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NCERT Solutions for Class 7 Maths Chapter 10: Algebraic Expressions — Free PDF

Master terms, coefficients, like and unlike terms, and addition and subtraction of algebraic expressions.

CBSEClass 7
The SparkEd Authors (IITian & Googler)15 March 202635 min read
NCERT Solutions Class 7 Maths Chapter 10 Algebraic Expressions — SparkEd

Chapter 10 Overview: Algebraic Expressions

This chapter introduces you to the formal language of algebra — expressions made up of variables, constants, and arithmetic operations. If you have been solving simple equations in Chapter 4, you already know that variables like xx and yy represent unknown numbers. Chapter 10 teaches you how to build, classify, and manipulate algebraic expressions systematically.

Algebraic expressions are the building blocks of all higher mathematics. Every equation you will solve in Classes 8, 9, and 10 is built from expressions. Learning to add, subtract, and evaluate them correctly is a skill you will use thousands of times in your math career.

The chapter has 4 exercises with a total of around 20 problems. Exercise 10.1 covers terms, factors, and coefficients. Exercise 10.2 teaches you to identify and group like and unlike terms. Exercise 10.3 focuses on addition and subtraction of expressions. Exercise 10.4 deals with finding the value of an expression by substituting given values for the variables.

Key Concepts and Definitions

Variable: A letter (like xx, yy, aa, bb) that represents an unknown or changeable number. In the expression 3x+53x + 5, xx is a variable.

Constant: A fixed number that does not change. In 3x+53x + 5, the number 55 is a constant. Also, 33 is a constant multiplied by xx.

Algebraic Expression: A combination of variables, constants, and operations (+,,×,÷+, -, \times, \div). Examples: 3x+53x + 5, 2a27b+12a^2 - 7b + 1, x2+3y\dfrac{x}{2} + 3y.

Term: Each part of an expression separated by ++ or - signs. In 4x23xy+74x^2 - 3xy + 7, the terms are 4x24x^2, 3xy-3xy, and 77.

Factor: The numbers and variables that are multiplied together to form a term. In 5xy5xy, the factors are 55, xx, and yy.

Coefficient: The numerical part of a term. In 7x2y-7x^2y, the coefficient is 7-7. In xx (which is 1x1 \cdot x), the coefficient is 11. The coefficient includes the sign.

Like Terms: Terms that have the same variable parts (same variables raised to the same powers). Examples: 3x23x^2 and 5x2-5x^2; 7ab7ab and 2ab2ab. Only like terms can be added or subtracted.

Unlike Terms: Terms with different variable parts. Examples: 3x3x and 3x23x^2; 5xy5xy and 5xz5xz.

Types of Expressions:
- Monomial: 11 term (e.g., 5x25x^2, 3ab-3ab, 77)
- Binomial: 22 terms (e.g., 3x+73x + 7, a2b2a^2 - b^2)
- Trinomial: 33 terms (e.g., x2+3x+2x^2 + 3x + 2)
- Polynomial: A general expression with one or more terms (monomials, binomials, and trinomials are all types of polynomials).

Exercise 10.1 — Terms, Factors, and Coefficients (Solved)

**Q1. Identify the terms in 4x23xy+5y+74x^2 - 3xy + 5y + 7.**

The terms are: 4x24x^2, 3xy-3xy, 5y5y, and 77.

Note: The sign belongs to the term. The second term is 3xy-3xy (not just 3xy3xy).

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**Q2. Identify the coefficient of xx in each:**

(a) 5x+35x + 3 — coefficient of xx is 55.

(b) 7x2+2x-7x^2 + 2x — coefficient of xx is 22; coefficient of x2x^2 is 7-7.

(c) x+1x + 1 — coefficient of xx is 11 (often not written explicitly, but it is there).

(d) x+4-x + 4 — coefficient of xx is 1-1.

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Q3. Classify the following as monomials, binomials, or trinomials:

  • 5x25x^2Monomial (11 term)
    - 3x+73x + 7Binomial (22 terms)
    - 4x23x+14x^2 - 3x + 1Trinomial (33 terms)
    - 2xy2xyMonomial
    - a+b+c+da + b + c + dPolynomial with 44 terms (none of the specific names apply)

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**Q4. Write two like terms for 3x2y3x^2y.**

Like terms have the same variable part. Examples: 5x2y-5x^2y and x2yx^2y (or 12x2y\dfrac{1}{2}x^2y).

