NCERT Solutions for 7th Class Maths: Chapter 12-Algebraic Expressions
NCERT Solutions for 7th Class Maths: Chapter 12-Algebraic Expressions

Class 7: Maths Chapter 12 solutions. Complete Class 7 Maths Chapter 12 Notes.

NCERT Solutions for 7th Class Maths: Chapter 12-Algebraic Expressions

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

Exercise 12.1

1. Get the algebraic expressions in the following cases using variables, constants, and arithmetic operations: 

(i) Subtraction of z from y.

(ii) One-half of the sum of numbers x and y.

(iii) The number z multiplied by itself.

(iv) One-fourth of the product of numbers p and q.

(v) Numbers x and y both squared and added.

(vi) Number 5 added to three times the product of m and n.

(vii) A product of numbers y and z subtracted from 10.

(viii) Sum of numbers a and b subtracted from their product.

Answer

(i) y – z

(ii) (x + y)/2

(iii) z2

(iv) pq/4

(v) x2 + y2

(vi) 3mn + 5

(vii) 10 – yz

(viii) ab – (a + b) 

2. (i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram: 

(a) x – 3

(b) 1 + x + x2

(c) y – y3

(d) 5xy2 + 7x2y

(e) -ab + 2b2 – 3a2

Answer

(ii) Identify the terms and factors in the expressions given below: 

(a) -4x + 5

(b) -4x + 5y 

(c) 5y + 3y2

(d) xy + 2x2y2

(e) pq + q

(f) 1.2ab – 2.4b + 3.6a

(a) -4x+ 5

Terms: -4x,5

Factors: -4,x ; 5

(b) -4x + 5y

Terms: -4x, 5y

Factors: -4,x ; 5,y

(c) 5y + 3y2

Terms: 5y,3y2

Factors: 5, y ; 3,y,y

(d) xy+2x2y2

Terms: xy,2x2y2 

Factors: x,y ; 2x,x,y,y

(e) pq+q

Terms: pq,q

Factors: p,q ; q

(f) 1.2ab-2.4b+3.6a

Terms: 1,2ab.-2.4b,3 6a

Factors: 1.2.a.b ; -2.4,6 ; 3.6,a

(h) 0.1p2+0.2q2

Terms: 0.1 p2,0.2q2

Factors: 0. 1,p,p, ; 0.2, q,q

3. Identify the numerical coefficients of terms (other than constants) in the following expressions: 

(i) 5 – 3t2

(ii) 1 + t + t2 + t2

(iii) x + 2xy + 3y

(iv) 100m + 1000n

(v) -p2q2 + 7pq

(vi) 1.2a + 0.8b

(vii) 3.14 r2

(viii) 2(l+b)

(ix) 0.1y + 0.01y2

Answer

S.No.ExpressionTermsNumerical Coefficient
(i)5-3t2-3t1-3
(ii)1+t+t2+t3t1
t21
t31
(iii)x + 2xy + 3yx1
2xy2
3y3
(iv)100m+1000n100 m100
1000 n1000
(v)-p2q2+7 pq-p2q2-1
7 pq7
(vi)1.2a+0.8b1.2 a1.2
0.8b0.8
(vii)3.14 r23.14 r23.14
(viii)2 (l + b) = 2l+ 2b2l2
2b2
(ix)0.1y + 0.01y20.1y0.1
0.01y20.01

4. (a) Identify terms which contain x and give the coefficient of x. 

(i) y2x + y

(ii) 13y2 – 8yx

(iii) x + y + 2

(iv) 5 + z + zx

(v) 1 + x + xy

(vi) 12xy2 + x25

(vii) 7x + xy2

(b) Identify terms which contain y2 and give the coefficient of y2.

