NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Question

Here you will find NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions, which are essential for understanding the chapter's basic concepts. You can get a PDF of Chapter 12 Algebraic Expressions NCERT Solutions for Class 7 Maths that will make learning fun and help you finish the syllabus in less time. These NCERT Solutions will assist you in developing your own homework answers and will also make you aware of the difficulty of the questions. Class 7 NCERT Solutions will help you solve difficult problems in a given exercise. Chapter 12 NCERT Solutions will be extremely beneficial in exam preparation and understanding of concepts.

• How are Expression Formed
• Terms of An Expression
• Like and Unlike Terms
• Monomials, Binomials, Trinomials and Polynomials
• Addition and Subtraction of Algebraic Expressions
• Finding The Value of An Expression
• Using Algebraic Expressions Formulas and Rules

NCERT Solutions Class 7 Maths provided by Sarthak is one the great way to learn about all the topics related to Algebraic Expressions and Identities. All the topics are explained in the step-wise method. As suggested by our mentors it is the best way to learn, solve, revise, complete homework, making assignments.

Answer 1

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

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 numbers m and n.

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

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

Answer:

(i) Subtraction of z from y.

= Y – z

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

= 1/2 (x + y)

= (x + y)/2

(iii) The number z multiplied by itself.

= z × z

= z²

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

= 1/4 (p × q)

= pq/4

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

= x² + y²

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

= 3mn + 5

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

= 10 – (y × z)

= 10 – yz

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

= (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 + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) – ab + 2b² – 3a²

(II) Identify terms and factors in the expressions given below.

(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2 ab – 2.4 b + 3.6 a
(g) 3/4 x + 1/4
(h) 0.1 p² + 0.2 q²

Answer:

(I)

(II) Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.

In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.

Factors are defined as numbers we can multiply together to get another number.

Sl. No.	Expression	Terms	Factors
(a)	– 4x + 5	-4x 5	-4, x 5
(b)	– 4x + 5y	-4x 5y	-4, x 5, y
(c)	5y + 3y2	5y 3y2	5, y 3, y, y
(d)	xy + 2x2y2	xy 2x2y2	x, y 2, x, x, y, y
(e)	pq + q	pq q	P, q Q
(f)	1.2 ab – 2.4 b + 3.6 a	1.2ab -2.4b 3.6a	1.2, a, b -2.4, b 3.6, a
(g)	3/4 x + 1/4	3/4 x 1/4	3/4, x 1/4
(h)	0.1 p2 + 0.2 q2	0.1p2 0.2q2	0.1, p, p 0.2, q, q

Answer 2

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

(i) 5 – 3t²

(ii) 1 + t + t² + t³

(iii) x + 2xy + 3y

(iv) 100m + 1000n

(v) – p²q² + 7pq

(vi) 1.2 a + 0.8 b

(vii) 3.14 r²

(viii) 2 (l + b)

(ix) 0.1 y + 0.01 y²

Answer:

Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.

In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.

A coefficient is a number used to multiply a variable (2x means 2 times x, so 2 is a coefficient). Variables on their own (without a number next to them) actually have a coefficient of 1 (x is really 1x).

Sl. No.	Expression	Terms	Coefficients
(i)	5 – 3t2	– 3t2	-3
(ii)	1 + t + t2 + t3	t t2 t3	1 1 1
(iii)	x + 2xy + 3y	x 2xy 3y	1 2 3
(iv)	100m + 1000n	100m 1000n	100 1000
(v)	– p2q2 + 7pq	-p2q2 7pq	-1 7
(vi)	1.2 a + 0.8 b	1.2a 0.8b	1.2 0.8
(vii)	3.14 r2	3.142	3.14
(viii)	2 (l + b)	2l 2b	2 2
(ix)	0.1 y + 0.01 y2	0.1y 0.01y2	0.1 0.01

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 + 25
(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:

(a)

Sl. No.	Expression	Terms	Coefficient of x
(i)	y2x + y	y2x	y2
(ii)	13y2 – 8yx	– 8yx	-8y
(iii)	x + y + 2	x	1
(iv)	5 + z + zx	x zx	1 z
(v)	1 + x + xy	xy	y
(vi)	12xy2 + 25	12xy2	12y2
(vii)	7x + xy2	7x xy2	7 y2

(b)

Sl. No.	Expression	Terms	Coefficient of y2
(i)	8 – xy2	– xy2	– x
(ii)	5y2 + 7x	5y2	5
(iii)	2x2y – 15xy2 + 7y2	– 15xy2 7y2	– 15x 7

Answer 3

5. Classify into monomials, binomials and trinomials.

(i) 4y – 7z

(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:

(i) 4y – 7z

Binomial.

An expression which contains two unlike terms is called a binomial.

(ii) y²

Monomial.

An expression with only one term is called a monomial.

(iii) x + y – xy

Trinomial.

An expression which contains three terms is called a trinomial.