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**Q5. Identify the terms and their factors in 3a2+5ab7b+23a^2 + 5ab - 7b + 2.**

TermFactors
3a23a^23,a,a3, a, a
5ab5ab5,a,b5, a, b
7b-7b7,b-7, b
2222 (constant)

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Exercise 10.2 — Like and Unlike Terms (Solved)

Like terms have the same variables raised to the same powers. Only the coefficients can differ.

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Q1. Group the like terms:
3x,  5y,  2x,  7,  3y,  4x,  103x, \; 5y, \; -2x, \; 7, \; -3y, \; 4x, \; 10

Group 1 (xx terms): 3x,2x,4x3x, -2x, 4x

Group 2 (yy terms): 5y,3y5y, -3y

Group 3 (constants): 7,107, 10

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**Q2. Simplify by combining like terms: 7x3x+5xx7x - 3x + 5x - x.**

(73+51)x=8x(7 - 3 + 5 - 1)x = 8x

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**Q3. Simplify: 3x2+5x2x2+x43x^2 + 5x - 2x^2 + x - 4.**

Combine x2x^2 terms: 3x22x2=x23x^2 - 2x^2 = x^2.

Combine xx terms: 5x+x=6x5x + x = 6x.

Constants: 4-4.

Result=x2+6x4\text{Result} = x^2 + 6x - 4

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**Q4. Are 5xy5xy and 5yx5yx like terms?**

Yes! xy=yxxy = yx (multiplication is commutative), so 5xy5xy and 5yx5yx are like terms. They can be combined.

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**Q5. Are 3x23x^2 and 3x3x like terms?**

No! The powers of xx are different (x2x^2 vs xx). These are unlike terms and cannot be combined.

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**Q6. Simplify: 2a2b+3ab2a2b+5ab22a^2b + 3ab^2 - a^2b + 5ab^2.**

a2ba^2b terms: 2a2ba2b=a2b2a^2b - a^2b = a^2b.

ab2ab^2 terms: 3ab2+5ab2=8ab23ab^2 + 5ab^2 = 8ab^2.

Result=a2b+8ab2\text{Result} = a^2b + 8ab^2

Note: a2ba^2b and ab2ab^2 are NOT like terms (different variable parts).

Exercise 10.3 — Addition and Subtraction of Expressions (Solved)

**Q1. Add 3x+53x + 5 and 4x24x - 2.**

(3x+5)+(4x2)=3x+4x+52=7x+3(3x + 5) + (4x - 2) = 3x + 4x + 5 - 2 = 7x + 3

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**Q2. Add 2x2+3x+12x^2 + 3x + 1 and x25x+4x^2 - 5x + 4.**

(2x2+3x+1)+(x25x+4)(2x^2 + 3x + 1) + (x^2 - 5x + 4)

=(2+1)x2+(35)x+(1+4)= (2 + 1)x^2 + (3 - 5)x + (1 + 4)

=3x22x+5= 3x^2 - 2x + 5

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**Q3. Subtract 2x23x+12x^2 - 3x + 1 from 5x2+4x75x^2 + 4x - 7.**

(5x2+4x7)(2x23x+1)(5x^2 + 4x - 7) - (2x^2 - 3x + 1)

=5x2+4x72x2+3x1= 5x^2 + 4x - 7 - 2x^2 + 3x - 1

=3x2+7x8= 3x^2 + 7x - 8

Important: When subtracting, change the sign of every term in the expression being subtracted.

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**Q4. Subtract 3a+4b2c3a + 4b - 2c from 7ab+5c7a - b + 5c.**

(7ab+5c)(3a+4b2c)(7a - b + 5c) - (3a + 4b - 2c)

=7ab+5c3a4b+2c= 7a - b + 5c - 3a - 4b + 2c

=4a5b+7c= 4a - 5b + 7c

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**Q5. Add: 5p2+3p75p^2 + 3p - 7, 2p2p+42p^2 - p + 4, and p2+6p1-p^2 + 6p - 1.**

Using the column method:

p2p^2 terms: 5p2+2p2+(p2)=6p25p^2 + 2p^2 + (-p^2) = 6p^2

pp terms: 3p+(p)+6p=8p3p + (-p) + 6p = 8p

Constants: 7+4+(1)=4-7 + 4 + (-1) = -4

Result=6p2+8p4\text{Result} = 6p^2 + 8p - 4

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**Q6. What should be added to 3x25x+23x^2 - 5x + 2 to get 7x2+x67x^2 + x - 6?**

Let the required expression be EE.

