RS Agarwal Solution | Class 6th | Chapter-8 | Algebraic Expressions | Edugrown

Exercise 8A

Page No 130:

Question 1:

Write the following using literals, numbers and signs of basic operations:
(i) x increased by 12
(ii) y decreased by 7
(iii) The difference of a and b, when a > b
(iv) The product of x and y added to their sum
(v) One-third of x multiplied by the sum of a and b
(vi) 5 times x added to 7 times y
(vii) Sum of x and the quotient of y by 5
(viii) x taken away from 4
(ix) 2 less than the quotient of x by y
(x) x multiplied by itself
(xi) Twice x increased by y
(xii) Thrice x added to y squared
(xiii) x minus twice y
(xiv) x cubed less than y cubed
(xv) The quotient of x by 8 is multiplied by y

ANSWER:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (y – 7).
(iii) The difference of a and b, when a>b is (a – b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
      So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is x3x3.
The sum of a and b is (a + b).
      ∴∴ One-third of x multiplied by the sum of a and b = x3×(a+b)=x(a+b)3 x3×(a+b)=x(a+b)3 
(vi) 5 times x added to 7 times y = (5×x)+(7×y),  which is equal to 5x+7y.(5×x)+(7×y),  which is equal to 5x+7y.

(vii) Sum of x and the quotient of y by 5 is x+y5x+y5.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is xy−2xy-2.
(x) x multiplied by itself is x×x=x2x×x=x2.
(xi) Twice x increased by y is (2×x)+y = 2x+y(2×x)+y = 2x+y.
(xii) Thrice x added to y squared is (3×x)+(y×y)=3x+y2(3×x)+(y×y)=3x+y2.
(xiii) x minus twice y is x−(2×y)=x−2yx-(2×y)=x-2y.
(xiv) x cubed less than y cubed is (y×y×y)−(x×x×x)=y3−x3.(y×y×y)-(x×x×x)=y3-x3.
(xv) The quotient of x by 8 is multiplied by y is x8×y=xy8x8×y=xy8.

Page No 130:

Question 2:

Ranjit scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?

ANSWER:

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

Page No 130:

Question 3:

Write the following in the exponential form:
(i) b × b × b × … 15 times
(ii) y × y × y × … 20 times
(iii) 14 × a × a × a × a × b × b × b
(iv) 6 × x × x × y × y
(v) 3 × z × z × z × y × y × x

ANSWER:

(i) b × b × b × … 15 times = b15b15
(ii) y × y × y × … 20 times = y20y20
(iii) 14 × a × a × a × a × b × b × b = 14×(a×a×a×a)×(b×b×b) =14a4b314×(a×a×a×a)×(b×b×b) =14a4b3
(iv) 6 × x × x × y × y = 6×(x×x)×(y×y)=6x2y26×(x×x)×(y×y)=6x2y2
(v) 3 × z × z × z × y × y × x = 3×(z×z×z)×(y×y)×x=3z3y2x3×(z×z×z)×(y×y)×x=3z3y2x

Page No 130:

Question 4:

Write down the following in the product form:
(i) x²y⁴
(ii) 6y⁵
(iii) 9xy²z
(iv) 10a³b³c³

ANSWER:

(i) x2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×yx2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×y
(ii) 6y5=6×(y×y×y×y×y)=6×y×y×y×y×y6y5=6×(y×y×y×y×y)=6×y×y×y×y×y
(iii) 9xy2z=9×x×(y×y)×z=9×x×y×y×z9xy2z=9×x×(y×y)×z=9×x×y×y×z
(iv) 10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c

Page No 132:

Question 1:

If a = 2 and b = 3, find the value of
(i) a + b
(ii) a² + ab
(ii) ab − a²
(iv) 2a − 3b
(v) 5a² − 2ab
(vi) a³ − b³

ANSWER:

(i) a+b
    Substituting a = 2 and b = 3 in the given expression:
    2+3 = 5

(ii) a2+aba2+ab
    Substituting a = 2 and b = 3 in the given expression:
    (2)2+(2×3)=4+6=10(2)2+(2×3)=4+6=10