(iv) 100

Monomial.

An expression with only one term is called a monomial.

(v) ab – a – b

Trinomial.

An expression which contains three terms is called a trinomial.

(vi) 5 – 3t

Binomial.

An expression which contains two unlike terms is called a binomial.

(vii) 4p²q – 4pq²

Binomial.

An expression which contains two unlike terms is called a binomial.

(viii) 7mn

Monomial.

An expression with only one term is called a monomial.

(ix) z² – 3z + 8

Trinomial.

An expression which contains three terms is called a trinomial.

(x) a² + b²

Binomial.

An expression which contains two unlike terms is called a binomial.

(xi) z² + z

Binomial.

An expression which contains two unlike terms is called a binomial.

(xii) 1 + x + x²

Trinomial.

An expression which contains three terms is called a trinomial.

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

(i) 1, 100

(ii) –7x, (5/2)x

(iii) – 29x, – 29y

(iv) 14xy, 42yx

(v) 4m²p, 4mp²

(vi) 12xz, 12x²z²

Answer:

(i) 1, 100

Like term.

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

(ii) –7x, (5/2)x

Like term.

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

(iii) – 29x, – 29y

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

(iv) 14xy, 42yx

Like term.

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

(v) 4m²p, 4mp²

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

(vi) 12xz, 12x²z²

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

7. Identify like terms in the following.

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

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

Answer:

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

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

They are,

– xy², 2xy²

– 4yx², 20x²y

8x², – 11x², – 6x²

7y, y

– 100x, 3x

– 11yx, 2xy

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

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

They are,

10pq, – 7qp, 78qp

7p, 2405p

8q, – 100q

– p²q², 12q²p²

– 23, 41

– 5p², 701p²

13p²q, qp²

Answer 4

8. Simplify combining like terms.

(i) 21b – 32 + 7b – 20b

(ii) – z² + 13z² – 5z + 7z³ – 15z

(iii) p – (p – q) – q – (q – p)

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

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

(vi) (3y² + 5y – 4) – (8y – y² – 4)

Answer:

(i) 21b – 32 + 7b – 20b

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

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)

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

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

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)

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

9. Add:

(i) 3mn, – 5mn, 8mn, – 4mn

(ii) t – 8tz, 3tz – z, z – t

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

(iv) a + b – 3, b – a + 3, a – b + 3

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

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

(vii) 4x²y, – 3xy², –5xy², 5x²y

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

(ix) ab – 4a, 4b – ab, 4a – 4b

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

Answer:

(i) 3mn, – 5mn, 8mn, – 4mn

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

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

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

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

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

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

Answer 5

(vii) 4x²y, – 3xy², –5xy², 5x²y

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

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

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

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

10. Subtract:

(i) –5y² from y²

(ii) 6xy from –12xy

(iii) (a – b) from (a + b)

(iv) a (b – 5) from b (5 – a)

(v) –m² + 5mn from 4m² – 3mn + 8

(vi) – x² + 10x – 5 from 5x – 10

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

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

Answer:

(i) –5y² from y²

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

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)

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)

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

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

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

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

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

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

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

Answer:

(a) 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) 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

Answer 6

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

Answer:

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

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

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

Answer:

(a) 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) 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

14. If m = 2, find the value of:

(i) m – 2

(ii) 3m – 5

(iii) 9 – 5m

(iv) 3m² – 2m – 7

(v) (5m/2) – 4

Answer:

(i) m – 2

From the question, it is given that m = 2

Then, substitute the value of m in the question.

= 2 -2

= 0

(ii) 3m – 5

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

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) 3m² – 2m – 7

From the question, it is given that m = 2

Then, substitute the value of m in the question.

= (3 × 2²) – (2 × 2) – 7

= (3 × 4) – (4) – 7

= 12 – 4 -7

= 12 – 11

= 1

(v) (5m/2) – 4

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

15. If p = – 2, find the value of:

(i) 4p + 7

(ii) – 3p² + 4p + 7

(iii) – 2p³ – 3p² + 4p + 7

Answer:

(i) 4p + 7

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) – 3p² + 4p + 7

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) – 2p³ – 3p² + 4p + 7

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

16. 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

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

From the question, it is given that x = -1

Then, substitute the value of x in the question.

= – (-1) + 2

= 1 + 2

= 3

(iii) x² + 2x + 1

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) 2x² – x – 2

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

Answer 7

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

(i) a² + b²

(ii) a² + ab + b²

(iii) a² – b²

Answer:

(i) a² + b²

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) a² + ab + b²

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

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

= 2² + (2 × -2) + (-2)²

= 4 + (-4) + (4)

= 4 – 4 + 4

= 4

(iii) a² – b²

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

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

= 2² – (-2)²

= 4 – (4)

= 4 – 4

= 0

18. 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

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) 2a² + b² + 1

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

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

= (2 × 0²) + (-1)² + 1

= 0 + 1 + 1

= 2

(iii) 2a²b + 2ab² + ab

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

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

= (2 × 0² × -1) + (2 × 0 × (-1)²) + (0 × -1)

= 0 + 0 +0

= 0

(iv) a² + ab + 2

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

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

= (0²) + (0 × (-1)) + 2

= 0 + 0 + 2

= 2

19. 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)

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

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)

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

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

20. 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) 3x – 5 – x + 9

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

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

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

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

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

Answer 8

21. (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) From the question, it is given that z = 10

We have,

= z³ – 3z + 30

Then, substitute the value of z in the equation.

