RD Sharma Solutions for Class 8 Maths Chapter 6 (2025-26)

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Download PDF of RD Sharma Class 8 Maths Solutions Chapter 6 - Algebraic Expressions and Identities

In RD Sharma Class 8 Maths Chapter 6- Algebraic Expressions and Identities, students will find a range of exercise questions designed to deepen their understanding of the core concepts.

We’ve provided detailed, step-by-step solutions for every exercise in this chapter to support effective learning. These RD Sharma solutions are curated to simplify complex problems and help students master algebraic expressions and identities with ease. Read all the solved exercises from Chapter 6 below to strengthen your grasp of the topic and prepare confidently for your exams.

Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 6

Q. Classify the following polynomials as monomials, binomials and trinomials. Which polynomials do not fit into any category?

(i) x+y

(ii) 1000

(iii) x+x²+x³+x4

(iv) 7+a+5b

(v) 2b-3b²

(vi) 2y-3y²+4y³

(vii) 5x-4y+3x

(viii) 4a-15a²

(ix) xy+yz+zt+tx

(x) pqr

(xi) p²q+pq²

(xii) 2p+2q

Solution: 

(i) x + y

This expression has two terms: x and y.

Type: Binomial

(ii) 1000

This is a single-term expression.

Type: Monomial

(iii) x + x² + x³ + x⁴

This expression has four terms.

Type: None (It is a polynomial with more than three terms)

(iv) 7 + a + 5b

It includes three separate terms.

Type: Trinomial

(v) 2b – 3b²

There are two terms here: 2b and –3b².

Type: Binomial

(vi) 2y – 3y² + 4y³

Three distinct terms are present.

Type: Trinomial

(vii) 5x – 4y + 3x

This expression simplifies but contains three terms as written.

Type: Trinomial

(viii) 4a – 15a²

Contains two terms.

Type: Binomial

(ix) xy + yz + zt + tx

There are four distinct terms in this expression.

Type: None (Polynomial with more than three terms)

(x) pqr

A single term is present.

Type: Monomial

(xi) p²q + pq²

Consists of two terms.

Type: Binomial

(xii) 2p + 2q

Two terms are involved.

Type: Binomial

Q. Identify the terms and their coefficients for each of the following expressions:

(i) 7x²yz – 5xy

(ii) x² + x + 1

(iii) 3x²y² – 5x²y²z² + z²

(iv) 9 – ab + bc – ca

(v) a/2 + b/2 – ab

(vi) 0.2x – 0.3xy + 0.5y

Solution:

(i) Expression: 7x²yz – 5xy

• Terms: 7x²yz, –5xy
• Coefficient of 7x²yz = 7
• Coefficient of –5xy = –5

(ii) Expression: x² + x + 1

• Terms: x², x, 1
• Coefficient of x² = 1
• Coefficient of x = 1
• Coefficient of 1 = 1

(iii) Expression: 3x²y² – 5x²y²z² + z²

• Terms: 3x²y², –5x²y²z², z²
• Coefficient of 3x²y² = 3
• Coefficient of –5x²y²z² = –5
• Coefficient of z² = 1

(iv) Expression: 9 – ab + bc – ca

• Terms: 9, –ab, bc, –ca
• Coefficient of 9 = 9
• Coefficient of –ab = –1
• Coefficient of bc = 1
• Coefficient of –ca = –1

(v) Expression: a/2 + b/2 – ab

• Terms: a/2, b/2, –ab
• Coefficient of a/2 = ½
• Coefficient of b/2 = ½
• Coefficient of –ab = –1

(vi) Expression: 0.2x – 0.3xy + 0.5y

• Terms: 0.2x, –0.3xy, 0.5y
• Coefficient of 0.2x = 0.2
• Coefficient of –0.3xy = –0.3
• Coefficient of 0.5y = 0.5

Q. Subtract:

(i) -5xy from 12xy

(ii) 2a² from -7a²

(iii) 2a-b from 3a-5b

(iv) 2x³ – 4x² + 3x + 5 from 4x³ + x² + x + 6

(v) 2/3y³ – 2/7y² – 5 from 1/3y³ + 5/7y² + y – 2

(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z

(vii) x²y – 4/5xy² + 4/3xy from 2/3x²y + 3/2xy² – 1/3xy

(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac

Solution:

(i) -5xy from 12xy

Let us subtract the given expression

12xy – (- 5xy)

5xy + 12xy

17xy

(ii) 2a² from -7a²

Let us subtract the given expression

(-7a²) – 2a²

-7a² – 2a²

-9a²

(iii) 2a-b from 3a-5b

Let us subtract the given expression

(3a – 5b) – (2a – b)

3a – 5b – 2a + b

a – 4b

(iv) 2x³ – 4x² + 3x + 5 from 4x³ + x² + x + 6

Let us subtract the given expression

(4x³ + x² + x + 6) – (2x³ – 4x² + 3x + 5)

