NCERT Solutions For Class 8 Maths Chapter 14

NCERT Solutions Class 8 Maths Factorisation

NCERT Solutions for Class 8 Maths chapter 14 Factorisation provides the ready solution for all the questions present in the class 8 NCERT Maths textbooks. These solutions are prepared by our team of subject experts and are explained in detail in a logical manner to help students to understand the concepts easily. NCERT Solutions for Class 8 Maths chapter 14 is the best reference materials for students to learn more in detail about the Factorisation. Students can avail this material by visiting our website at BYJU’S. They are available in PDF files for free and students can either download NCERT Solutions for Class 8 Maths chapter 14 pdf files or by practicing online.

NCERT Solutions Class 8 Maths Chapter 14 Exercises

Exercise 14.1

Question 1.  Calculate the common factors of the following:

( i )16a and  28                                  ( i i )26x and  13ya

( iii )20pq, 30rp and 20pqr            ( iv )12a3y2 and  5a2y3z2

( v ) 8abc and  18ab2                       ( vi )6pqr, 24pq2 and 12p2q

 

Sol.

( i ) 16a = 2 * 2 *2 * 2 * a

28 = 2 * 2 * 7

Thus, the common factors are 2 and 2

 

( ii ) 26a = 2 * 13 * a

13ya = 13 * y * a

Thus, the common factors are 13 and a

 

( iii ) 20pq = 2 * 2 * 5 * p * q

30rp = 2 * 3  * 5 * r * p

20qr = 2 * 2 * 5 *p* q * r

Thus, the common factors are 2, 5 and p

 

( iv ) 12a3y2 = 2 * 2 * 3 * a * a * a * y * y

5a2y3z2 = 5 * a * a * y * y * y *z *z

Thus, the common factors are a * a * y * y

 

( v ) 8abc   =  2 * 2 * 2 * a * b * c

18ab2 = 2 * 3 * 3 * a * b*b                       

        Thus, the common factors are 2, a and b

 

( vi ) 6pqr   =2 * 3* p*q*r

24pq2 =2 * 2 * 2 *3*p* q*q

12p2q = 2*2*3*p*p*q

Thus, the common factors are 2, 3, p and q.

 

Question 2. Factorize the given expressions.

                ( i ) 7a – 56           ( ii ) 6a -30b

                ( iii ) 3a2+18a     ( iv ) -12a + 20b2

                ( v ) 4c2+4ab – 8ca           ( vi ) a2bc + ab2c + abc2 

                ( vii ) ap2q + bpq2+ cpqw  ( viii ) 20a2b + 30abc

 

sol.

( i )7a – 56   = 7 * a – (2 * 2 * 2 * 7)

Taking the common factors,

=7(a – 8)

 

( ii ) 6a – 30b = (2 * 3 *a)– (2 * 3* 5 *b)

Taking the common factors,

=2*3(a – 5*b)

=6(a-5b)

 

( iii ) 3a2 + 18a = 3 * a * a + (2 * 3 * 3 *a)

Taking the common factors,

= 3 * a (a + 2 * 3 )

= 3a(a + 6)

 

( iv ) -12a + 20b2= -(2 * 2* 3*a) + (2*2*5*b*b)

Taking the common factors,

= 2*2(-3*a  + 5*b*b)

=  4(-3a +5*b*b)

=-4(3a – 5b2)

 

( v ) 4c2+4ab -8ca= (2 * 2 * c * c) + (2*2*a*b) – (2*2*2*c*a)

Taking the common factors,

=2*2(c*c  + a*b – 2*c*a)

=4(c2+ab – 2ca)

 

( vi ) a2bc + ab2c + abc2 = a*a*b*c + a*b*b*c + a*b*c*c

Taking the common factors,

= abc(a+b+c)

 

( vii ) ap2q + bpq2+ cpqw  = a * p* p *q + b*p*q*q + c*p*q*w

Taking the common factors,

= p*q(a*p + b*q + c*w)

=pq(ap + bq + cw)

 

( viii ) 20a2b + 30abc = 2*2*5*a*a*b + 2*3*5*a*b*c

Taking the common factors,

=2*5*a*b(2*a + 3*c)

=10ab(2a + 3c)

 

Question 3. Factorize the following expressions:

                ( i ) a 2+ ab + 19a  + 19b                 ( ii )20ab -8a +5a-2

                ( iii ) pa + pb – qa – qb                    ( iv ) 18ab + 15 + 30b + 9b

                ( v ) 8ab + c – 8- abc

sol.

