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 for their exam preparations. Students can either practice online or download these NCERT Solutions for Class 7 Maths and practice different types of questions.

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**1. Simplify combining like terms:**

**(i) 21b â€“ 32 + 7b â€“ 20b**

**Solution:-**

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

Then,

= (21b + 7b â€“ 20b) â€“ 32

= b (21 + 7 â€“ 20) â€“ 32

= b (28 â€“ 20) â€“ 32

= b (8) â€“ 32

= 8b â€“ 32

**(ii) â€“ z ^{2} + 13z^{2} â€“ 5z + 7z^{3} â€“ 15z**

**Solution:-**

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

Then,

= 7z^{3} + (-z^{2} + 13z^{2}) + (-5z â€“ 15z)

= 7z^{3 }+ z^{2 }(-1 + 13) + z (-5 â€“ 15)

= 7z^{3 }+ z^{2} (12) + z (-20)

= 7z^{3 }+ 12z^{2} â€“ 20z

**(iii) p â€“ (p â€“ q) â€“ q â€“ (q â€“ p)**

**Solution:-**

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

Then,

= p â€“ p + q â€“ q â€“ q + p

= p â€“ q

**(iv) 3a â€“ 2b â€“ ab â€“ (a â€“ b + ab) + 3ab + b â€“ a**

**Solution:-**

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

Then,

= 3a â€“ 2b â€“ ab â€“ a + b â€“ ab + 3ab + b â€“ a

= 3a â€“ a â€“ a â€“ 2b + b + b â€“ ab â€“ ab + 3ab

= a (1 â€“ 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)

= a (1 â€“ 2) + b (-2 + 2) + ab (-2 + 3)

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

= a + ab

**(v) 5x ^{2}y â€“ 5x^{2} + 3yx^{2} â€“ 3y^{2} + x^{2} â€“ y^{2} + 8xy^{2} â€“ 3y^{2}**

**Solution:-**

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

Then,

= 5x^{2}y + 3yx^{2} â€“ 5x^{2} + x^{2} â€“ 3y^{2} â€“ y^{2} â€“ 3y^{2}

= x^{2}y (5 + 3) + x^{2} (- 5 + 1) + y^{2} (-3 â€“ 1 -3) + 8xy^{2}

= x^{2}y (8) + x^{2} (-4) + y^{2} (-7) + 8xy^{2}

= 8x^{2}y â€“ 4x^{2} â€“ 7y^{2} + 8xy^{2}

**(vi) (3y ^{2} + 5y â€“ 4) â€“ (8y â€“ y^{2} â€“ 4)**

**Solution:-**

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

Then,

= 3y^{2} + 5y â€“ 4 â€“ 8y + y^{2} + 4

= 3y^{2} + y^{2} + 5y â€“ 8y â€“ 4 + 4

= y^{2} (3 + 1) + y (5 â€“ 8) + (-4 + 4)

= y^{2} (4) + y (-3) + (0)

= 4y^{2} â€“ 3y

**2. Add:**

**(i) 3mn, â€“ 5mn, 8mn, â€“ 4mn**

**Solution:-**

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

Then, we have to add the like terms

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

= 3mn â€“ 5mn + 8mn â€“ 4mn

= mn (3 â€“ 5 + 8 â€“ 4)

= mn (11 â€“ 9)

= mn (2)

= 2mn

**(ii) t â€“ 8tz, 3tz â€“ z, z â€“ t**

**Solution:-**

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

Then, we have to add the like terms

= t â€“ 8tz + (3tz â€“ z) + (z â€“ t)

= t â€“ 8tz + 3tz â€“ z + z â€“ t

= t â€“ t â€“ 8tz + 3tz â€“ z + z

= t (1 â€“ 1) + tz (- 8 + 3) + z (-1 + 1)

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

= â€“ 5tz

**(iii) â€“ 7mn + 5, 12mn + 2, 9mn â€“ 8, â€“ 2mn â€“ 3**

**Solution:-**

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

Then, we have to add the like terms

= â€“ 7mn + 5 + 12mn + 2 + (9mn â€“ 8) + (- 2mn â€“ 3)

