4  5 x  6 x 2  8 x3  6 x3  7 x2  5 x  3  (8  6) x3  (6  7) x 2  (5  5) x  7  14 x3  13x 2  10 x  7 (v) Solution: Term to be subtracted  x3  2 x 2 y  6 xy 2  y 3 = Changing the sign of each term of the expression gives  x  2 x 2 y  6 xy 2  y 3. 3 = On adding:  ( y  3 xy 2  4 x 2 y )  ( x 3  2 x 2 y  6 xy 2  y 3 ) 3  y 3  3 xy 2  4 x 2 y  x 3  2 x 2 y  6 xy 2  y 3   x 3  (2  4) x 2 y  (6  3) xy 2  (1  1) y 3   x 3  6 x 2 y  9 xy 2  2 y 3 (vi) Solution: Term to be subtracted  11x 2 y 2  7 xy  6 www.vedantu.com 14  = Changing the sign of each term of the expression gives 11x 2 y 2  7 xy  6. = On adding:  (9 x 2 y 2  6 xy  9)  (11x 2 y 2  7 xy  6)  9 x 2 y 2  6 xy  9  11x 2 y 2  7 xy  6  (9  11) x 2 y 2 (7  6) xy  15  20 x 2 y 2  13xy  15 (vii) Solution: Term to be subtracted = −2a + b + 6d = Changing the sign of each term of the expression gives 2a-b-6d. = On adding: = (5a − 2b -3c)+(2a-b-6d) = 5a − 2b -3c +2a-b-6d = (5+2)a+(− 2-1)b -3c -6d = 7a − 3b-3c -6d Question 6: (i) Solution: Rearranging and collecting the like terms  (2  6) p 3  (3  2) p 2  (4  8  6) p  5  2  8  4 p 2  p 2  2 p  1 (ii) Solution: Rearranging and collecting the like terms  (2  6) x 2  (?1  5) xy  (6  4) x  (4  3) y  8 x 2  4 xy  2 x  y (iii) Solution: Rearranging and collecting the like terms  (1  1) x 4  (?6  7) x3  5 x 2  (2  1) x  7  2  0  x3  5 x 2  x  5  x3  5 x 2  x  5 Question 7: Solution: Adding:  (3x 2  5 x  2)  (5 x 2  8 x  6) = Rearranging and collecting the like terms: www.vedantu.com 15   (3  5) x 2  (5  8) x  2  6  2 x 2  13 x  8 Subtract 4 x 2  9 x  7 from  2 x 2  13x  8. =Change the sign of each term of the expression that is to be subtracted and then add. = Term to be subtracted  4 x 2  9 x  7 = Changing the sign of each term of the expression gives 4 x 2  9 x  7. = On adding:  (2 x 2  13 x  8)  (42 x 2  9 x  7)  2 x 2  13 x  8  42 x 2  9 x  7  (2  4)2 x 2  (13  9) x  8  7  6 x 2  4 x  1 Question 8: Solution: A  7 x 2  5 xy  9 y 2 B  4 x 2  xy  5 y 2 C  4 y 2  3x 2  6 xy Substituting the values of A, B and C in A+B+C:  (7 x 2  5 xy  9 y 2 )  (4 x 2  xy  5 y 2 )  (4 y 2  3x 2  6 xy )  7 x 2  5 xy  9 y 2 ? 4 x 2  xy  5 y 2  4 y 2  3x 2  6 xy = Rearranging and collecting the like terms:  (7  4  3) x 2  (5  1  6) xy  (9  5  4) y 2  (0) x 2  (0) xy  (0) y 2 0 Question 9: Solution: Let the expression to be added be X. www.vedantu.com 16   (5 x 2  2 x 2  6 x  7)  X  ( x3  3x3  x  1) X  ( x3  3x3  x  1)  (5 x 2  2 x3  6 x  7) Changing the sign of each term of the expression that is to be subtracted and then adding: X  ( x3  3 x3  x  1)  (5 x 2  2 x 3  6 x  7) X  x3  3x3  x  1  5 x 2  2 x3  6 x  7 = Rearranging and collecting the like terms: X  (1  5) x 3  (3  2) x 2  (1  6) x  1  7 X  4 x 3  5 x 2  7 x  6 So, 4 x 3  5 x 2  7 x  6 must be added to 4 x 3  5 x 2  6 x  7 to get the sum as x3  3x 2  x  