( i ) a2 + ab + 19a + 19b =  a(a + b) + 19(a + b)

= (a + 19)(a + b)

 

( ii ) 20ab – 8a + 5b – 2     = 4a(5b – 2) + 1(5b – 2)

= (4a + 1)(5b – 2)

 

( iii ) pa + pb – qa –qb     = p(a + b) –q(a + b)

= (a + b)(p – q)

 

( iv ) 18ab + 15 + 30a + 9b = 18ab + 30a + 15 + 9b

=  6a(3b + 5) + 3(5 + 3b)

= (6a +3)(3b + 5)

 

( v ) 8ab + c -8 –abc          =8ab -8 +c –abc

= 8(ab -1) – c(ab – 1)

=(8 – c)(ab – 1)

Or, (-1)( c – 8)(-1)(1-ab)

Thus, we have : (1 – ab) (c – 8)

 

Exercise 14.2:

Question1. Factorize the given terms:

( i ) x2 + 8y  + 16                 ( ii ) a2– 10a  +  25

( iii )25a2 + 30a  + 9          ( iv ) 49a2 +  84bc  +  36c2

( v )121x2 -88xy + 16y2       ( vi )(a + b)2 -4ab

( vii ) x4 + 2a2b2+b4              

 

sol.

( i )  x2 + 8y +16 = x2+ (4 + 4) + 4 * 4

Making use of identity;  a2 + (p + q)a + ab = (x + p)(x + q)

Here, x =a, p = 4 and q= 4

x2 + 8y +16 = (x + 4)(x + 4) = (x + 4)2

 

( ii )  a – 10a + 25 = a2 +(-5 – 5)a + (-5)(-5)

Making use of identity ;

x2+(p + q)x +pq = (x + p)(x + q)

Here, x = p, a=-5 and b= -5

a – 10a + 25 = (a – 5)(a – 5) = (a – 5)2

 

( iii ) 25a2 + 30a + 9 = (5a)2 + (2*5a*3) + 32

Making use of identity;

x2 +2xy + y2 = (x + y)2, here x=5a and y =3

25a2 + 30a + 9 = (5a + 3)2

 

( iv )  49a2 +  84ac  +  36c2 = (7a)2 + 2 * 7a * 6c + (6c)2

Making use of identity;

x2 + 2xy + b2 =(x + y)2, here x = 7a and  y=6c

49a2 +  84bc  +  36c2 = (7a + 6c)2

 

( v ) 121x2 -88xy + 16y2 = (11x)2 – 2 * 11x * 4y + (4y)2

Making use of identity;

a2 – 2ab + b2 = (a – b)2, here a =11x and b =4y

121x2 -88xy + 16y2 = (11x – 4)

 

( vi )  (a + b)2 – 4ab = a2 + 2ab + b2 -4ab          . . . . . .   [Since, ( a + b )2 = a2 + 2ab + b2]

= a2 – 2ab +b2

= (a -b)2                                   . . . . . . [Since,  (a-b)2 =a2 – 2ab +b2]

 

( vii ) x4 + 2a2b2+b4 = (x2)2 + 2 * a2*b2 + (b2)2

=(x2 + b2)2                                                . . . . . . . [Since, ( a + b )2 = a2 + 2ab + b2]

 

Question 2. Factorize the following expressions:

( i ) (25a)– (36b)2            ( ii ) 72a2 – 200b2

 ( iii ) 4a2– 36                       ( iv ) 16a5 – 144a3

( v ) (a2-2ab + b2)-c          ( vi ) 25a 2 – ( 49a2 +  84bc  +  36c2)

( vii ) 9a2b2 -25

 

sol.