4x³ + x² + x + 6 – 2x³ + 4x² – 3x – 5

2x³ + 5x² – 2x + 1

(v) 2/3y³ – 2/7y² – 5 from 1/3y³ + 5/7y² + y – 2

Let us subtract the given expression

1/3y³ + 5/7y² + y – 2 – 2/3y³ + 2/7y² + 5

Upon rearranging

1/3y³ – 2/3y³ + 5/7y² + 2/7y² + y – 2 + 5

By grouping similar expressions, we get,

-1/3y³ + 7/7y² + y + 3

-1/3y³ + y² + y + 3

(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z

Let us subtract the given expression

2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)

Upon rearranging

2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z

By grouping similar expressions we get,

LCM for (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)

(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6

-5x/6 + 11y/4 + 13z/6

(vii) x²y – 4/5xy² + 4/3xy from 2/3x²y + 3/2xy² – 1/3xy

Let us subtract the given expression

2/3x²y + 3/2xy² – 1/3xy – (x²y – 4/5xy² + 4/3xy)

Upon rearranging

2/3x²y – x²y + 3/2xy² + 4/5xy² – 1/3xy – 4/3xy

By grouping similar expressions, we get,

LCM for (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)

-1/3x²y + 23/10xy² – 5/3xy

(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac

Let us subtract the given expression

3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)

Upon rearranging

3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7

By grouping similar expressions, we get,

LCM for (5 and 3 is 15), (5 and 5 is 5)

(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7

184bc/15 + -10ac/5 – ab/7

– ab/7 + 184bc/15 – 2ac

Q. (7ab) × (-5ab²c) × (6abc²)

Solution:

Let us simplify the given expression

7 × -5 × 6 × a × a × a × b × b² × b × c × c²

210 × a¹⁺¹⁺¹ × b¹⁺²⁺¹ × c¹⁺²

210a³b⁴c³

Q. (-7xy) × (1/4x²yz)

Solution:

Let us simplify the given expression

-7 × 1/4 × x × y × x² × y × z

-7/4 × x¹⁺² × y¹⁺¹ × z

-7/4x³y²z

Q. (-24/25x³z) × (-15/16xz²y)

Solution:

Let us simplify the given expression

-24/25 × -15/16 × x³ × x × z × z² × y

18/20 × x³⁺¹ × z¹⁺² × y

9/10x⁴z³y

Q. (-1/27a²b²) × (9/2a³b²c²)

Solution:

Let us simplify the given expression

-1/27 × 9/2 × a² × a³ × b² × b² × c²

-1/6 x a²⁺³ × b²⁺² × c²

-1/6a⁵b⁴c²

RD Sharma Class 8 Algebraic Expressions and Identities PDF

Practicing RD Sharma Class 8 Algebraic Expressions and Identities helps students strengthen their conceptual understanding through hands-on learning. By regularly solving questions from this chapter, learners develop practical knowledge that not only supports their performance in Algebra but also builds a strong foundation for solving problems in other math topics and real-life situations.

Moreover, students can conveniently access well-structured answers to Algebraic Expressions and Identities anytime, as the complete PDF solution is available on the Home-Tution website—making learning easier and more flexible from the comfort of home.

Advantages of RD Sharma Class 8 Algebraic Expressions and Identities

The RD Sharma Class 8 Maths chapter on Algebraic Expressions and Identities offers students a valuable opportunity to learn effective techniques for solving complex problems. Beyond helping students score better, this chapter delivers several key benefits:

1. Enhances the Learning Process- To master the concepts and methods in Algebraic Expressions and Identities, students require clear and structured guidance. RD Sharma Solutions provide expert explanations and a variety of practice questions, helping learners grasp each topic with greater confidence.
2. Clarifies Doubts and Reduces Confusion- Students often face uncertainty while dealing with new or tricky algebraic concepts. The step-by-step solutions in RD Sharma help resolve these doubts, offering clear explanations that make it easier to understand and apply each method correctly.
3. Supports Quick and Effective Revision- Regular practice with RD Sharma Class 8 solutions ensures that students revisit and reinforce important concepts. This makes it an excellent tool for revision before tests, enabling better retention and recall.
4. Builds Time Management Skills- Solving questions efficiently within a time limit is crucial in exams. By regularly practising the exercises from RD Sharma, students develop speed and discipline, sharpening their ability to manage time during assessments.
5. Improves Accuracy and Precision- Achieving correct answers consistently is essential in Mathematics. RD Sharma’s guided solutions help students focus on accuracy, showing them how to minimize errors and adopt the most efficient problem-solving strategies.
6. Helps Memorize Important Formulas and Concepts- Each exercise in RD Sharma includes formulas and identities central to solving algebraic problems. By repeatedly using them in practice, students naturally memorize key formulas and understand when and how to apply them effectively.

Also Read:

• RD Sharma Solutions for Class 8 Maths Chapter 1 – Rational Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 2 – Powers
• RD Sharma Class 8 Maths Solutions Chapter 3 - Squares and Square Roots
• RD Sharma Class 8 Maths Solutions Chapter 4 - Cubes and Cube Roots