1. Question 10: Solution: P  a 2  b 2  2ab Q  a 2  4b  6ab 2 R  b2  6 S  a 2  4ab T  2a 2  b 2  ab  a Adding P, Q, R and S : P  Q  R  S  (a 2  b  2ab)  (a 2  4b 2  6ab)  (b 2  6)  (a 2  4ab) 2  a 2  b 2  2ab  a 2  4b 2  6ab  b 2  6  a 2  4ab = Rearranging and collecting the like terms:  (1  1  1)a 2  (?1  4  1)b 2  (2  6  4) ab  6 P  Q  R  S  3a 2  4b 2  8ab  6 To find P  Q  R  S - T , subtract T  (-2a 2  b 2 - ab  a) from P  Q  R  S  (3a 2  4b 2 -8ab  6). = On changing the sign of each term of the expression that is to be subtracted and then adding: =Term to be subtracted = 2a 2  b 2  ab  a = Changing the sign of each term of the expression gives 2a 2  b 2  ab  a. Now add: www.vedantu.com 17   (3a 2  4b 2  8ab  6)  (2a 2  b 2  ab  a)  3a 2  4b 2  8ab  6  2a 2  b 2  ab  a  (3  2)a 2  (4  1)b 2  (8  1)ab  a  6 P  Q  R  S  T  5a 2  3b 2  7ab  a  6 Question 11: Solution: Let the expression to be subtracted be X.  (a 3  4a 3  5a  6)  X  (a 2  2a  1) X  (a 3  4a 2  5a  6)  (a 2  2a  1) Since '-' sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis. X  a 3  4a 2  5a  6  a 2  2a  1 Rearranging and collecting the like terms: X  a 3  (4  1)a 2  (5  2)a  6  1 X  a 3  5a 2  7a  7 So, a 3  5a 2  7 a  7 must be subtracted from a 3  4a 2  5a  6 to obtain a 2  2a  1 Question 12: Solution: = To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c. Change the sign of each term of the expression that is to be subtracted and then add. =Term to be subtracted = 2a − 3b + c = Changing the sign of each term of the expression gives -2a + 3b - c. On adding: www.vedantu.com 18   a  2b  3c    2a  3b  c   a  2b  3c  2a  3b  c  1  2  a   2  3 b    3  1 c   a  5b  4c Question 13: Solution: = To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z. = Change the sign of each term of the expression that is to be subtracted and then add. = Term to be subtracted = 2x − 4y − z = Changing the sign of each term of the expression gives -2x + 4y + z. On adding:  x  2 y  3z    2 x  4 y  z   x  2 y  3z  2 x  4 y  z  1  2  x   2  4  y   3  1 z   x  2 y  4z Question 14: Solution: = To calculate how much does 3 x 2  5 x  6 exceed x 3  x 2  4 x  1 , we have to subtract x 3  x 2  4 x  1 from 3 x 2  5 x  6. = Change the sign of each term of the expression that is to be subtracted and then add. = Term to be subtracted  x3  x 2  4 x  1 = Changing the sign of each term of the expression gives  x 2  x 2  4 x  1. On adding: www.vedantu.com 19  (3 x 2  5 x  6)  ( x 3  x 2  4 x  1)  3x 2  5 x  6  x3  x 2  4 x  1   x3  (3  1) x 2  (5  4) x  6  1   x3  4 x 2  9 x  7 Question 15: Solution: Add 5 x  4 y  6 z and  8 x  y  2 z.  5x  4 y  6 z    8 x  y  2 z   5x  4 y  6 z  8x  y  2 z   5  8 x   4  1 y  6  2 z   3x  3 y  4 z Adding 12 x  y  3 z and  3 x  5 y  8 z :  12 x  y  3 z    3 x  5 y  8 z   12 x  y  3 z  3 x  5 y  8 z  12  3 x   