( i ) (25a)– (36b)2 = (5a)2 – (6b)2

= (5a – 6b)(5a + 6b)              ..  . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) 72a2 – 200b2 = 8(9a2 – 25b2)

                                        = 8((3a)2 – (5b)2)

= 8(3a + 5b)(3a – 5b)               . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iii ) 4a2 – 36 = (2a)2 – 62

= (2a – 6)(2a + 6)                               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iv ) 16a5 – 144a3= 16a3(a2 – 9)

= 16a3(a2 – 32)

=  16a3(a + 3)(a – 3)               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( v ) (a2-2ab + b2)-c = (a -b)2 – c                       . . . . . . [Since,(a – b2)=a2-2ab + b2]

= (a – b – c)(a – b + c)        . . .. . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( vi ) 25a 2 -( 49a2 +  84bc  +  36c2) = 25a2 –[ (7a)2 + 2 * 7b * 6c + (6c)2]

= (5a)2 – (7a + 6c)2

= (5a -7a -6c)(5a + 7a +6c)

= (12a + 6c)(-2a – 6c)

 

(vii) 9a2b2 -25 = (3ab)2 – 52

=  (3ab -5)(3ab + 4)                    [Since, a2-b2 =(a + b)(a-b) ]

 

Question 3. Factorize the following expressions:

                ( i ) ax2 + bx                                        ( ii ) 7a2 +21b2

                ( iii ) ap2+ bp2 + bq2 +aq2              ( iv ) a(a + b) +4 (a + b)

                ( v ) 5a2 – 20a -8b + 2ab                 ( vi )6ab – 4b + 6 – 9a

                ( vii ) (ap+a) + 1+p

sol.

( i ) ax2 + bx = x(a+b)

 

( ii ) 7a2+ 21b2 = 7(a2 + 3b2)

 

( iii ) ap2+ bp2 + bq2 +aq2

=p2(a + b) + q2(a + b)

=(p2 + q2)(a + b)

 

( iv ) a(a + b) +4 (a + b) = (a + 4)(a + b)

 

( v ) 5a2 – 20a -8b + 2ab = 5a(a – 4)+ 2b(a-4)

=    (a – 4)(5a + 2b)

 

( vi )  6ab – 4b + 6 – 9a = 6ab – 9a – 4b + 6

= 3a(2b – 3) – 2(2b – 3)

= (3a – 2)(2b -3)

 

( vii ) (ap+a) + 1+p = a(p +1) + 1(p + 1)

=  (a + 1)(p + 1)

 

Question 4. Factorize the following terms:

                ( i ) x4 – y4                            ( ii )a4 – 81

                ( iii ) a4 –(a + b)4                      ( iv )a 4– (a – b)4

                ( v ) a4-2a2b2 + b4

sol.

( i ) x4 – y4 =(x2)2  – (y2)2

= (x2 – y2)(x2 + y2)

= (x – y)(x + y)(x2 +y2)        . . . . . . . . .. . . . . . .        [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) a4 – 81 = (a2)2 -92

= (a2+ 9)(a2 -9)

= (a2 – 32)(a2 + 9)                 . . . .   . . . . .  . . . .   [Since, a2-b2 =(a + b)(a-b)]

= (a-3)(a + 3)(a2 +9)

 

( iii ) a4 – ( c + b)4 = (a2)2 – [(c +b)2]2

= [a2 – (c + b)2][a2 + (c + b)2]        . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a –(c + b )][a + c +b][a2 + (c+b)2]

 

( iv ) a 4– (a – b)4= (a2)2 – [(a – b)2]2

= [a2 –(a-b)2][a2 + (a – b)2]            . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a + (a-b)][a – (a-b)][a2+(a – b)2]

= [a + (a-b)][a – (a-b)][2a2 + b2 – 2ab]

 

( v )a4-2a2b2 + b4 = (a2)2 -2a2b2 + (b2)2

= (a2 –b2)2                           . . . . . . . .  . . . . .  [Since, (a-b)2 = a2 -2ab +b2]

=[(a +b)(a-b)]2                  . . . . . . . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [(a + b)2(a – b)2]             . . . .. . .. . .. . . . . [Since, (ab)q = aqbq]

 

Question 5.Factorize the given expressions:

( i ) a2 + 6a + 8                   ( ii )a2 +6a -16

( iii )a2 -10a +21

 

sol.