(iii) ab−a2ab-a2
    Substituting a = 2 and b = 3 in the given expression:
     (2×3)−(2)2=6−4=2(2×3)-(2)2=6-4=2

(iv) 2a-3b
    Substituting a = 2 and b = 3 in the given expression:
    (2×2)−(3×3)=4−9=−5(2×2)-(3×3)=4-9=-5

(v) 5a2−2ab5a2-2ab
    Substituting a=2 and b=3 in the given expression:
    5×(2)2−2×2×3=5×4−12=20−12=85×(2)2-2×2×3=5×4-12=20-12=8

(vi) a3−b3a3-b3
    Substituting a=2 and b=3 in the given expression:
    23−33=2×2×2−3×3×3=8−27=−1923-33=2×2×2-3×3×3=8-27=-19

Page No 132:

Question 2:

If x = 1, y = 2 and z = 5, find the value of
(i) 3x − 2y + 4z
(ii) x² + y² + z²
(iii) 2x² − 3y² + z²
(iv) xy + yz − zx
(v) 2x² y − 5yz + xy²
(vi) x³ − y³ − z³

ANSWER:

(i) 3x-2y+4z
    Substituting x = 1, y = 2 and z = 5 in the given expression:
    3×(1)−2×(2)+4×(5)=3−4+20=193×(1)-2×(2)+4×(5)=3-4+20=19

(ii) x2+y2+z2 x2+y2+z2 
    Substituting x = 1, y = 2 and  z = 5 in the given expression:
    12+22+52=(1×1)+(2×2)+(5×5)=1+4+25=3012+22+52=(1×1)+(2×2)+(5×5)=1+4+25=30

(iii) 2×2−3y2+z22x2-3y2+z2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2−3×(2)2+52=2×(1×1)−3×(2×2)+(5×5)=2−12+25=152×(1)2-3×(2)2+52=2×(1×1)-3×(2×2)+(5×5)=2-12+25=15

(iv) xy+yz−zxxy+yz-zx
     Substituting x = 1, y = 2 and z = 5 in the given expression:
      (1×2)+(2×5)−(5×1)=2+10−5=7(1×2)+(2×5)-(5×1)=2+10-5=7

(v) 2x2y−5yz+xy22x2y-5yz+xy2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2×2−5×2×5+1×(2)2=4−50+4=−422×(1)2×2-5×2×5+1×(2)2=4-50+4=-42

(vi) x3−y3−z3x3-y3-z3
    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=−13213-23-53=(1×1×1)-(2×2×2)-(5×5×5)=1-8-125=-132

Page No 132:

Question 3:

If p = −2, q = −1 and r = 3, find the value of
(i) p² + q² − r²
(ii) 2p² − q² + 3r²
(iii) p − q − r
(iv) p³ + q³ + r³ + 3pqr
(v) 3p²q + 5pq² + 2pqr
(vi) p⁴ + q⁴ − r⁴

ANSWER:

(i) p2+q2−r2p2+q2-r2
     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(-2)2+(-1)2-(3)2=(-2×-2)+(-1×-1)-(3×3)⇒4+1-9=-4

(ii) 2p2−q2+3r22p2-q2+3r2
     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=342×(-2)2-(-1)2+3×(3)2=2×(-2×-2)-(-1×-1)+3×(3×3)⇒8-1+27=34

(iii) p−q−rp-q-r
      Substituting p = -2, q = -1 and r = 3 in the given expression:
     (−2)−(−1)−(3)=−2+1−3=−4(-2)-(-1)-(3)=-2+1-3=-4

(iv) p3+q3+r3+3pqrp3+q3+r3+3pqr
    Substituting p = -2, q = -1 and r = 3 in the given expression:

   (−2)3+(−1)3+(3)3+3×(−2×−1×3)=(−2×−2×−2)+(−1×−1×−1)+(3×3×3)+3×(6)=(−8)+(−1)+(27)+18=36(-2)3+(-1)3+(3)3+3×(-2×-1×3)=(-2×-2×-2)+(-1×-1×-1)+(3×3×3)+3×(6)=(-8)+(-1)+(27)+18=36
   
