**Exercise 12.1**

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

**(i) Subtraction of z from y.**

**Solution:-**

= Y – z

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

**Solution:-**

= ½ (x + y)

= (x + y)/2

**(iii) The number z multiplied by itself.**

**Solution:-**

= z × z

= z^{2}

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

**Solution:-**

= ¼ (p × q)

= pq/4

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

**Solution:-**

= x^{2 }+ y^{2}

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

**Solution:-**

= 3mn + 5

**(vii) Product of numbers y and z subtracted from 10.**

**Solution:-**

= 10 – (y × z)

= 10 – yz

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

**Solution:-**

= (a × b) – (a + b)

= ab – (a + b)

**2. (i) Identify the terms and their factors in the following expressions**

**Show the terms and factors by tree diagrams.**

(a) x – 3

(b) 1 + x + x^{2 }

(c) y – y^{3 }

(d) 5xy^{2 } + 7x^{2 }y

(e) -ab + 2b^{2 } – 3a^{2 }

Solution:

(ii) Identify terms and factors in the expression

given below:

(a) -4x + 5

(b) -4x + 5y

(c) 5y + 3y^{2}

(d) xy + 2x^{2}y^{2}

(e) pq + q

(f) 1.2ab – 2.4b + 3.6a

(g) 34x+14

(h) 0.1p^{2} + 0.2q^{2}

Solution:

Question 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) 12 xy2 + 25

(vii) 7x + xy2

Solution:

(b) Identify terms which contain y^{2} and give the coefficients of y^{2}.

(i) 8 – xy^{2}

(ii) 5y^{2} + 7x

(iii) 2x^{2}y – 15xy^{2} + 7y^{2}

Question 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

Solution:

(i)4y – 7z – Binomial

(ii) y2 – Monomial

(iii) x + y – xy – Trinomial

(iv) 100 Monomial

(v) ab – a – b – Trinomial

(vi) 5 – 3t – Binomial

(vii) 4p2q – 4pq2 – Binomial

(viii) 7mn – Monomial

(ix) z2 -3z + 8 – Trinomial

(x) a2 + b2 – Binomial

(xi) z2 + z – Binomial

(xii) 1 + x + x2 – Trinomial

Question 6.

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

(i) 1, 100

(ii) -7x, 52x

(iii) -29x, -29y

(iv) 14xy, 42yx

(v) 4m2p, 4mp2

(vi) 12xz, 12 x2y2

Solution:

(i) 1, 100 – Like

(ii) -7x, 52x – Like

(iii) -29x, -29y – Unlike

(iv) 14xy, 42yx – Like

(v) 4m2p, 4mp2 – Unlike

(vi) 12xz, 12x2z2 – Unlike

Question 7.

Identify like terms in the following:

(a)-xy2, -4yx2, 8x2, 2xy2, 7y2, -11x2, -100x, -11yx, 20x2y, -6x2, y, 2xy, 3x

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

Solution:

(a) Like terms are:

(i) -xy2, 2xy2

(ii) -4yx2, 20x2y

(iii)8x2, -11x2, -6x2

(iv) 7y, y

(v) -100x, 3x

(vi) -11yx, 2xy

(b) Like terms are:

(i) 10pq, – 7qp, 78qp

(ii) 7p, 2405p

(iii) 8q, -100q

(iv) -p2q2, 12 q2p2

(v) -23, 41

(vi) -5p2, 701p2

## Exercise 12.2

1. Simplify combining like terms:

(i) 21b – 32 + 7b – 20b

Solution:-

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

Solution:-

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 + 12z2 – 20z

(iii) p – (p – q) – q – (q – p)

Solution:-

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

Solution:-

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

Solution:-

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)

Solution:-

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

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 – 5mn + 8mn – 4mn

= mn (3 – 5 + 8 – 4)

= mn (11 – 9)

= mn (2)

= 2mn

(ii) t – 8tz, 3tz – z, z – t

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

When term 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 term 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 term 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 term 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 term 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 term 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 term 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 term 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

** Exercise 12.3**

**1. If m = 2, find the value of:**

**(i) m – 2**

**Solution:-**

From the question it is given that m = 2

Then, substitute the value of m in the question

= 2 -2

= 0

**(ii) 3m – 5**

**Solution:-**

From the question it is given that m = 2

Then, substitute the value of m in the question

= (3 × 2) – 5

= 6 – 5

= 1

**(iii) 9 – 5m**

**Solution:-**

From the question it is given that m = 2

Then, substitute the value of m in the question

= 9 – (5 × 2)