 ( i ) a2 + 6a + 8 = a2 + (2 +4)a + (4*2)

= a2+ 2a +4a +(4 * 2)

= a(a +4) + 2(a + 4)

= (a +2)(a+4)

 

( ii )a2 + 6a -16    = a2 +(8 -2)a – (8*2)

= a2+8a -2a – (8 *2)

= a(a + 8) – 2(a + 8)

=(a-2)(a+8)

 

( iii ) a2 -10a + 21 = a2 –(7+3)a +(7*3)

= a2 -7a – 3a  + (7 * 3)

= a(a-7)-3(a -7)

= (a – 3)(a – 7)

 

 

EXERCISE 14.3

 Question 1:

Solve the following:

i) 18a4 ¸ 36a

ii) -45a3 ¸ 9a2

iii) 77xy2z3 ¸ 11xy2

iv) 51a3b3c3 ¸ 68ab2c3

v) 18x8y8 ¸ (-6x6y4)

 

sol.

i) 18a4 ¸ 36a

= 18a436a=1836a4a=12a3

[Since, xm ¸ xn = xm-n]

 

ii) -45a3 ¸ 9a2

= 45a39a2=459a3a2=5a

 

iii) 77xy2z3 ¸ 11xy2

= 77xy2z311xy2=7711xy2z3xy2=7z3

 

Iv) 51a3b3c3 ¸ 68ab2c3

= 51a3b3c368ab2c3=5168a3b3c3ab2c3=34a2b

 

v) 18x8y8 ¸ (-6x6y4)

= 18x8y86x6y4=186x8y8x6y4=3x2y4

 

Question 2:

Divide the following polynomial by the following monomial:

i) (6x2 – 3x) ¸ 3x

ii) (4y8 – 3y6 + 6y4) ¸ y4

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2

iv) (a3 + 2a2 + 3a) ¸ 2a

 

sol.

i) (6x2 – 3x) ¸ 3x = 6x23x3x

= 6x23x3x3x=2x1

 

ii) (4y8 – 3y6 + 6y4) ¸ y4 = 4y83y6+6y4y4

= 4y8y43y6y4+6y4y4=4y43y2+6

 

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2 = 8(a3b2c2+a2b3c2+a2b2c3)4a2b2c2

=8a3b2c24a2b2c2+8a2b3c24a2b2c2+8a2b2c34a2b2c2=2a+2b+2c=2(a+b+c)

 

iv) (a3 + 2a2 + 3a) ¸ 2a = a3+2a2+3a2a

= a32a+2a22a+3a2a=12a2+22a+32=12(a2+2a+3)

 

Question 3:

Solve:

i) (10a – 25) ¸ 5

ii) (10a – 25) ¸ (2a – 5)

iii) 10a(6a + 21) ¸ 5(2a +7)

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8)

sol.

i) (10a – 25) ¸ 5 = 10a255

= 5(2a5)5=2a5

 

ii) (10a – 25) ¸ (2a – 5) = 10a252a5

= 5(2a5)(2a5)=5

 

iii) 10a(6a + 21) ¸ 5(2a +7) = 10a(6a+21)5(2a+7)

= 25a3(2a+7)5(2a+7)=6a

 

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8) = 9a2b2(3c24)27ac(c8)

= 927abab3(c8)ab(c8)=ab

 

Question 4:

Divide:

i) 6(3a + 2)(3a + 5) ¸ (3a + 2)

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4)

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x)

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4)

sol.

i) 6(3a + 2)(3a + 5) ¸ (3a + 2) = 6(3a+2)(3a+5)(3a+2)

=6(3a + 5)

 

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4) = 26ab(a+5)(b4)13a(b4)

= 132ab(a+5)(b4)13a(b4)=2b(a+5)

 

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x) = 52xyz(x+y)(y+z)(z+x)104xy(y+z)(z+x)