(v) 3p2q+5pq2+2pqr3p2q+5pq2+2pqr
     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×−2)×(−1)+5×(−2)×(−1×−1)+2×(−2×−1×3)=−12−10+12=−103×(-2)2×(-1)+5×(-2)×(-1)2+2×(-2×-1×3)=3×(-2×-2)×(-1)+5×(-2)×(-1×-1)+2×(-2×-1×3)=-12-10+12=-10

(vi) p4+q4−r4p4+q4-r4
     Substituting p = -2, q = -1 and r = 3 in the given expression:
    (−2)4+(−1)4−(3)4=(−2×−2×−2×−2)+(−1×−1×−1×−1)−(3×3×3×3)=16+1−81=−64(-2)4+(-1)4-(3)4=(-2×-2×-2×-2)+(-1×-1×-1×-1)-(3×3×3×3)=16+1-81=-64
  

Page No 132:

Question 4:

Write the coefficient of
(i) x in 13x
(ii) y in −5y
(iii) a in 6ab
(iv) z in −7xz
(v) p in −2pqr
(vi) y² in 8xy²z
(vii) x³ in x³
(viii) x² in −x²

ANSWER:

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y² in 8xy²z is 8xz.
(vii) Coefficient of x³ in  x³ is 1.
(viii) Coefficient of x² in -x² is -1.

Page No 132:

Question 5:

Write the numerical coefficient of
(i) ab
(ii) −6bc
(iii) 7xyz
(iv) −2x³y²z

ANSWER:

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x³y²z is -2.

Page No 132:

Question 6:

Write the constant term of
(i) 3x² + 5x + 8
(ii) 2x² − 9
(iii) 4y2−5y+354y2-5y+35
(iv) z3−2z2+z−83z3-2z2+z-83

ANSWER:

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x² + 5x + 8, the constant term is 8.
(ii) In the expression 2x² − 9, the constant term is -9.
(iii) In the expression  4y2−5y+354y2−5y+35, the constant term is 3535.
(iv) In the expression z3−2z2+z−83z3−2z2+z−83 , the constant term is −83-83.

Page No 132:

Question 7:

Identify the monomials, binomials and trinomials in the following:
(i) −2xyz
(ii) 5 + 7x³y³z³
(iii) −5x³
(iv) a + b − 2c
(v) xy + yz − zx
(vi) x⁵
(vii) ax³ + bx² + cx + d
(viii) −14
(ix) 2x + 1

ANSWER:

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

Page No 133:

Question 8:

Write all the terms of the algebraic expressions:
(i) 4x⁵ − 6y⁴ + 7x²y − 9
(ii) 9x³ − 5z⁴ + 7z³ y − xyz

ANSWER:

(i) Expression 4x⁵ − 6y⁴ + 7x²y − 9 has four terms, namely 4x⁵ ,-6y⁴ , 7x²y and -9.
(ii) Expression 9x³ − 5z⁴ + 7z³ y − xyz has four terms, namely 9x³ , -5z⁴ , 7z³ y and -xyz.

Page No 133:

Question 9:

Identify the like terms in the following:
(i) a² , b², −2a², c², 4a
(ii) 3x, 4xy, −yz,12zy3x, 4xy, -yz,12zy
(iii) −2xy², x²y, 5y²x, x²z
(iv) abc, ab²c, acb², c²ab, b²ac, a²bc, cab²

ANSWER:

The terms that have same literals are called like terms.
(i)  a² and 2a² are like terms.
(ii) −yz and 12zy-yz and 12zy are like terms.
(iii) −2xy² and 5y²x are like terms.
(iv) ab²c , acb² , b²ac and cab² are like terms.

