NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions are provided here. Students can check for the solutions whenever they are facing difficulty while solving the questions from NCERT Solutions for Class 7 Maths Chapter 12.

The NCERT Solutions for Chapter 12 are available in PDF format so that students can download and learn offline as well. These books are one of the top materials when it comes to providing a question bank to practice. The topics covered in the chapter are as follows.

• How Are Expressions 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 for Class 7 Maths Chapter 12 Algebraic Expressions

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Exercise 12.1 Solutions

Exercise 12.2 Solutions

Exercise 12.3 Solutions

Exercise 12.4 Solutions

Access answers to Maths NCERT Solutions for Class 7 Chapter 12 – Algebraic Expressions

Exercise 12.1 Page: 234

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.

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).

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:-

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

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

Solution:-

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

Exercise 12.2 Page: 239

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

(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 Page: 242

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

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

Exercise 12.4 Page: 246

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.

Solution:-

(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

(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.

Disclaimer:

Dropped Topics – 12.6 Addition and subtraction of algebraic expressions, 12.8 Using algebraic expressions–formulas and rules