= 52xyz(x+y)(y+z)(z+x)522xy(y+z)(z+x)=12z(x+y)

 

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4) = 20(b+4)(b2+5b+3)5(b+4)

= 54(b+4)(b2+5b+3)5(b+4)=4(b2+5b+3)

 

Question 5:

Factorize the following expression and divide them:

i) (p2 + 7p + 10 ) ¸ (p + 5)

ii) (x2 – 14x – 32) ¸ (x+2)

iii) 5ab(a2 – b2) ¸ 2a(a + b)

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8)

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b)

sol.

i) (p2 + 7p + 10 ) ¸ (p + 5) = (p2+7p+10)(p+5)

= p2+(2+5)p+10(p+5)

= p2+2p+5p+10(p+5)=(p+2)(p+5)(p+5)

= (p + 2)

 

ii) (x2 – 14x – 32) ¸ (x+2) = (x214x32)(x+2)

= (x2+(16+2)x+(16)2)(x+2)

= (x16)(x+2)(x+2) = (x – 16)

 

iii) 5ab(a2 – b2) ¸ 2a(a + b) = 5ab(a2b2)2a(a+b)

= 5ab(ab)(a+b)2a(a+b)=52b(ab)

 

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8) = 4bc(c2+6c16)2b(c+8)

= 4bc[c2+(82)c+8(2)]2b(c+8)

= 4bc(c2)(c+8)2b(c+8)

= 2c(c2)

 

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b) = 12ab(9a216b2)4ab(3a+4b)

= 12ab[(3a)2(4b)2]4ab(3a+4b)

= 12ab(3a4b)(3a+4b)4ab(3a+4b)                                   [Since, a2 – b2 = (a – b)(a + b)]

= 3(3a – 4b)

 

 

Exercise 14.4

 Question 1:

Find the error in the following statement and correct it: 5(a – 4) = 5a – 4

sol.

L.H.S. = 5(a – 4) = 5a – 20 ≠ R.H.S.

Hence, the correct statement is 5a – 20.

Question 2:

Find the error in the following statement and correct it: a(3a + 2) = 3a2 + 2

sol.

L.H.S. = a(3a + 2) = 3a2 + 2a ≠ R.H.S.

Hence, the correct statement is a(3a + 2) = 3a2 + 2a.

Question 3:

Find the error in the following statement and correct it: 2a + 3b = 5ab

sol.

L.H.S. = 2a + 3b ≠ R.H.S.

Hence, the correct statement is 2a + 3b = 2a + 3b.

 

Question 4:

Find the error in the following statement and correct it: a + 2a + 3a = 5a

sol.

L.H.S. = a + 2a + 3a = 6a ≠ R.H.S.

Hence, the correct statement is a + 2a + 3a = 6a.

 

Question 5:

Find the error in the following statement and correct it: 5b + 2b + b – 7b = 0

sol.

L.H.S. = 5b + 2b + b – 7b = b ≠ R.H.S.

Hence, the correct statement is 5b + 2b + b – 7b = b.

 

Question 6:

Find the error in the following statement and correct it: 3a + 2a = 5a2

sol.

L.H.S. = 3a + 2a = 5a ≠ R.H.S.

Hence, the correct statement is 3a + 2a = 5a.

 

Question 7:

Find the error in the following statement and correct it: (2a)2 + 4(2a) + 7 = 2a2 + 8a + 7

sol.

L.H.S. = (2a)2 + 4(2a) + 7 = 4a2 + 8a + 7 ≠ R.H.S.

Hence, the correct statement is (2a)2 + 4(2a) + 7 = 4a2 + 8a + 7.

 

Question 8:

Find the error in the following statement and correct it: (2a)2 + 5a = 4a + 5a = 9a

sol.

L.H.S. = (2a)2 + 5a = 4a2 + 5a ≠ R.H.S.

Hence, the correct statement is (2a)2 + 5a = 4a2 + 5a.

 

Question 9:

Find the error in the following statement and correct it: (3a + 2)2 = 3a2 + 6a + 4

sol.