Page No 134:

Exercise 8B

Question 1:

Add:
(i) 3x, 7x
(ii) 7y, −9y
(iii) 2xy, 5xy, −xy
(iv) 3x, 2y
(v) 2x², − 3x², 7x²
(vi) 7xyz, − 5xyz, 9xyz, −8xyz
(vii) 6a³, − 4a³, 10a³, −8a³
(viii) x² − a², −5x² + 2a², −4x² + 4a²

ANSWER:

  (i) Required sum = 3x + 7x
     = (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
     = (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
    = (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x² + (− 3x²) + 7x²
    =(2-3+7)x² = 6x²

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
       = (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a³ +(− 4a³) + 10a³ +( −8a³)
     =(6-4+10-8)a³ = 4a³
 
(viii) Required sum = x² − a² + (−5x² + 2a²) +( −4x² + 4a² )
       Rearranging and collecting the like terms =  x² -5x² -4x² -a² + 2a² +4a²
      = (1-5-4)x² +(-1+2+4)a²
     = -8x² + 5a²

Page No 134:

Question 2:

Add the following:
(i)    x − 3y − 2z  5x + 7y − z− 7x − 2y + 4z                            x – 3y – 2z  5x + 7y – z- 7x – 2y + 4z                         

(ii)    m2 − 4m + 5− 2m2 + 6m − 6  − m2 − 2m − 7                               m2 – 4m + 5- 2m2 + 6m – 6  – m2 – 2m – 7                            

(iii)   2×2 − 3xy + y2− 7×2 − 5xy − 2y2  4×2 + xy − 6y2                                     2×2 – 3xy + y2- 7×2 – 5xy – 2y2  4×2 + xy – 6y2                                   

(iv)   4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx                               4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx                             

ANSWER:

(i)
     x −  3y    −2z  5 x + 7y    −  z−7x − 2y + 4z−x   +2y   +  z     x –  3y    -2z  5 x + 7y    –  z-7x – 2y + 4z-x   +2y   +  z

(ii)
          m2 − 4m + 5   − 2m2 + 6m − 6     − m2 − 2m − 7       −2m2  +0×m−8= −2m2  + 0 −8 = −2m2−8          m2 – 4m + 5   – 2m2 + 6m – 6     – m2 – 2m – 7       -2m2  +0×m-8= -2m2  + 0 -8 = -2m2-8

(iii)
     2×2 − 3xy   + y2− 7×2 − 5xy − 2y2     4×2 + xy  − 6y2                                      −x2−7xy  −7y2     2×2 – 3xy   + y2- 7×2 – 5xy – 2y2     4×2 + xy  – 6y2                                      -x2-7xy  -7y2

(iv)
     4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx−3xy   −6yz −3zx     4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx-3xy   -6yz -3zx

Page No 134:

Question 3:

Add:
(i) 3a − 2b + 5c, 2a + 5b − 7c, − a − b + c
(ii) 8a − 6ab + 5b, −6a − ab − 8b, −4a + 2ab + 3b
(iii) 2x³ − 3x² + 7x − 8, −5x³ + 2x² − 4x + 1, 3 − 6x + 5x² − x³
(iv) 2x² − 8xy + 7y² − 8xy², 2xy² + 6xy − y² + 3x², 4y² − xy − x² + xy²
(v) x³ + y³ − z³ + 3xyz, − x³ + y³ + z³ − 6xyz, x³ − y³ − z³ − 8xyz
(vi) 2 + x − x² + 6x³, −6 −2x + 4x² −3x³, 2 + x², 3 − x³ + 4x − 2x²

ANSWER:

(i) 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) Sum of the given expressions
 = (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
  =(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
  = -2a-5ab+0 = -2a – 5ab

(iii) Sum of the given expressions
    = (2x³ − 3x² + 7x − 8) + (−5x³ + 2x² − 4x + 1) + ( 3 − 6x + 5x² − x³ )
      Rearranging and collecting the like terms
     =2x³−5x³ − x³ − 3x² + 2x² + 5x² +7x-4x-6x-8+1+3
     = (2-5-1)x³ +(-3+2+5)x²+(7-4-6)x-4
     = -4x³ +4x²-3x-4


(iv) Sum of the given expressions
= (2x² − 8xy + 7y² − 8xy²)+( 2xy² + 6xy − y² + 3x²)+( 4y² − xy − x² + xy² )
     Rearranging and collecting the like terms
  = 2x² +3x² − x²   + 7y² − y² +4y² − 8xy + 6xy − xy− 8xy² +2xy² + xy²
  = (2 +3− 1)x²   + (7 − 1 +4)y² + (-8 + 6 −1)xy + (− 8 +2 +1)xy²
 = 4x²   + 10y² − 3xy -5xy²