(i) 8 – xy2

(ii) 5y2 + 7x

(iii) 2x2y – 15xy2 + 7y2

Answer

S.No.ExpressionTerm with factor xCoefficient of x
(i)y2x + yy2 xy
(ii)13y2 -8yx-8 yx-8 y
(iii)x + y + 2x1
(iv)5 + z + zxzxz
(v)1 + x + xyx1
xy1
(vi)12xy2 +2512 xy212y2
(vii)7x+xy2xy2y2
7x7
S. No. Expression Term containing y2Coefficient of y2
(i)8-xy2-xy2-x
(ii)5y+7x5y25
(iii)2x2y-15xy2 + 7y2-15xy2-15x
7y27

5. Classify into monomials, binomials and trinomials: 

(i) 4y – 7x

(ii) y2

(iii) x + y – xy

(iv) 100

(v) ab – a – b

(vi) 5 – 3t

(vii) 4p2q – 4pq2

(viii) 7mn

(ix) z2 – 3z + 8

(x) a2 + b2

(xi) z2 + z

(xii) 1 + x + x2

Answer

S.No.ExpressionType of Polynomial
(i)4y-7zBinomial
(ii)y2Monomial
(iii)x+y-xyTrinomial
(iv)100Monomial
(v)ab-a-bTrinomial
(vi)5-3tBinomial
(vii)4p2q-4pq2Binomial
(viii)7mnMonomial
(ix)z2-3z + 8Trinomial
(x)a2 + b2Binomial
(xi)z2 +zBinomial
(xii)1 + x + x2Trinomial

6. State whether a given pair of terms is of like or unlike terms: 

(i) 1,100

(ii) 

(iii) -29x, -29y

(iv) 14xy, 42 yx

(v) 4m2p, 4mp2

(vi) 12xz, 12x2 z2

Answer

S.No.Pair of termsLike / Unlike terms
(i)1, 100Like terms
(ii)Like terms
(iii)-29x,-29yUnlike terms
(iv)14xy,42yxLike terms
(v)4m2p,4mp2Unlike terms
(vi)12xz,12x2z2Unlike terms

7. Identify like terms in the following: 

(a) -xy2, -4yx2, 8x2, 2xy2, 7y,  -11×2  – 100x, – 11yx, 20x2y, -6x2, y, 2xy, 3x

Answer

(i) -xy2,2 xy2

(ii) -4yx2 , 20x2y

(iii) 8x2,-11x2,-6x2 

(iv) 7y, y

(v) -100x, 3x

(vi) -11yx, 2xy

(b) 10pq, 7p, 8q, -p2q2, -7qp, -100q, -23, 12q2p2, -5p2, 41,2405 p, 78qp, 13p2q, qp2, 701p2

Answer

(i) 10 pq – 7 pq,78 pq

(ii) 7p, 2405 p

(iii) 8q,- 100q

(iv) -p2q2, 12p2q2

(v) -12,41

(vi) -5p2,701p2 

(vii) 13 p2q,qp2

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

Exercise 12.2

1. Simplify combining like terms: 

(i) 21b – 32 + 7b – 20b

Answer

When term have the same algebraic factors, they are like terms.

Then,

= (21b + 7b – 20b) – 32

= b (21 + 7 – 20) – 32

= b (28 – 20) – 32

= b (8) – 32

= 8b – 32

(ii) – z2 + 13z2 – 5z + 7z3 – 15z

Answer

When term have the same algebraic factors, they are like terms.

Then,

= 7z3 + (-z2 + 13z2) + (-5z – 15z)

= 7z3 + z2 (-1 + 13) + z (-5 – 15)

= 7z3 + z2 (12) + z (-20) + 7z3 

= 7z3 + 12z2 – 20z + 7z3

(iii) p – (p – q) – q – (q – p)

Answer

When term have the same algebraic factors, they are like terms.

Then,

= p – p + q – q – q + p

= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a

Answer

When term have the same algebraic factors, they are like terms.

Then,

= 3a – 2b – ab – a + b – ab + 3ab + b – a

= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)

= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)

= a (1) + b (0) + ab (1)

= a + ab

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2 

Answer

When term have the same algebraic factors, they are like terms.