= â€“ 7mn + 5 + 12mn + 2 + 9mn â€“ 8 â€“ 2mn â€“ 3

= â€“ 7mn + 12mn + 9mn â€“ 2mn + 5 + 2 â€“ 8 â€“ 3

= mn (-7 + 12 + 9 â€“ 2) + (5 + 2 â€“ 8 â€“ 3)

= mn (- 9 + 21) + (7 â€“ 11)

= mn (12) â€“ 4

= 12mn â€“ 4

**(iv) a + b â€“ 3, b â€“ a + 3, a â€“ b + 3**

**Solution:-**

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

Then, we have to add the like terms

= a + b â€“ 3 + (b â€“ a + 3) + (a â€“ b + 3)

= a + b â€“ 3 + b â€“ a + 3 + a â€“ b + 3

= a â€“ a + a + b + b â€“ b â€“ 3 + 3 + 3

= a (1 â€“ 1 + 1) + b (1 + 1 â€“ 1) + (-3 + 3 + 3)

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

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

= a + b + 3

**(v) 14x + 10y â€“ 12xy â€“ 13, 18 â€“ 7x â€“ 10y + 8xy, 4xy**

**Solution:-**

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

Then, we have to add the like terms

= 14x + 10y â€“ 12xy â€“ 13 + (18 â€“ 7x â€“ 10y + 8xy) + 4xy

= 14x + 10y â€“ 12xy â€“ 13 + 18 â€“ 7x â€“ 10y + 8xy + 4xy

= 14x â€“ 7x + 10yâ€“ 10y â€“ 12xy + 8xy + 4xy â€“ 13 + 18

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

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

= 7x + 5

**(vi) 5m â€“ 7n, 3n â€“ 4m + 2, 2m â€“ 3mn â€“ 5**

**Solution:-**

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

Then, we have to add the like terms

= 5m â€“ 7n + (3n â€“ 4m + 2) + (2m â€“ 3mn â€“ 5)

= 5m â€“ 7n + 3n â€“ 4m + 2 + 2m â€“ 3mn â€“ 5

= 5m â€“ 4m + 2m â€“ 7n + 3n â€“ 3mn + 2 â€“ 5

= m (5 â€“ 4 + 2) + n (-7 + 3) â€“ 3mn + (2 â€“ 5)

= m (3) + n (-4) â€“ 3mn + (-3)

= 3m â€“ 4n â€“ 3mn â€“ 3

**(vii) 4x ^{2}y, â€“ 3xy^{2}, â€“5xy^{2}, 5x^{2}y**

**Solution:-**

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

Then, we have to add the like terms

= 4x^{2}y + (-3xy^{2}) + (-5xy^{2}) + 5x^{2}y

= 4x^{2}y + 5x^{2}y â€“ 3xy^{2} â€“ 5xy^{2}

= x^{2}y (4 + 5) + xy^{2 }(-3 â€“ 5)

= x^{2}y (9) + xy^{2} (- 8)

= 9x^{2}y â€“ 8xy^{2}

**(viii) 3p ^{2}q^{2} â€“ 4pq + 5, â€“ 10 p^{2}q^{2}, 15 + 9pq + 7p^{2}q^{2}**

**Solution:-**

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

Then, we have to add the like terms

= 3p^{2}q^{2} â€“ 4pq + 5 + (- 10p^{2}q^{2}) + 15 + 9pq + 7p^{2}q^{2}

= 3p^{2}q^{2} â€“ 10p^{2}q^{2} + 7p^{2}q^{2} â€“ 4pq + 9pq + 5 + 15

= p^{2}q^{2} (3 -10 + 7) + pq (-4 + 9) + (5 + 15)

= p^{2}q^{2 }(0) + pq (5) + 20

= 5pq + 20

**(ix) ab â€“ 4a, 4b â€“ ab, 4a â€“ 4b**

**Solution:-**

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

Then, we have to add the like terms

= ab â€“ 4a + (4b â€“ ab) + (4a â€“ 4b)

= ab â€“ 4a + 4b â€“ ab + 4a â€“ 4b

= ab â€“ ab â€“ 4a + 4a + 4b â€“ 4b

= ab (1 -1) + a (4 â€“ 4) + b (4 â€“ 4)

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

= 0

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**(x) x ^{2} â€“ y^{2} â€“ 1, y^{2} â€“ 1 â€“ x^{2}, 1 â€“ x^{2} â€“ y^{2}**