(v) Sum of the given expressions
 = (x³ + y³ − z³ + 3xyz)+(− x³ + y³ + z³ − 6xyz)+(x³ − y³ − z³ − 8xyz)
      Rearranging and collecting the like terms
   = x³ -x³ + x³ + y³ + y³ − y³ -z³ + z³ − z³ + 3xyz-6xyz-8xyz
  = (1-1+1)x³ + (1+1-1)y³ + (-1+1-1)z³ +(3-6-8)xyz
  = x³ + y³ − z³ -11xyz

(vi) Sum of the given expressions
  = (2 + x − x² + 6x³)+(−6 −2x + 4x² −3x³)+( 2 + x²)+( 3 − x³ + 4x − 2x² )
   Rearranging and collecting the like terms
  = 6x³ −3x³− x³− x² +4x²+ x²− 2x²+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x³+(-1+4+1-2)x²+(1-2+4)x+1
= 2x³+2x²+3x+1

Page No 135:

Question 4:

Subtract:
(i) 5x from 2x
(ii) −xy from 6xy
(iii) 3a from 5b
(iv) −7x from 9y
(v) 10x² from −7x²
(vi) a² − b² from b² − a²

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) 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)  Term to be subtracted = -xy
 Changing the sign of each term of the expression gives xy.
 On adding:

   6xy+xy
   = (6+1)xy
   = 7xy

(iii) Term to be subtracted = 3a
 Changing the sign of each term of the expression gives -3a.
 On adding:

   5b+(-3a)
   = 5b-3a

(iv) Term to be subtracted = -7x
 Changing the sign of each term of the expression gives 7x.
On adding:
9y+7x


(v) Term to be subtracted = 10x²
 Changing the sign of each term of the expression gives -10x².
On adding:
−7x² + (-10x²) = −7x² −10x²
   = (−7−10)x²
   = −17x²

(vi) Term to be subtracted = a² − b²
 Changing the sign of each term of the expression gives -a² + b².
On adding:
   b² − a² + (-a² + b²) = b² − a² -a² + b²
   = (1+1)b²  +(−1-1) a²
   = 2b² − 2a²

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Question 5:

Subtract:
(i) 5a + 7b − 2c from 3a − 7b + 4c
(ii) a − 2b − 3c from −2a + 5b − 4c
(iii) 5x² − 3xy + y² from 7x² − 2xy − 4y²
(iv) 6x³ − 7x² + 5x − 3 from 4 − 5x + 6x² − 8x³
(v) x³ + 2x²y + 6xy² − y³ from y³ − 3xy² − 4x²y
(vi) −11x²y² + 7xy −6 from 9x²y² −6xy + 9
(vii) −2a + b + 6d from 5a − 2b − 3c

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) 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

(ii) 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) Term to be subtracted = 5x² − 3xy + y²
 Changing the sign of each term of the expression gives  -5x² + 3xy – y².

   On adding:
(7x² − 2xy − 4y²)+(-5x² + 3xy – y²) = 7x² − 2xy − 4y²-5x² + 3xy – y²
   = (7-5)x² +(−2+3)xy +(−4-1)y²
   = 2x² +xy − 5y²


(iv) Term to be subtracted = 6x³ − 7x² + 5x − 3
 Changing the sign of each term of the expression gives  -6x³ + 7x² – 5x + 3.
  On adding:
  (4 − 5x + 6x² − 8x³)+(-6x³ + 7x² – 5x + 3) = 4 − 5x + 6x² − 8x³-6x³ + 7x² – 5x + 3
   = (-8-6)x³ +(6+7)x² +(-5- 5)x + 7
   = -14x³ + 13x² – 10x + 7

(v) Term to be subtracted = x³ + 2x²y + 6xy² − y³
 Changing the sign of each term of the expression gives  -x³ – 2x²y – 6xy² + y³.
  On adding:
(y³ − 3xy² − 4x²y)+(-x³ – 2x²y – 6xy² + y³) = y³ − 3xy² − 4x²y-x³ – 2x²y – 6xy² + y³
   = -x³ +(- 2-4)x²y +(-6-3)xy² + (1+1)y³
   = -x³ – 6x²y – 9xy² + 2y³