Then,

= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2

= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2

= x2y (8) + x2 (-4) + y2 (-7) + 8xy2

= 8x2y – 4x2 – 7y2 + 8xy2 

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)

Answer

When term have the same algebraic factors, they are like terms.

Then,

= 3y2 + 5y – 4 – 8y + y2 + 4

= 3y2 + y2 + 5y – 8y – 4 + 4

= y2 (3 + 1) + y (5 – 8) + (-4 + 4)

= y2 (4) + y (-3) + (0)

= 4y2 – 3y.

2. Add: 

(i) 3mn, – 5mn, 8mn, – 4mn

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 3mn + (-5mn) + 8mn + (- 4mn)

= 3mn – 5mn + 8mn – 4mn

= mn (3 – 5 + 8 – 4)

= mn (11 – 9)

= mn (2)

= 2mn

(ii) t – 8tz, 3tz – z, z – t

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= t – 8tz + (3tz – z) + (z – t)

= t – 8tz + 3tz – z + z – t

= t – t – 8tz + 3tz – z + z

= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)

= t (0) + tz (- 5) + z (0)

= – 5tz 

(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)

= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3

= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3

= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)

= mn (- 9 + 21) + (7 – 11)

= mn (12) – 4

= 12mn – 4

(iv) a + b – 3, b – a + 3, a – b + 3

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= a + b – 3 + (b – a + 3) + (a – b + 3)

= a + b – 3 + b – a + 3 + a – b + 3

= a – a + a + b + b – b – 3 + 3 + 3

= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)

= a (2 -1) + b (2 -1) + (-3 + 6)

= a (1) + b (1) + (3)

= a + b + 3

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy

= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy

= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18

= x (14 – 7) + y (10 – 10) + xy(-12 + 8 + 4) + (-13 + 18)

= x (7) + y (0) + xy(0) + (5)

= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 5m – 7n + (3n – 4m + 2) + (2m – 3mn – 5)

= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5

= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5

= m (5 – 4 + 2) + n (-7 + 3) – 3mn + (2 – 5)

= m (3) + n (-4) – 3mn + (-3)

= 3m – 4n – 3mn – 3

(vii) 4x2y, – 3xy2, –5xy2, 5x2y

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 4x2y + (-3xy2) + (-5xy2) + 5x2y

= 4x2y + 5x2y – 3xy2 – 5xy2

= x2y (4 + 5) + xy2 (-3 – 5)

= x2y (9) + xy2 (- 8)

= 9x2y – 8xy2

(viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 3p2q2 – 4pq + 5 + (- 10p2q2) + 15 + 9pq + 7p2q2

= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15

= p2q2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)

= p2q2 (0) + pq (5) + 20

= 5pq + 20

(ix) ab – 4a, 4b – ab, 4a – 4b

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= ab – 4a + (4b – ab) + (4a – 4b)

= ab – 4a + 4b – ab + 4a – 4b

= ab – ab – 4a + 4a + 4b – 4b

= ab (1 -1) + a (4 – 4) + b (4 – 4)

= ab (0) + a (0) + b (0)

= 0

(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2 

Answer

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= x2 – y2 – 1 + (y2 – 1 – x2) + (1 – x2 – y2)

= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2 

= x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1

= x2 (1 – 1- 1) + y2 (-1 + 1 – 1) + (-1 -1 + 1)

= x2 (1 – 2) + y2 (-2 +1) + (-2 + 1)

= x2 (-1) + y2 (-1) + (-1)

= -x2 – y2 -1

3. Subtract: 

(i) -5y2 from y2

Answer

y2 – (-5y2) = y2 + 5y2 = 6y2

(ii) 6xy from -12xy

Answer

-12xy -(6xy) = -12xy – 6xy = -18xy

(iii) (a – b) from (a + b)

Answer

(a + b)-(a -b) = a + b -a + b

= a – a + b + b = 2b

(iv) a (b – 5) from b (5 – a)