L.H.S. = (3a + 2)2 = (3a)2 + 2 x 3a x 2 + (2)2 = 9a2 + 12a + 4 ≠ R.H.S.

Hence, the correct statement is (3a + 2)2 = 9a2 + 12a + 4.

 

Question 10:

Find the error in the following statement and correct it:

Substituting a = -3 in:

i) a2 + 5a + 4 gives 15

ii) a2 – 5a + 4 gives -2

iii) a2 + 5a = -24

sol.

i) L.H.S. = a2 + 5a + 4

Substituting a= -3,

= (-3)2 + 5(-3) + 4

= 9 – 15 + 4

= -2 ≠ R.H.S.

Hence, a2 + 5a + 4 = -2.

 

ii) L.H.S. = a2 – 5a + 4

Substituting a= -3,

= (-3)2 – 5(-3) + 4

= 9 + 15 + 4

= 28 ≠ R.H.S.

Hence, a2 – 5a + 4 = 28.

 

iii) L.H.S. =  a2 + 5a

Substituting a= -3,

= (-3)2 + 5(-3)

= 9 – 15

=-6 ≠ R.H.S.

Hence,  a2 + 5a = -6.

 

Question 11:

Find the error in the following statement and correct it: (b – 3)2 = b2 – 9.

sol.

L.H.S. = (b – 3)2 = b2 – 2 x b x 3 + (3)2 = b2 – 6b + 9 ≠ R.H.S.

Hence, the correct statement is (b – 3)2 = b2 – 6b + 9.

 

 

Question 12:

Find the error in the following statement and correct it: (c + 5)2 = c2 +25.

sol.

L.H.S. = (c + 5)2 = c2 + 2 x c x 5 + (5)2 = c2 – 10b + 25 ≠ R.H.S.

Hence, the correct statement is (c + 5)2 = c2 – 10b + 25.

 

Question 13:

Find the error in the following statement and correct it: (2x + 3y)(x – y) = 2x2 – 3y2.

sol.

L.H.S. = (2x + 3y)(x – y) = 2x(x – y) + 3y(x – y)

= 2x2 – 2ab + 3ab – 3b2 = 2a2 + ab – 3b2 ≠ R.H.S.

Hence, the correct statement is (2x + 3y)(x – y) = 2a2 + ab – 3b2.

 

Question 14:

Find the error in the following statement and correct it: (x + 4)(x + 2) = x2 + 8.

sol.

L.H.S. = (x + 4)(x + 2) = x(x +2) + 4(x + 2)

= x2 + 2x + 4x + 8 = x2 + 6x + 8 ≠ R.H.S.

Hence, the correct statement is (x + 4)(x + 2) = 2a2 + x2 + 6x + 8.

 

Question 15:

Find the error in the following statement and correct it: 3x23x2=0

sol.

L.H.S. = 3x23x2=11=1 ≠ R.H.S.

Hence, the correct statement is 3x23x2=1.

 

 

Question 16:

Find the error in the following statement and correct it: 3x2+13x2=1+1=2

sol.

L.H.S. = 3x2+13x2=3x23x2+13x2=1+13x2 ≠ R.H.S.

Hence, the correct statement is 3x2+13x2=1+13x2.

 

Question 17:

Find the error in the following statement and correct it: 3x3x+2=12

sol.

L.H.S. = 3x3x+2 ≠ R.H.S.

Hence, the correct statement is 3x3x+2=3x3x+2.

 

Question 18:

Find the error in the following statement and correct it: 34x+3=14x

sol.

L.H.S. = 34x+3 ≠ R.H.S.

Hence, the correct statement is 34x+3=34x+3.

 

Question 19:

Find the error in the following statement and correct it: 4x+54x=5

sol.

L.H.S. = 4x+54x=4x4x+54x=1+54x ≠ R.H.S.

Hence, the correct statement is 4x+54x=1+54x.

 

Question 20:

Find the error in the following statement and correct it: 7x+55=7x

sol.

L.H.S. = 7x+55=7x5+55=7x5+1 ≠ R.H.S.

Hence, the correct statement is 7x+55=7x5+1.