(vi) Term to be subtracted = −11x²y² + 7xy −6
 Changing the sign of each term of the expression gives  11x²y² -7xy +6.
  On adding:
(9x²y² −6xy + 9)+(11x²y² -7xy +6) = 9x²y² −6xy + 9+11x²y² -7xy +6
   = (9+11)x²y² (-7−6)xy + 15
   = 20x²y² −13xy +15

(vii) 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

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Question 6:

Simplify:
(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴

ANSWER:

(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
   Rearranging and collecting the like terms
 = (2-6)p³ +(−3+2)p² + (4-8+6)p − 5-2+8
 = -4p³ −p² +2p +1

(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
     Rearranging and collecting the like terms
  = (2+6)x² +(−1+5) xy + (6-4)x +(− 4+3)y
 = 8x² + 4xy + 2x − y

(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴
       Rearranging and collectingthe like terms
    = (1-1)x⁴ +(− 6+7)x³ + 5x² +(2-1)x-7+ 2
    = 0 +  x³ + 5x² +x-5
    = x³ + 5x² +x-5

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Question 7:

From the sum of 3x² − 5x + 2 and −5x² − 8x + 6, subtract 4x² − 9x + 7.

ANSWER:

Adding:
(3x² − 5x + 2) + (−5x² − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x² +(− 5-8)x + 2 +6
= -2x² − 13x + 8

Subtract 4x² − 9x + 7 from -2x² − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x² − 9x + 7
Changing the sign of each term of the expression gives -4x² + 9x – 7.
On adding:
   ( -2x² − 13x + 8 )+(-4x² + 9x – 7 )    = -2x² − 13x + 8 -4x² + 9x – 7
   = ( -2-4)x² +(−13+9)x + 8 -7
      = -6x² − 4x + 1

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Question 8:

If A = 7x² + 5xy − 9y², B = −4x² + xy + 5y² and C = 4y² − 3x² − 6xy then show that A + B + C = 0.

ANSWER:

A = 7x² + 5xy − 9y²
B = −4x² + xy + 5y² 
C = 4y² − 3x² − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x² + 5xy − 9y²)+(−4x² + xy + 5y²)+(4y² − 3x² − 6xy)
= 7x² + 5xy − 9y²−4x² + xy + 5y²+4y² − 3x² − 6xy

Rearranging and collecting the like terms:
(7-4-3)x² + (5+1-6)xy +(−9+5+4)y²
= (0)x² + (0)xy + (0)y²
= 0
⇒A+B+C=0⇒A+B+C=0

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Question 9:

What must be added to 5x³ − 2x² + 6x + 7 to make the sum x³ + 3x² − x + 1?

ANSWER:

Let the expression to be added be X.
(5x³ − 2x² + 6x + 7)+X = (x³ + 3x² − x + 1)
X = (x³ + 3x² − x + 1) – (5x³ − 2x² + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x³ + 3x² − x + 1) + (-5x³ + 2x² – 6x – 7)
X = x³ + 3x² − x + 1-5x³ + 2x² – 6x – 7

Rearranging and collecting the like terms:
X = (1-5)x³ + (3+2)x² +(−1-6) x + 1-7
X = -4x³ + 5x² − 7x -6

So, -4x³ + 5x² − 7x -6 must be added to 5x³ − 2x² + 6x + 7 to get the sum as x³ + 3x² − x + 1.

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Question 10:

Let      P = a² − b² + 2ab,      Q = a² + 4b² − 6ab,      R = b² + 6,      S = a² − 4ab      and      T = −2a² + b² − ab + a. Find P + Q + R + S − T.