Answer

= b (5 – a)-a (b -5)

= 5b – ab – ab + 5a

= 5b – 2ab+5a

= 5a + 5b -2ab

(v) -m2 + 5mn from 4m2 – 3mn + 8

Answer

= 4m2 – 3mn + 8 – (- m2 + 5mn)

= 4m2 – 3mn + 8 + m2 – 5mn

= 4m2 + m2 – 3mn – 5mn + 8

= 5m2 – 8mn + 8

(vi) -x2 +10x – 5 from 5x-10

Answer

= 5x – 10 – (-x2 + 10x – 5)

= 5x – 10 + x2 – 10x + 5

= x2 + 5x – 10x – 10 + 5

= x2 – 5x – 5

(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2

Answer

= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)

= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2 

= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2 

= 10ab – 7a2 – 7b2

(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Answer

= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)

= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 

= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq

= 8p2 + 8q2 – 5pq

4. (a) What should be added to x2 +xy+y2 to obtain 2x2 +3xy ?

Answer

Let p should be added.

Then according to question,

x2 + xy + y2 + p = 2x2 + 3xy

⇒ p = 2x2 + 3xy – (x2 + xy + y2)

⇒ p = 2x2 + 3xy – x2 – xy – y2

⇒ p = 2x2 – x2– y2 +3xy – xy

⇒ p = x2 – y2 + 2xy

Hence, x2 – y2 + 2xy should be added.

(b) What should be subtracted from 2a + 8b+10 to get -3a + 7b + 16?

Answer

Let q should be subtracted.

Then according to question, 2a + 8b + 10-q = -3a + 7b + 16

⇒ -q = -3a +7b + 16 – (2a + 8b + 10)

⇒ -q = -3a + 7b + 16 – 2a – 8b – 10

⇒ – q = -3a – 2a + 7b – 8b + 16 – 10

⇒ -q = -5a – b + 6

⇒ q = – (- 5a – b + -6)

⇒ q = 5a + b – 6

5. What should be taken away from 3x2– 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20 ? 

Answer

Let q should be subtracted.

Then according to question,

3x2 – 4y2 + 5xy + 20 -q = -x2 – y2 + 6xy + 20

⇒ q = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20)

⇒ q = 3x2 – 4y2 + 5xy+ 20 + x2 + y2 – 6xy – 20

⇒ q = 3x2 + x2 – 4y2 +y2 + 5xy – 6xy + 20-20

⇒ q =  4x2 – 3y2 –  xy + 0

Hence, 4x2 -3y2 -xy should be subtracted.

6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.

Answer

First we have to find out the sum of 3x – y + 11 and – y – 11

= 3x – y + 11 + (-y – 11)

= 3x – y + 11 – y – 11

= 3x – y – y + 11 – 11

= 3x – 2y

Now, subtract 3x – y – 11 from 3x – 2y

= 3x – 2y – (3x – y – 11)

= 3x – 2y – 3x + y + 11

= 3x – 3x – 2y + y + 11

= -y + 11

(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and -x2 + 2x + 5.

Answer

First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2 

= 4 + 3x + (5 – 4x + 2x2)

= 4 + 3x + 5 – 4x + 2x2 

= 4 + 5 + 3x – 4x + 2x2 

= 9 – x + 2x2 

= 2x2 – x + 9     … [equation 1]

Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5.

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

1. If m = 2, find the value of: 

(i) m – 2    

(ii) 3m – 5    

(iii) 9 – 5m

(iv) 3m2 – 2m – 7

(v) 

Answer

(i) m – 2 = 2 – 2    [Putting m = 2]

= 0

(ii) 3m – 5 = 3 x 2 – 5     [Putting m = 2]

= 6 – 5 = 1

(iii) 9 – 5m = 9 – 5 x 2    [Putting m = 2]

= 9 – 10 = – 1

(iv) 3m2 – 2m – 7

= 3(2)2 – 2 (2) – 7        [Putting m = 2]