= 9 – 10

= – 1

**(iv) 3m ^{2} – 2m – 7**

**Solution:-**

From the question it is given that m = 2

Then, substitute the value of m in the question

= (3 × 2^{2}) – (2 × 2) – 7

= (3 × 4) – (4) – 7

= 12 – 4 -7

= 12 – 11

= 1

**(v) (5m/2) – 4**

**Solution:-**

From the question it is given that m = 2

Then, substitute the value of m in the question

= ((5 × 2)/2) – 4

= (10/2) – 4

= 5 – 4

= 1

**2. If p = – 2, find the value of:**

**(i) 4p + 7**

**Solution:-**

From the question it is given that p = -2

Then, substitute the value of p in the question

= (4 × (-2)) + 7

= -8 + 7

= -1

**(ii) – 3p ^{2} + 4p + 7**

**Solution:-**

From the question it is given that p = -2

Then, substitute the value of p in the question

= (-3 × (-2)^{2}) + (4 × (-2)) + 7

= (-3 × 4) + (-8) + 7

= -12 – 8 + 7

= -20 + 7

= -13

**(iii) – 2p ^{3} – 3p^{2} + 4p + 7**

**Solution:-**

From the question it is given that p = -2

Then, substitute the value of p in the question

= (-2 × (-2)^{3}) – (3 × (-2)^{2}) + (4 × (-2)) + 7

= (-2 × -8) – (3 × 4) + (-8) + 7

= 16 – 12 – 8 + 7

= 23 – 20

= 3

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

**(i) 2x – 7**

**Solution:-**

From the question it is given that x = -1

Then, substitute the value of x in the question

= (2 × -1) – 7

= – 2 – 7

= – 9

**(ii) – x + 2**

**Solution:-**

From the question it is given that x = -1

Then, substitute the value of x in the question

= – (-1) + 2

= 1 + 2

= 3

**(iii) x ^{2} + 2x + 1**

**Solution:-**

From the question it is given that x = -1

Then, substitute the value of x in the question

= (-1)^{2} + (2 × -1) + 1

= 1 – 2 + 1

= 2 – 2

= 0

**(iv) 2x ^{2} – x – 2**

**Solution:-**

From the question it is given that x = -1

Then, substitute the value of x in the question

= (2 × (-1)^{2}) – (-1) – 2

= (2 × 1) + 1 – 2

= 2 + 1 – 2

= 3 – 2

= 1

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

**(i) a ^{2} + b^{2}**

**Solution:-**

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= (2)^{2} + (-2)^{2}

= 4 + 4

= 8

**(ii) a ^{2} + ab + b^{2}**

**Solution:-**

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= 2^{2} + (2 × -2) + (-2)^{2}

= 4 + (-4) + (4)

= 4 – 4 + 4

= 4

**(iii) a ^{2} – b^{2}**

**Solution:-**

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= 2^{2} – (-2)^{2}

= 4 – (4)

= 4 – 4

= 0

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

**(i) 2a + 2b**

**Solution:-**

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 0) + (2 × -1)

= 0 – 2

= -2

**(ii) 2a ^{2} + b^{2} + 1**

**Solution:-**

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 0^{2}) + (-1)^{2} + 1

= 0 + 1 + 1

= 2

**(iii) 2a ^{2}b + 2ab^{2} + ab**

**Solution:-**

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 0^{2} × -1) + (2 × 0 × (-1)^{2}) + (0 × -1)

= 0 + 0 +0

= 0

**(iv) a ^{2} + ab + 2**

**Solution:-**

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (0^{2}) + (0 × (-1)) + 2

= 0 + 0 + 2

= 2

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

**(i) x + 7 + 4 (x – 5)**

**Solution:-**

From the question it is given that x = 2

We have,

= x + 7 + 4x – 20

= 5x + 7 – 20

Then, substitute the value of x in the equation

= (5 × 2) + 7 – 20

= 10 + 7 – 20

= 17 – 20

= – 3

**(ii) 3 (x + 2) + 5x – 7**

**Solution:-**

From the question it is given that x = 2

We have,

= 3x + 6 + 5x – 7

= 8x – 1

Then, substitute the value of x in the equation

= (8 × 2) – 1

= 16 – 1

= 15

**(iii) 6x + 5 (x – 2)**

**Solution:-**

From the question it is given that x = 2

We have,

= 6x + 5x – 10

= 11x – 10

Then, substitute the value of x in the equation

= (11 × 2) – 10

= 22 – 10

= 12

**(iv) 4(2x – 1) + 3x + 11**

**Solution:-**

From the question it is given that x = 2

We have,

= 8x – 4 + 3x + 11

= 11x + 7

Then, substitute the value of x in the equation

= (11 × 2) + 7

= 22 + 7

= 29

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

**(i) 3x – 5 – x + 9**

**Solution:-**

From the question it is given that x = 3

We have,

= 3x – x – 5 + 9

= 2x + 4

Then, substitute the value of x in the equation

= (2 × 3) + 4

= 6 + 4

= 10

**(ii) 2 – 8x + 4x + 4**

**Solution:-**

From the question it is given that x = 3

We have,

= 2 + 4 – 8x + 4x

= 6 – 4x

Then, substitute the value of x in the equation

= 6 – (4 × 3)