ANSWER:

P = a² − b² + 2ab
Q = a² + 4b² − 6ab
R = b² + 6
S = a² − 4ab 
T = −2a² + b² − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a² − b² + 2ab)+(a² + 4b² − 6ab)+(b² + 6)+(a² − 4ab )
= a² − b² + 2ab+a² + 4b² − 6ab+b² + 6+a² − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a² +(−1+4+1) b² + (2-6-4)ab+6
P+Q+R+S = 3a² +4b² – 8ab+6

To find P + Q + R + S − T, subtract T = (−2a² + b² − ab + a) from P+Q+R+S = (3a² +4b² – 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a² + b² − ab + a
Changing the sign of each term of the expression gives 2a² – b² + ab – a.
 Now add:
(3a² +4b² – 8ab+6)+(2a² – b² + ab – a) = 3a² +4b² – 8ab+6+2a² – b² + ab – a
= (3+2)a² +(4-1) b² +(-8+1) ab – a+6
 
P + Q + R + S − T = 5a² +3b² -7 ab – a+6

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Question 11:

What must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1?

ANSWER:

Let the expression to be subtracted be X.
(a³ − 4a² + 5a − 6)-X = (a² − 2a + 1)
X = (a³ − 4a² + 5a − 6)- (a² − 2a + 1)
Since ‘-‘ sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a³ − 4a² + 5a − 6- a² + 2a – 1
Rearranging and collecting the like terms:
X = a³ +(− 4-1)a² + (5+2)a − 6 – 1
X = a³ −5a² + 7a − 7
So, a³ −5a² + 7a − 7 must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1.

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Question 12:

How much is a + 2a − 3c greater than 2a − 3b + c?

ANSWER:

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:
   (a + 2b − 3c )+(-2a + 3b – c )   
   = a + 2b − 3c -2a + 3b – c
   = (1-2)a + (2+3)b +(− 3-1)c
   = -a + 5b − 4c

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Question 13:

How much less than x − 2y + 3z is 2x − 4y − z?

ANSWER:

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 − 2y + 3z)+(-2x + 4y + z )   
   = x − 2y + 3z-2x + 4y + z 
   = (1-2)x +(−2+4)y + (3+1)z
   = -x + 2y + 4z

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Question 14:

By how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1?

ANSWER:

To calculate how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1, we have to subtract x³ − x² + 4x − 1 from 3x² − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = x³ − x² + 4x − 1
Changing the sign of each term of the expression gives -x³ + x² – 4x + 1.
On adding:
(3x² − 5x + 6)+(-x³ + x² – 4x + 1 )   
   = 3x² − 5x + 6-x³ + x² – 4x + 1
   = -x³ + (3+1)x² +(-5-4)x+6 + 1
   = -x³ +4 x² – 9x + 7

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Question 15:

Subtract the sum of 5x − 4y + 6z and −8x + y − 2z from the sum of 12x − y + 3z and −3x + 5y − 8z.

ANSWER:

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -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 +4y -5z)+(3x + 3y – 4z )   
   = 9x +4y -5z+3x + 3y – 4z   
   = (9+3)x +(4+3)y + (-5-4)z
  = 12x +7y -9z

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Question 16:

By how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2?

ANSWER:

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.

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 − 3y + 4z )+(-2x – 5y + 6z – 2 )   
   = 2x − 3y + 4z-2x – 5y + 6z – 2   
   = (2-2)x + (-3-5)y +(4+6)z-2
   = 0-8y+10z-2
   = -8y+10z-2

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Question 17:

By how much does 1 exceed 2x − 3y − 4?

ANSWER:

To calculate how much does 1 exceed 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)+(-2x+3y+4 )   
   = 1-2x+3y+4
   = 5-2x+3y

Page No 136:

Exercise 8C

Question 1:

Simplify:
a − (b − 2a)

ANSWER:

a – (b – 2a)
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

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Question 2:

Simplify:
4x − (3y − x + 2z)

ANSWER:

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

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Question 3:

Simplify:
(a² + b² + 2ab) − (a² + b² −2ab)

ANSWER:

(a² + b² + 2ab) − (a² + b² − 2ab)
Here, ‘−’ sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
 a² + b² + 2ab − a² − b² +2ab

Rearranging and collecting the like terms:
a² − a² +b² − b² + 2ab + 2ab
=(1 − 1)a² + (1− 1)b² + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

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Question 4:

Simplify:
−3(a + b) + 4(2a − 3b) − (2a − b)

ANSWER:

−3(a + b) + 4(2a − 3b) − (2a − b)
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 + (4××2a )−(4××3b) − 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

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Question 5:

Simplify:
−4x² + {(2x² − 3) − (4 − 3x²)}

ANSWER:

