Chapter 10: Algebraic Expressions

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

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

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

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

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

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