=3 × 4 – 2 × 2 – 7

 = 12-4-7

 = 12- 11 = 1

(v)  [Putting m = 2]

= 5 – 4 = 1

2. If p = -2, find the value of:

(i) 4p + 7

(ii) – 3p2 + 4p + 7

(iii) -2p3 – 3p2 +4/7 + 7

Answer

(i) 4p + 7 = 4 (- 2) + 7    [Putting p= -2]

= -8 + 7 = -1

(ii) -3p2+4p + 7

= -3 (-2)2+ 4 (-2) + 7    [Putting p = – 2]

= – 3 × 4 – 8 + 7

= – 12 – 8 + 7

= -20 + 7 = -13

(iii) – 2p3 – 3p2 +4p + 7

= – 2 (-2)3 – 3(-2)2 + 4 (-2) + 7     [Putting p = – 2]

= -2 ×(-8)-3 ×4 -8 + 7

= 16-12-8 + 7 

= -20 + 23 = 3

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

3. Find the value of the following expressions, when x = -1:

(i) 2x – 7

(ii) -x + 2

(iii) x2 + 2x  + 1

(iv) 2x2– x – 2

Answer

(i) 2x – 7 = 2 (-1) – 7      [Putting x= – 1]

= – 2 – 7 = – 9

(ii) – x + 2 = – (-1) + 2     [Putting x= – 1]

= 1 + 2 = 3

(iii) x2 + 2 x + 1 = (-1)2 + 2 (-1) + 1    [Putting x= – 1] 

= 1 – 2 + 1

= 2 – 2 = 0

(iv) 2x2– x – 2 = 2 (-1)2 – (-1) – 2     [Putting x= – 1] 

= 2x 1 + 1-2

= 2 + 1 – 2

= 3 – 2 = 1

4. If a = 2,b = -2, find the value of: 

(i) a2 + b2 

(ii) a2+ab + b2

(iii) a2 – b2

Answer

(i) a2 + b2 ( 2)2 + (- 2)2    [Putting a = 2. b = – 2 ]

= 4 + 4 = 8

(ii) a2+ab + b2 

= (2) + ( 2) (- 2) +(-2)2   [Putting a = 2. b = – 2 ]

= 4 – 4 + 4 = 4

(iii) a2 – b2 = (2)2 – (-2)2  [Putting a = 2,b = – 2]

= 4 – 4 = 0

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

5. When a = 0, b = -1, find the value of the given expressions: 

(i) 2a + 2b

(ii) 2a2+b2+1

(iii) 2a2b + 2ab2 +ab

(iv) a2+ab+2

Answer

(i) 2a + 2b = 2 (0) + 2 (-1)    [Putting a – 0,b = – 1]

= 0 – 2 = -2  

(ii) 2a2 + b2 + 1 = 2 (0)2 + (-1)2 + 1      [Putting a – 0,b = – 1]

= 2 x 0 + 1+ 1 = 0 + 2 = 2

(iii) 2a2b + 2ab2 + ab = 2(0)2 (-1) + 2 (0 )(-1)2 + (0 )(-1)     [Putting a – 0,b = – 1]

= 0 + 0 + 0 = 0

(iv) a2 +ab + 2 – (0)2 + (0) (-1) + 2   [Putting a – 0,b = – 1]

= 0 + 0 + 2 = 2

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

6. Simplify the expressions and find the value if x is equal to 2: 

(i) x + 7 + 4 (x- 5)

(ii) 3 (x + 2) + 5x – 7

(iii) 6x + 5 (x – 2)

(iv) 4 (2x – 1) + 3x + 11

Answer

(i) x + 7 + 4(x- 5) = x + 7 + 4x – 20 = x + 4 x + 7 – 20

= 5 x – 13 = 5 x 2 – 13                            [Putting x = 2]

= 10-13 = -3

(ii) 3 (x+ 2) + 5x – 7 = 3x + 6 + 5x -7 = 3x + 5x + 6 – 7

= 8x – 1 = 8 x 2-1                    [Putting x = -1]