−4x² + {(2x² − 3) − (4 − 3x²)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x² + {(2x² − 3) − (4 − 3x²)}
= −4x² + {2x² − 3 − 4 + 3x²}
= −4x² + {5x² − 7}
= −4x² + 5x² − 7
= x² − 7

Page No 136:

Question 6:

Simplify:
−2(x² − y² + xy) −3(x² + y² − xy)

ANSWER:

−2(x² − y² + xy) −3(x² + y² − xy)
Here a ‘−’ sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x² +2y² − 2xy −3x² − 3y² + 3xy
= (−2 − 3)x² +(2 − 3)y² + (− 2 + 3)xy
= −5x² − y² + xy

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Question 7:

Simplify:
a − [2b − {3a − (2b − 3c)}]

ANSWER:

a − [2b − {3a − (2b − 3c)}]
We 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 + 3c

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Question 8:

Simplify:
−x + [5y − {x − (5y − 2x)}]

ANSWER:

−x + [5y − {x − (5y − 2x)}]
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]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

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Question 9:

Simplify:
86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]

ANSWER:

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

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Question 10:

Simplify:
12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]

ANSWER:

12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + x³+ 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + 7x³} − 3x]
= 12x − [3x³ + 5x² − 7x² + 4 − 3x − 7x³ − 3x]
= 12x − [ − 2x² + 4 − 4x³ − 6x]
= 12x + 2x² − 4 + 4x³ + 6x
= 4x³ + 2x² +18x-4

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Question 11:

Simplify:
5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a

ANSWER:

5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
= 5a − [a² − {2a − 2a² + 8a³ − 3a³ + 15a² + 9a}] −8a
= 5a − [a² − {5a³ + 13a² + 11a}] − 8a
= 5a − [a² − 5a³ − 13a² −11a] − 8a
= 5a − [ − 5a³ − 12a² − 11a] − 8a
= 5a + 5a³ + 12a² + 11a − 8a
= 5a³ + 12a² + 8a

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Question 12:

Simplify:
3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

ANSWER:

3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]
= 3 − [x − {2y − 5x – y + 3 + 2x²} − x² + 3y]
= 3 − [x − {y − 5x + 3 + 2x²} − x² + 3y]
= 3 − [x − y + 5x − 3 − 2x² − x² + 3y]
= 3 − [ 6x − 3 − 3x² + 2y]
= 3 − 6x + 3 + 3x² − 2y
= 3x² − 2y − 6x+6

Page No 137:

Question 13:

Simplify:
xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

ANSWER:

xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]
= xy − [yz − zx − {yx − 3y + xz − xy + zy}]
= xy − [yz − zx − {− 3y + xz + zy}]  (∵xy=yx)(∵xy=yx)
= xy − [yz − zx + 3y – xz – zy] 
= xy − [− 2zx + 3y ] (∵ yz=zy, zx=xz)(∵ yz=zy, zx=xz)
= xy + 2zx − 3y 

Page No 137:

Question 14:

Simplify:
2a − 3b − [3a − 2b − {a − c − (a − 2b)}]

ANSWER:

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
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 − c

Page No 137:

Question 15:

Simplify:
−a − [a + {a + b − 2a − (a − 2b)} − b]

ANSWER:

-a − [a + {a + b − 2a − (a − 2b)} − b]
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

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Question 16:

Simplify:
2a − [4b − {4a − (3b − 2a + 2b)}]2a – [4b – {4a – (3b – 2a + 2b)}]

ANSWER:

2a-[4b-{4a-(3b-2a+2b¯¯¯¯¯¯¯¯¯)}]2a-[4b-{4a-(3b-2a+2b¯)}]
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-(3b-2a+2b¯)}]
= 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

Page No 137:

Question 17:

Simplify:
5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]

ANSWER:

5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
= 5x − [4y − {7x − 3z + 2y + 4z − 3x − 9y + 6z}]
= 5x − [4y − {4x + 7z − 7y}]
= 5x − [4y − 4x − 7z + 7y]
= 5x − [11y − 4x − 7z ]
= 5x − 11y + 4x + 7z
= 9x − 11y + 7z