Algebraic Expressions – NCERT Solutions for Class 7

Algebraic Expressions deals with solving mathematical expressions consisting of arithmetic operations to be evaluated by students. The exercise present in your NCERT mathematics textbook is designed to test students’ understanding of the same. 

 

SimplyAcad is offering the best NCERT Solutions for Class 7 Maths Chapter 10: Algebraic Expressions for consistent practice and revision to boost confidence in students. The solutions are explained in a step by step manner organised structurally for a hassle free experience. 

 

Following are the list of topics discussed in the Chapter 10: Algebraic Expressions

 

• Expressions Formation
• Different 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 of Class 7 Chapter 10 Algebraic Expressions Exercise 10.1 

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

= z2

(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:-

= x2 + y2

(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

Solution:-

Expression: x – 3

Terms: x, -3

Factors: x; -3

(b) 1 + x + x2

Solution:-

Expression: 1 + x + x2

Terms: 1, x, x2

Factors: 1; x; x,x

(c) y – y3

Solution:-

Expression: y – y3

Terms: y, -y3

Factors: y; -y, -y, -y

(d) 5xy2 + 7x2y

Solution:-

Expression: 5xy2 + 7x2y

Terms: 5xy2, 7x2y

Factors: 5, x, y, y; 7, x, x, y

(e) – ab + 2b2 – 3a2

Solution:-

Expression: -ab + 2b2 – 3a2

Terms: -ab, 2b2, -3a2

Factors: -a, b; 2, b, b; -3, a, a

(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) ¾ x + ¼

(h) 0.1 p2 + 0.2 q2

Solution:-

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)	¾ x + ¼	¾ x ¼	¾, x ¼
(h)	0.1 p2 + 0.2 q2	0.1p2 0.2q2	0.1, p, p 0.2, q, q

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

Solution:-

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

Solution:-

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) Identify terms which contain y2 and give the coefficient of y2.

(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2

Solution:-

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

5. Classify into monomials, binomials and trinomials.

(i) 4y – 7z

Solution:-

Binomial.

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

(ii) y2

Solution:-

Monomial.

An expression with only one term is called a monomial.

(iii) x + y – xy

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(iv) 100

Solution:-

Monomial.

An expression with only one term is called a monomial.

(v) ab – a – b

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(vi) 5 – 3t

Solution:-

Binomial.

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

 

(vii) 4p2q – 4pq2

Solution:-

Binomial.

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

 

(viii) 7mn

Solution:-

Monomial.

An expression with only one term is called a monomial.

(ix) z2 – 3z + 8

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(x) a2 + b2

Solution:-

Binomial.

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

(xi) z2 + z

Solution:-

Binomial.

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

(xii) 1 + x + x2

Solution:-

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

Solution:-

Like term.

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

 

(ii) –7x, (5/2)x

Solution:-

Like term.

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

 

(iii) – 29x, – 29y

Solution:-

Unlike terms.

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

(iv) 14xy, 42yx

Solution:-

Like term.

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

 

(v) 4m2p, 4mp2

Solution:-

Unlike terms.

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

(vi) 12xz, 12x2z2

Solution:-

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

Solution:-

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

Solution:-

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

NCERT Solutions of Class 7 Chapter 10 Algebraic Expressions Exercise 10.2 

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

131

(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

65

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

60

(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

NCERT Solutions of Class 7 Chapter 10 Algebraic Expressions Exercise 10.3

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) 3m2 – 2m – 7

Solution:-

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

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) – 3p2 + 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) – 2p3 – 3p2 + 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) x2 + 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) 2x2 – 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) a2 + b2

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) a2 + ab + b2

Solution:-

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

Solution:-

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

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) 2a2 + b2 + 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 × 02) + (-1)2 + 1

= 0 + 1 + 1

= 2

(iii) 2a2b + 2ab2 + 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 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)

= 0 + 0 +0

= 0

(iv) a2 + 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.

= (02) + (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 z3 – 3(z – 10).

Solution:-

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) If p = – 10, find the value of p2 – 2p – 100

Solution:-

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

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

Solution:-

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

36

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

2(a2 + ab) + 3 – ab

Solution:-

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

NCERT Solutions of Class 7 Chapter 10 Algebraic Expressions Exercise 10.4 

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

Solution:-

(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

2. 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	–

Solution:-

(i) From the table (2n – 1)

Then, 100th term =?

Where n = 100

= (2 × 100) – 1

= 200 – 1

= 199

(ii) From the table (3n + 2)

5th term =?

Where n = 5

= (3 × 5) + 2

= 15 + 2

= 17

Then, 10th term =?

Where n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Then, 100th term =?

Where n = 100

= (3 × 100) + 2

= 300 + 2

= 302

(iii) From the table (4n + 1)

5th term =?

Where n = 5

= (4 × 5) + 1

= 20 + 1

= 21

Then, 10th term =?

Where n = 10

= (4 × 10) + 1

= 40 + 1

= 41

Then, 100th term =?

Where n = 100

= (4 × 100) + 1

= 400 + 1

= 401

(iv) From the table (7n + 20)

5th term =?

Where n = 5

= (7 × 5) + 20

= 35 + 20

= 55

Then, 10th term =?

Where n = 10

= (7 × 10) + 20

= 70 + 20

= 90

Then, 100th term =?

Where n = 100

= (7 × 100) + 20

= 700 + 20

= 720

(v) From the table (n2 + 1)

5th term =?

Where n = 5

= (52) + 1

= 25+ 1

= 26

Then, 10th 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	–