**Solution:-**

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

Then, we have to add the like terms

= x^{2} â€“ y^{2} â€“ 1 + (y^{2} â€“ 1 â€“ x^{2}) + (1 â€“ x^{2} â€“ y^{2})

= x^{2} â€“ y^{2} â€“ 1 + y^{2} â€“ 1 â€“ x^{2} + 1 â€“ x^{2} â€“ y^{2}

= x^{2} â€“ x^{2} â€“ x^{2} â€“ y^{2} + y^{2} â€“ y^{2 }â€“ 1 â€“ 1 + 1

= x^{2} (1 â€“ 1- 1) + y^{2} (-1 + 1 â€“ 1) + (-1 -1 + 1)

= x^{2} (1 â€“ 2) + y^{2}Â (-2 +1) + (-2 + 1)

= x^{2} (-1) + y^{2}Â (-1) + (-1)

= -x^{2} â€“ y^{2} -1

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**3. Subtract:**

**(i) â€“5y ^{2} from y^{2}**

**Solution:-**

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

Then, we have to subtract the like terms

= y^{2} â€“ (-5y^{2})

= y^{2}Â + 5y^{2}

= 6y^{2}

**(ii) 6xy from â€“12xy**

**Solution:-**

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

Then, we have to subtract the like terms

= -12xy â€“ 6xy

= â€“ 18xy

**(iii) (a â€“ b) from (a + b)**

**Solution:-**

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

Then, we have to subtract the like terms

= (a + b) â€“ (a â€“ b)

= a + b â€“ a + b

= a â€“ a + b + b

= a (1 â€“ 1) + b (1 + 1)

= a (0) + b (2)

= 2b

**(iv) a (b â€“ 5) from b (5 â€“ a)**

**Solution:-**

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

Then, we have to subtract the like terms

= b (5 -a) â€“ a (b â€“ 5)

= 5b â€“ ab â€“ ab + 5a

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

= 5a + 5b â€“ 2ab

**(v) â€“m ^{2} + 5mn from 4m^{2} â€“ 3mn + 8**

**Solution:-**

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

Then, we have to subtract the like terms

= 4m^{2} â€“ 3mn + 8 â€“ (- m^{2} + 5mn)

= 4m^{2} â€“ 3mn + 8 + m^{2} â€“ 5mn

= 4m^{2} + m^{2} â€“ 3mn â€“ 5mn + 8

= 5m^{2 }â€“ 8mn + 8

**(vi) â€“ x ^{2} + 10x â€“ 5 from 5x â€“ 10**

**Solution:-**

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

Then, we have to subtract the like terms

= 5x â€“ 10 â€“ (-x^{2} + 10x â€“ 5)

= 5x â€“ 10 + x^{2} â€“ 10x + 5

= x^{2} + 5x â€“ 10x â€“ 10 + 5

= x^{2} â€“ 5x â€“ 5

**(vii) 5a ^{2} â€“ 7ab + 5b^{2} from 3ab â€“ 2a^{2} â€“ 2b^{2}**

**Solution:-**

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

Then, we have to subtract the like terms

= 3ab â€“ 2a^{2} â€“ 2b^{2} â€“ (5a^{2} â€“ 7ab + 5b^{2})

= 3ab â€“ 2a^{2} â€“ 2b^{2} â€“ 5a^{2 }+ 7ab â€“ 5b^{2}

= 3ab + 7ab â€“ 2a^{2} â€“ 5a^{2} â€“ 2b^{2} â€“ 5b^{2}

= 10ab â€“ 7a^{2} â€“ 7b^{2}

**(viii) 4pq â€“ 5q ^{2} â€“ 3p^{2} from 5p^{2} + 3q^{2} â€“ pq**

**Solution:-**

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

Then, we have to subtract the like terms

= 5p^{2} + 3q^{2} â€“ pq â€“ (4pq â€“ 5q^{2} â€“ 3p^{2})