= (10)³ – (3 × 10) + 30

= 1000 – 30 + 30

= 1000

(ii) From the question, it is given that p = -10

We have,

= p² – 2p – 100

Then, substitute the value of p in the equation.

= (-10)² – (2 × (-10)) – 100

= 100 + 20 – 100

= 20

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

Answer:

From the question, it is given that x = 0

We have,

2x² + x – a = 5

a = 2x² + x – 5

Then, substitute the value of x in the equation.

a = (2 × 0²) + 0 – 5

a = 0 + 0 – 5

a = -5

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

2(a² + ab) + 3 – ab

Answer:

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

We have,

= 2a² + 2ab + 3 – ab

= 2a² + ab + 3

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

= (2 × 5²) + (5 × (-3)) + 3

= (2 × 25) + (-15) + 3

= 50 – 15 + 3

= 53 – 15

= 38

24. 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

.

Answer:

(a) From the question, it is given that the number of segments required to form n digits of the kind

 

is (5n + 1)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 1)

= (25 + 1)

= 26

The number of segments required to form 10 digits = ((5 × 10) + 1)

= (50 + 1)

= 51

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 1)

= 501

(b) From the question, it is given that the number of segments required to form n digits of the kind

is (3n + 1)

Then,

The number of segments required to form 5 digits = ((3 × 5) + 1)

= (15 + 1)

= 16

The number of segments required to form 10 digits = ((3 × 10) + 1)

= (30 + 1)

= 31

The number of segments required to form 100 digits = ((3 × 100) + 1)

= (300 + 1)

= 301

(c) From the question, it is given that the number of segments required to form n digits of the kind

is (5n + 2)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 2)

= (25 + 2)

= 27

The number of segments required to form 10 digits = ((5 × 10) + 2)

= (50 + 2)

= 52

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 2)

= 502

Answer 9

25. Use the given algebraic expression to complete the table of number patterns.

S. No.	Expression	Terms
		1st	2nd	3rd	4th	5th	…	10th	…	100th	…
(i)	2n – 1	1	3	5	7	9	–	19	–	–	–
(ii)	3n + 2	5	8	11	14	–	–	–	–	–	–
(iii)	4n + 1	5	9	13	17	–	–	–	–	–	–
(iv)	7n + 20	27	34	41	48	–	–	–	–	–	–
(v)	n2 + 1	2	5	10	17	–	–	–	–	10001	–

Answer:

(i) From the table (2n – 1)

Then, 100^th term =?

Where n = 100

= (2 × 100) – 1

= 200 – 1

= 199

(ii) From the table (3n + 2)

5^th term =?

Where n = 5

= (3 × 5) + 2

= 15 + 2

= 17

Then, 10^th term =?

Where n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Then, 100^th term =?

Where n = 100

= (3 × 100) + 2

= 300 + 2

= 302

(iii) From the table (4n + 1)

5^th term =?

Where n = 5

= (4 × 5) + 1

= 20 + 1

= 21

Then, 10^th term =?

Where n = 10

= (4 × 10) + 1

= 40 + 1

= 41

Then, 100^th term =?

Where n = 100

= (4 × 100) + 1

= 400 + 1

= 401

(iv) From the table (7n + 20)

5^th term =?

Where n = 5

= (7 × 5) + 20

= 35 + 20

= 55

Then, 10^th term =?

Where n = 10

= (7 × 10) + 20

= 70 + 20

= 90

Then, 100^th term =?

Where n = 100

= (7 × 100) + 20

= 700 + 20

= 720

Answer 10

(v) From the table (n² + 1)

5^th term =?

Where n = 5

= (5²) + 1

= 25+ 1

= 26

Then, 10^th term =?

Where n = 10

= (10²) + 1

= 100 + 1

= 101

So, the table is completed below.

S. No.	Expression	Terms
		1st	2nd	3rd	4th	5th	…	10th	…	100th	…
(i)	2n – 1	1	3	5	7	9	–	19	–	199	–
(ii)	3n + 2	5	8	11	14	17	–	32	–	302	–
(iii)	4n + 1	5	9	13	17	21	–	41	–	401	–
(iv)	7n + 20	27	34	41	48	55	–	90	–	720	–
(v)	n2 + 1	2	5	10	17	26	–	101	–	10001	–