= 16 – 1 = 15

(iii) 6x + 5 (x – 2) = 6x + 5x -10 = 11x – 10

= 11 x 2 – 10                      [Putting x = -1]

= 22 – 10 = 12

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x +11 = 8x + 3a – 4 + 11

= 11a + 7 = 11 x 2 + 7 [Putting x = – 1]

= 22+7 = 29

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

7. Simplify these expressions and find their values if x = 3,a = -1, b = – 2 :

(i) 3x – 5 – x + 9

(ii) 2 – 8x + 4x + 4

(iii) 3a + 5 – 8a + 1

(iv) 10 – 3b – 4 – 5b

(v) 2a – 2b – 4 – 5 + a

Answer

(i) 3a – 5 – x + 9 = 3x – x – 5 + 9 = 2x + 4

= 2×3+4         [Putting a = 3]

= 6 + 4 = 10

(ii) 2 – 8x + 4x + 4 = – 8x + 4x + 2 + 4 = -4x + 6

= – 4 x 3 + 6     [Putting a = 3]

= -12 + 6 =12

(iii) 3a + 5 – 8a + 1 = 3a – 8a + 5 + 1 = – 5a + 6

= -5(- 1) + 6       [Putting a = – 1]

= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = – 3b – 5b + 10 – 4 = -8b+6

= -8 (-2)+ 6    [Putting b = -2]

= 16 + 6 = 22

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5

= 3a – 2b – 9 = 3 (-1)-2 (-2) -9    [Putting a = -1 , b = – 2]

= -3 + 4 -9 = -8

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

8. (i) If z = 10, find the value of z3 – 3 (z – 10).

(ii) If p = – 10, find the value of p2 – 2p – 100

Answer

(i) z3 -3(z-10) = (10)3-3(10 – 10)       [Putting z = 10]

= 1000 – 3 x 0 = 1000- 0

= 1000

(ii) p2 – 2p – 100 = (-10)2 – 2 (-10) – 100    (Putting p = – 10]

= 100+ 20 – 100 = 20

9. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0 ? 

Answer

Given: 2x2 + x – a = 5

⇒ 2 (0)2 + 0 – a = 5     [Putting x = 0]

⇒ 0 + 0 – a = 5

⇒ a = -5

Hence, the value of a is -5.

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

10. Simplify the expression and find its value when a = 5 and b = – 3: 2 (a2 + ab) + 3 – ab

Answer

Given 2 (a2 + ab) + 3 – ab

⇒ 2a2 + 2ab + 3 – ab

⇒ 2a2 + 2ab – ab + 3

⇒ 2a2 + ab + 3

⇒ 2 (5)2 + (5) (-3) + 3   [Putting a = 5 , b = -3]

⇒ 2 x 25 – 15 + 3

⇒ 50 – 15 + 3

⇒ 38.

NCERT 7th Maths Chapter 12, class 7 Maths Chapter 12 solutions

NCERT Solutions for 7th Class Maths: Chapter 12: Download PDF

NCERT Solutions for 7th Class Maths: Chapter 12-Algebraic Expressions

Download PDF: NCERT Solutions for 7th Class Maths: Chapter 12-Algebraic Expressions PDF

Chapter-wise NCERT Solutions Class 7 Maths

Chapter 1 Integers
Chapter 2 Fractions and Decimals
Chapter 3 Data Handling
Chapter 4 Simple Equations
Chapter 5 Lines and Angles
Chapter 6 The Triangle and its Properties
Chapter 7 Congruence of Triangles
Chapter 8 Comparing Quantities
Chapter 9 Rational Numbers
Chapter 10 Practical Geometry
Chapter 11 Perimeter and Area
Chapter 12 Algebraic Expressions
Chapter 13 Exponents and Powers
Chapter 14 Symmetry
Chapter 15 Visualising Solid Shapes


About NCERT


About NCERT

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website to learn more.