3x25x+2+E=7x2+x63x^2 - 5x + 2 + E = 7x^2 + x - 6

E=(7x2+x6)(3x25x+2)E = (7x^2 + x - 6) - (3x^2 - 5x + 2)

=7x2+x63x2+5x2=4x2+6x8= 7x^2 + x - 6 - 3x^2 + 5x - 2 = 4x^2 + 6x - 8

Exercise 10.4 — Finding the Value of an Expression (Solved)

To find the value of an expression, substitute the given values of the variables and evaluate using BODMAS.

**Q1. Find the value of 3x+23x + 2 when x=4x = 4.**

3(4)+2=12+2=143(4) + 2 = 12 + 2 = 14

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**Q2. Find the value of x2+2x3x^2 + 2x - 3 when x=2x = -2.**

(2)2+2(2)3=443=3(-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3

Note: (2)2=4(-2)^2 = 4 (positive), not 4-4. Be very careful with negative numbers!

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**Q3. Find the value of 2a2b+3ab2a^2b + 3ab when a=1a = 1 and b=1b = -1.**

2(1)2(1)+3(1)(1)=2(1)(1)+3(1)=2+(3)=52(1)^2(-1) + 3(1)(-1) = 2(1)(-1) + 3(-1) = -2 + (-3) = -5

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**Q4. If p=2p = -2, find the value of 4p23p+74p^2 - 3p + 7.**

4(2)23(2)+7=4(4)+6+7=16+6+7=294(-2)^2 - 3(-2) + 7 = 4(4) + 6 + 7 = 16 + 6 + 7 = 29

Note: 3×(2)=+6-3 \times (-2) = +6 (negative times negative is positive).

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**Q5. Find the value of x21x+1\dfrac{x^2 - 1}{x + 1} when x=3x = 3.**

3213+1=914=84=2\dfrac{3^2 - 1}{3 + 1} = \dfrac{9 - 1}{4} = \dfrac{8}{4} = 2

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**Q6. If a=2a = 2 and b=3b = 3, find the value of a2+b2+2aba^2 + b^2 + 2ab. Does it equal (a+b)2(a + b)^2?**

a2+b2+2ab=4+9+12=25a^2 + b^2 + 2ab = 4 + 9 + 12 = 25

(a+b)2=(2+3)2=52=25(a + b)^2 = (2 + 3)^2 = 5^2 = 25

Yes, they are equal! This is because a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2 is an algebraic identity (you will study this formally in Class 8, Chapter 8).

Additional Worked Examples

**Example 1. The perimeter of a rectangle with length ll and breadth bb is 2(l+b)2(l + b). If l=3x+2l = 3x + 2 and b=x1b = x - 1, find the perimeter.**

Solution:

P=2(l+b)=2[(3x+2)+(x1)]=2[4x+1]=8x+2P = 2(l + b) = 2[(3x + 2) + (x - 1)] = 2[4x + 1] = 8x + 2

If x=5x = 5: P=8(5)+2=42P = 8(5) + 2 = 42 units.

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**Example 2. What should be subtracted from 5a23a+95a^2 - 3a + 9 to get 2a2+a12a^2 + a - 1?**

Solution:
Let the expression to be subtracted be EE.

5a23a+9E=2a2+a15a^2 - 3a + 9 - E = 2a^2 + a - 1

E=(5a23a+9)(2a2+a1)E = (5a^2 - 3a + 9) - (2a^2 + a - 1)

=5a23a+92a2a+1=3a24a+10= 5a^2 - 3a + 9 - 2a^2 - a + 1 = 3a^2 - 4a + 10

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**Example 3. If x=2x = 2 and y=3y = -3, evaluate 3x2y2xy2+xy3x^2y - 2xy^2 + xy.**

Solution:

=3(4)(3)2(2)(9)+(2)(3)= 3(4)(-3) - 2(2)(9) + (2)(-3)

=3636+(6)= -36 - 36 + (-6)

=78= -78

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**Example 4. Simplify and then find the value for x=1x = 1: (4x2+3x2)+(x25x+6)(2x2+x3)(4x^2 + 3x - 2) + (x^2 - 5x + 6) - (2x^2 + x - 3).**

Solution:
Simplify first:

=4x2+3x2+x25x+62x2x+3= 4x^2 + 3x - 2 + x^2 - 5x + 6 - 2x^2 - x + 3

x2x^2 terms: 4+12=3x24 + 1 - 2 = 3x^2.
xx terms: 351=3x3 - 5 - 1 = -3x.
Constants: 2+6+3=7-2 + 6 + 3 = 7.

Result =3x23x+7= 3x^2 - 3x + 7.