= 6 – 12

= – 6

**(iii) 3a + 5 – 8a + 1**

**Solution:-**

From the question it is given that a = -1

We have,

= 3a – 8a + 5 + 1

= – 5a + 6

Then, substitute the value of a in the equation

= – (5 × (-1)) + 6

= – (-5) + 6

= 5 + 6

= 11

**(iv) 10 – 3b – 4 – 5b**

**Solution:-**

From the question it is given that b = -2

We have,

= 10 – 4 – 3b – 5b

= 6 – 8b

Then, substitute the value of b in the equation

= 6 – (8 × (-2))

= 6 – (-16)

= 6 + 16

= 22

**(v) 2a – 2b – 4 – 5 + a**

**Solution:-**

From the question it is given that a = -1, b = -2

We have,

= 2a + a – 2b – 4 – 5

= 3a – 2b – 9

Then, substitute the value of a and b in the equation

= (3 × (-1)) – (2 × (-2)) – 9

= -3 – (-4) – 9

= – 3 + 4 – 9

= -12 + 4

= -8

**8. (i) If z = 10, find the value of z ^{3} – 3(z – 10).**

**Solution:-**

From the question it is given that z = 10

We have,

= z^{3} – 3z + 30

Then, substitute the value of z in the equation

= (10)^{3} – (3 × 10) + 30

= 1000 – 30 + 30

= 1000

**(ii) If p = – 10, find the value of p ^{2} – 2p – 100**

**Solution:-**

From the question it is given that p = -10

We have,

= p^{2} – 2p – 100

Then, substitute the value of p in the equation

= (-10)^{2} – (2 × (-10)) – 100

= 100 + 20 – 100

= 20

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

**Solution:-**

From the question it is given that x = 0

We have,

2x^{2} + x – a = 5

a = 2x^{2} + x – 5

Then, substitute the value of x in the equation

a = (2 × 0^{2}) + 0 – 5

a = 0 + 0 – 5

a = -5

**10. Simplify the expression and find its value when a = 5 and b = – 3.**

**2(a ^{2} + ab) + 3 – ab**

**Solution:-**

From the question it is given that a = 5 and b = -3

We have,

= 2a^{2} + 2ab + 3 – ab

= 2a^{2} + ab + 3

Then, substitute the value of a and b in the equation

= (2 × 5^{2}) + (5 × (-3)) + 3

= (2 × 25) + (-15) + 3

= 50 – 15 + 3

= 53 – 15

= 38

**Exercise 12.4**

Question 1.

Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 3.

Solution:

(i) The number of line segments required to form

n digits is given by the expressions.

**6 = 5n + 1**

For 5 figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26

For 10 figures, the number of line segments = 5 × 10 + 1

= 50 + 1 = 51

For 100 figures, the number of line segments = 5 × 100 + 1

= 500 + 1 = 501

** 4 = 3n + 1**

For 5 figures, the number of line segments =3 ×5 + 1

= 15 + 1 = 16

For 10 figures, the number of line segments = 3 × 10 + 1

= 30 + 1 = 31

For 100 figures, the number of line segments = 3 × 100 + 1

= 300 + 1 = 301

**8 = 5n + 1**

For 5 figures, the number of line numbers = 5 × 5 + 2

= 25 + 2 = 27

For 10 figures, the number of line segments = 5 × 10 + 2

= 50 + 2 = 52

For 100 figures, the number of line segments = 5 × 100 + 2

= 500 + 2 = 502

Question 2.

Use the given algebraic expression to complete the table of number patterns:

Solution:

(i) Given expression is 2n – 1

For n = 100, 2 × 100 – 1

= 200 – 1 = 199

(ii) Given expression is 3n + 2

For n = 5, 3 × 5 + 2 = 15 + 2 = 17

For n = 10, 3 × 10 + 2 = 30 + 2 = 32

For n = 100, 3 × 100 + 2 = 300 + 2 = 302

(iii) Given expression is 4n + 1

For n = 5, 4 × 5 + 1 = 20 + 1 = 21

For n = 10, 4 × 10 + 1 = 40 + 1 = 41

For n = 100, 4 × 100 + 1 = 400 + 1 = 401

(iv) Given expression is 7n + 20

For n = 5, 7 × 5 + 20 = 35 + 20 = 55

For n = 10, 7 × 10 + 20 = 70 + 20 = 90

For n = 100, 7 × 100 + 20 = 700 + 20 = 720

(v) Given expression is n^{2} + 1

For n = 5, 5^{2} + 1 = 25 + 1 = 26

For n = 10, 10^{2} + 1 = 100 + 1 = 101