1  5  y   3  8  z  9x  4 y  5z Subtract -3x − 3y + 4z from 9x +4y -5z. = Change the sign of each term of the expression that is to be subtracted and then add. = Term to be subtracted = -3x − 3y + 4z = Changing the sign of each term of the expression gives 3x + 3y - 4z. On adding:  9x  4 y  5 z    3x  3 y  4 z   9 x  4 y  5 z  3x  3 y  4 z   9  3 x   4  3 y   5  4  z  12 x  7 y  9 z Question 16: = To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z. =Change the sign of each term of the expression that is to be subtracted and then add. www.vedantu.com 20  = Term to be subtracted = 2x + 5y − 6z + 2 = Changing the sign of each term of the expression gives -2x - 5y + 6z - 2. On adding:   2x  3 y  4 z    2 x  5 y  6 z  2   2x  3y  4z  2x  5 y  6z  2   2  2 x   3  5  y   4  6  z  2  0  8 y  10 z  2   8 y  10 z  2 Question 17: Solution: To calculate how much 1 exceeds 2x-3y-4, we have to subtract 2x-3y-4 from 1 =Change the sign of each term of the expression to be subtracted and then add. = Term to be subtracted = 2x-3y-4 = Changing the sign of each term of the expression gives -2x+3y+4. On adding:  1   2 x  3 y  4   1 2x  3y  4  5  2x  3y Page Number: - 136 Exercise: - 8D Simplify Question 1: Solution: Here, ‘-' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis. =a - b + 2a =3a - b www.vedantu.com 21 Question 2: Solution: = 4x − (3y − x + 2z) Here, '−' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis. = 4x − 3y + x − 2z = 5x − 3y − 2z Question 3: Solution: Here, '−' sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.  a 2  b 2  2ab  a 2  b 2  2ab = Rearranging and collecting the like terms:  a 2  a 2  b 2  b 2  2ab  2ab  (1  1)a (2)  (1  1)b 2  (2  2)ab  0  0  4ab  4ab Question 4: Solution: Here, '−' sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.   3a  3b   42a    43b   2a  b   3a  3b  8a  12b  2a  b = Rearranging and collecting the like terms:   3a  8a  2a  3b  12b  b   3  8  2  a   3  12  1 b  3a  14b Question 5: Solution: = We will first remove the innermost grouping symbol ( ) and then { }. www.vedantu.com 22  4 x 2  ( x 2  3)  (4  3 x 2 )  4 x 2  2 x 2  3  4  3 x 2   4 x 2  5 x 2  7  4 x 2  5 x 2  7  x 2  xy Question 6: Solution: Here a '−' sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.  2 x 2  2 y 2  2 xy  3x 2  3 y 2  3 xy  (2  3) x 2  (2  3) y 2  (2  3) xy  5 x 2  y 2  xyc Question 7: Solution: Will first remove the innermost grouping symbol ( ), followed by { } and then [ ]. ∴ a − [2b − {3a − (2b − 3c)}] = a − [2b − {3a − 2b + 3c}] = a − [2b − 3a + 2b − 3c] = a − [4b − 3a − 3c] = a − 4b + 3a + 3c = 4a − 4b + 3cy Question 8: Solution: We will first remove the innermost grouping symbol ( ), followed by { } and then [ ]. ∴ −x + [5y − {x − (5y − 2x)}] = −x + [5y − {x − 5y + 2x}] = −x + [5y − {3x − 5y}] = −x + [5y − 3x + 5y] www.vedantu.com 23  = −x + [10y − 3x] = −x + 10y − 3x = − 4x + 10y Question 9: Solution: 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}] = We will first remove the innermost grouping symbol ( ), followed by { } and then [ ]. ∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}] = 86 − [15x − 42x + 63 −2{10x − 10 + 15x}] = 86 − [15x − 42x + 63 −2{25x − 10}] = 86 − [15x − 42x + 63 −50x + 20] = 86 − [− 77x + 83] = 86 + 77x − 83 = 77x + 3 Question 10: Solution: We will first remove the innermost grouping symbol ( ), followed by { } and then [ ]. 12 x  [3 x 3  5 x 3  7 x 3  (4  3 x  x 3 )  6 x 3  3 x]  12 x  [3 x 3  5 x 2  7 x 2  4  3 x  x 3  6 x 3  3 x]  12 x  [3 x 3  5 x 2  7 x 2  4  3 x  7 x 3  3 x]  12 x  [3 x 3  5 x 2  7 x 2  4  3 x  7 x 3  3 x]  12 x  [ x 2  4  4 x 3  6 x]  12 x  x 2  4  4 x 2  6 x  4 x3  x 2  18 x  4a Question 11: Solution: Will first remove the innermost grouping symbol ( ), followed by { } and then [ ]. www.vedantu.com 24   5a  [a 2  2a(1  a  4a 2 )  3a( a 2  5a  3)]  8a  5a  [a 2  2a  2a 2  8a 2  3a 3  15a 2  9a]  8a  5a  [a 2  5a 3  13a 2  11a]  8a  5a  [a 2 ?5a 3  13a 2  11a ]  8a  5a  [5a 3  12a 2  11a ]  8a  5a  5a 3  12a 2  11a  8a  5a 3  12a 2  8a Question 12: Solution We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].  3   x  2 y  (5 x  y  3)  2 x 2   ( x 2  3 y )   3   x  2 y  5 x  y  3  2 x 2   x 2  3 y   3   x   y  5 x  3  2 x 2   x 2  3 y   3  [ x  y  5x  3  2 x2  x2  3 y]  3  [6 x  3  3 x 2  2 y ]  3  6 x  3  3x 2  2 y  3x 2  2 y  6 x  6 Question 13: Solution: We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].  xy   yz  zx  { yx   3 y  xz    xy  zy }  xy   yz  zx  { yx  3 y  xz  xy  zy}  xy   yz  zx  { 3 y  xz  zy}  xy   yz  zx  3 y  xz  zy   xy    2 zx  3y  xy  2 zx  3 y c Question 14: www.vedantu.com 25 Solution: We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].  2a  3b  3a  2b  {a  c   a  2b }  2a  3b  3a  2b  {a  c  a  2b}  2a  3b  3a  2b  { c  2b}  2a  3b  3a  2b  c  2b   2a  3b  3a  4b  c   2a  3b  3a  4b  c   a  b  cb Question 15: Solution: We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].   a   a  a b  2a   a  2b   b    a   a  a  b  2a  a  2b  b    a   a  3b  2a  b    a   a  3b  2a  b    a   2b  a    a  2b  a   2b Question 16: Solution: We will first remove the innermost grouping symbol bar bracket. Next, we will remove ( ), followed by { } and then [ ].  2a   4b  {4a   3b  2a  2b }  2a   4b  {4a   b  2a }  2a   4b  {4a  b  2a}  2a   4b  {6a  b}  2a   4b  6a  b   2a  5b  6a   2a  5b  6a  8a  5b www.vedantu.com 26 Question 17: Solution We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].  5 x   4 y  {7 x   3 z  2 y   4 z  3  x  3 y  2 z }  5 x   4 y  {7 x  3 z  2 y  4 z  3 x  9 y  6 z}  5 x   4 y  {4 x  7 z  7 y}  5x  4 y  4 x  7 z  7 y   5 x  11 y  4 x  7 z   5 x  11 y  4 x  7 z  9 x  11 y  7 z www.vedantu.com 27