= 5p^{2} + 3q^{2} â€“ pq â€“ 4pq + 5q^{2} + 3p^{2}

= 5p^{2} + 3p^{2} + 3q^{2} + 5q^{2} â€“ pq â€“ 4pq

= 8p^{2} + 8q^{2} â€“ 5pq

**4. (a) What should be added to x ^{2} + xy + y^{2} to obtain 2x^{2} + 3xy?**

**Solution:-**

Let us assume p be the required term

Then,

p + (x^{2} + xy + y^{2}) = 2x^{2} + 3xy

p = (2x^{2}Â + 3xy) â€“ (x^{2} + xy + y^{2})

p = 2x^{2} + 3xy â€“ x^{2} â€“ xy â€“ y^{2}

p = 2x^{2} â€“ x^{2} + 3xy â€“ xy â€“ y^{2}

p = x^{2}Â + 2xy â€“ y^{2}

**(b) What should be subtracted from 2a + 8b + 10 to get â€“ 3a + 7b + 16? **

**Solution:-**

Let us assume x be the required term

Then,

2a + 8b + 10 â€“ x = -3a + 7b + 16

x = (2a + 8b + 10) â€“ (-3a + 7b + 16)

x = 2a + 8b + 10 + 3a â€“ 7b â€“ 16

x = 2a + 3a + 8b â€“ 7b + 10 â€“ 16

x = 5a + b â€“ 6

**5. What should be taken away from 3x ^{2} â€“ 4y^{2} + 5xy + 20 to obtain â€“ x^{2} â€“ y^{2} + 6xy + 20?**

**Solution:-**

Let us assume a be the required term

Then,

3x^{2} â€“ 4y^{2} + 5xy + 20 â€“ a = -x^{2} â€“ y^{2} + 6xy + 20

a = 3x^{2} â€“ 4y^{2} + 5xy + 20 â€“ (-x^{2} â€“ y^{2} + 6xy + 20)

a = 3x^{2} â€“ 4y^{2} + 5xy + 20 + x^{2} + y^{2} â€“ 6xy â€“ 20

a = 3x^{2} + x^{2} â€“ 4y^{2} + y^{2} + 5xy â€“ 6xy + 20 â€“ 20

a = 4x^{2} â€“ 3y^{2} â€“ xy

**6. (a) From the sum of 3x â€“ y + 11 and â€“ y â€“ 11, subtract 3x â€“ y â€“ 11.**

**Solution:-**

First we have to find out the sum of 3x â€“ y + 11 and â€“ y â€“ 11

= 3x â€“ y + 11 + (-y â€“ 11)

= 3x â€“ y + 11 â€“ y â€“ 11

= 3x â€“ y â€“ y + 11 â€“ 11

= 3x â€“ 2y

Now, subtract 3x â€“ y â€“ 11 from 3x â€“ 2y

= 3x â€“ 2y â€“ (3x â€“ y â€“ 11)

= 3x â€“ 2y â€“ 3x + y + 11

= 3x â€“ 3x â€“ 2y + y + 11

= -y + 11

**(b) From the sum of 4 + 3x and 5 â€“ 4x + 2x ^{2}, subtract the sum of 3x^{2} â€“ 5x and **

**â€“x ^{2} + 2x + 5.**

**Solution:-**

First we have to find out the sum of 4 + 3x and 5 â€“ 4x + 2x^{2}

= 4 + 3x + (5 â€“ 4x + 2x^{2})

= 4 + 3x + 5 â€“ 4x + 2x^{2}

= 4 + 5 + 3x â€“ 4x + 2x^{2}

= 9 â€“ x + 2x^{2}

= 2x^{2} â€“ x + 9 â€¦ [equation 1]

Then, we have to find out the sum of 3x^{2} â€“ 5x and â€“ x^{2} + 2x + 5

= 3x^{2} â€“ 5x + (-x^{2} + 2x + 5)

= 3x^{2} â€“ 5x â€“ x^{2} + 2x + 5

= 3x^{2} â€“ x^{2} â€“ 5x + 2x + 5

= 2x^{2} â€“ 3x + 5 â€¦ [equation 2]

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

= 2x^{2 }â€“ x + 9 â€“ (2x^{2} â€“ 3x + 5)

= 2x^{2} â€“ x + 9 â€“ 2x^{2}Â + 3x â€“ 5

= 2x^{2} â€“ 2x^{2} â€“ x + 3x + 9 â€“ 5

= 2x + 4