NCERT Solutions for Class 7 Maths Exercise 12.2 Chapter 12 Algebraic Expressions

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 7 Maths Exercise 12.2 Chapter 12 Algebraic Expressions in simple PDF are available here. Adding and subtracting like terms and adding and subtracting general algebraic expressions are the two topics covered in this exercise of NCERT Solutions for Class 7 Maths Chapter 12. Subject experts prepare these solutions for algebraic expressions to help students gear up their exam preparations. Students can either practise online or download these NCERT Solutions for Class 7 Maths and practise different types of questions.

NCERT Solutions for Class 7 Maths Chapter 12 Perimeter and Area – Exercise 12.2

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Access Other Exercises of NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions

Exercise 12.1 Solutions

Exercise 12.3 Solutions

Exercise 12.4 Solutions

Access Answers to NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions Exercise 12.2

1. Simplify combining like terms:

(i) 21b – 32 + 7b – 20b

Solution:-

When terms 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

Solution:-

When terms 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 + 12z² – 20z

(iii) p – (p – q) – q – (q – p)

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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)

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

When terms 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

Solution:-

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

Then, we have to subtract the like terms

= y² – (-5y²)

= y² + 5y²

= 6y²

(ii) 6xy from –12xy

Solution:-

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

Then, we have to subtract the like terms

= -12xy – 6xy

= – 18xy

(iii) (a – b) from (a + b)

Solution:-

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

Then, we have to subtract the like terms

= (a + b) – (a – b)

= a + b – a + b

= a – a + b + b

= a (1 – 1) + b (1 + 1)

= a (0) + b (2)

= 2b

(iv) a (b – 5) from b (5 – a)

Solution:-

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

Then, we have to subtract the like terms

= b (5 -a) – a (b – 5)

= 5b – ab – ab + 5a

= 5b + ab (-1 -1) + 5a

= 5a + 5b – 2ab

(v) –m² + 5mn from 4m² – 3mn + 8

Solution:-

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

Then, we have to subtract the like terms

= 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

Solution:-

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

Then, we have to subtract the like terms

= 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

Solution:-

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

Then, we have to subtract the like terms

= 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

Solution:-

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

Then, we have to subtract the like terms

= 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?

Solution:-

Let us assume p be the required term

Then,

p + (x² + xy + y²) = 2x² + 3xy

p = (2x² + 3xy) – (x² + xy + y²)

p = 2x² + 3xy – x² – xy – y²

p = 2x² – x² + 3xy – xy – y²

p = x² + 2xy – y²

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

Solution:-

Let us assume x be the required term

Then,

2a + 8b + 10 – x = -3a + 7b + 16

x = (2a + 8b + 10) – (-3a + 7b + 16)

x = 2a + 8b + 10 + 3a – 7b – 16

x = 2a + 3a + 8b – 7b + 10 – 16

x = 5a + b – 6

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

Solution:-

Let us assume a be the required term

Then,

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

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

a = 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20

a = 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20

a = 4x² – 3y² – xy

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

Solution:-

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.

Solution:-

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

= 3x² – 5x + (-x² + 2x + 5)

= 3x² – 5x – x² + 2x + 5

= 3x² – x² – 5x + 2x + 5

= 2x² – 3x + 5 … [equation 2]

Now, we have to subtract equation (2) from equation (1)

= 2x² – x + 9 – (2x² – 3x + 5)

= 2x² – x + 9 – 2x² + 3x – 5

= 2x² – 2x² – x + 3x + 9 – 5

= 2x + 4

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NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 

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