For x=1x = 1: 3(1)3(1)+7=33+7=73(1) - 3(1) + 7 = 3 - 3 + 7 = 7.

Common Mistakes to Avoid

Mistake 1: Forgetting to change ALL signs when subtracting.
When subtracting (2x23x+1)(2x^2 - 3x + 1), you must change ALL three signs: 2x2+3x1-2x^2 + 3x - 1. Missing even one sign leads to a wrong answer.

Mistake 2: Combining unlike terms.
3x2+2x5x33x^2 + 2x \neq 5x^3 or 5x25x^2. You cannot combine x2x^2 and xx terms — they are unlike terms. Similarly, 5xy+3x8xy5xy + 3x \neq 8xy.

Mistake 3: Sign errors with negative numbers.
(2)2=4(-2)^2 = 4 (positive), NOT 4-4. Also, 3×(2)=+6-3 \times (-2) = +6, not 6-6. Be especially careful when substituting negative values.

**Mistake 4: Forgetting the coefficient 11 or 1-1.**
In the term xx, the coefficient is 11 (not zero). In ab-ab, the coefficient is 1-1. These hidden 11s are easy to forget.

**Mistake 5: Treating x2x^2 and 2x2x as like terms.**
The variable part must be EXACTLY the same. x2x^2 and xx have different powers, so they are unlike terms. 3xy3xy and 3x2y3x^2y are also unlike terms.

Exam Tips for Algebraic Expressions

1. When subtracting, change ALL signs of the expression being subtracted. This is the most common error in this chapter.
2. Arrange terms by decreasing powers of the variable for clarity: 3x2+2x53x^2 + 2x - 5 (not 2x5+3x22x - 5 + 3x^2).
3. Remember: x2x^2 and xx are NOT like terms. They cannot be combined.
4. When finding values, be careful with negative numbers: (2)2=4(-2)^2 = 4, not 4-4.
5. Use the column method for adding/subtracting expressions — write like terms in the same column for easy computation.
6. Check your answer by substituting a simple value (like x=1x = 1) into both the original and simplified expressions. They should give the same result.
7. The coefficient of xx in x+5-x + 5 is 1-1, not 00 or 11. Always include the sign.
8. Practice identifying like terms quickly — this skill is needed throughout Classes 8, 9, and 10.

Practice Questions with Answers

Q1. Add: 5a3b+25a - 3b + 2 and 2a+7b4-2a + 7b - 4.

Answer: (5a2a)+(3b+7b)+(24)=3a+4b2(5a - 2a) + (-3b + 7b) + (2 - 4) = 3a + 4b - 2.

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Q2. Subtract x2+3x5x^2 + 3x - 5 from 4x2x+24x^2 - x + 2.

Answer: (4x2x+2)(x2+3x5)=4x2x+2x23x+5=3x24x+7(4x^2 - x + 2) - (x^2 + 3x - 5) = 4x^2 - x + 2 - x^2 - 3x + 5 = 3x^2 - 4x + 7.

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Q3. Find the value of 2x23x+12x^2 - 3x + 1 when x=1x = -1.

Answer: 2(1)23(1)+1=2+3+1=62(-1)^2 - 3(-1) + 1 = 2 + 3 + 1 = 6.

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Q4. What should be added to 2p3q+72p - 3q + 7 to get 5p+q25p + q - 2?

Answer: (5p+q2)(2p3q+7)=3p+4q9(5p + q - 2) - (2p - 3q + 7) = 3p + 4q - 9.

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Q5. Simplify: 4x[3x(2xx)]4x - [3x - (2x - x)].

Answer: Inner bracket: 2xx=x2x - x = x. Then: 3xx=2x3x - x = 2x. Finally: 4x2x=2x4x - 2x = 2x.

Key Takeaways

  • Term: A product of factors (e.g., 3x2y3x^2y is a single term).
    - Coefficient: The numerical part of a term, including the sign (7-7 in 7x2-7x^2).
    - Like terms: Same variable parts (3x3x and 5x-5x). Unlike terms: different variable parts (3x3x and 3x23x^2).
    - Monomial =1= 1 term, Binomial =2= 2 terms, Trinomial =3= 3 terms.
    - Addition: Combine like terms by adding their coefficients.
    - Subtraction: Change the sign of every term being subtracted, then add.
    - Finding value: Substitute the given values and evaluate using BODMAS. Be careful with negative numbers.
    - This chapter lays the groundwork for Chapter 8 (Algebraic Expressions and Identities) in Class 8.

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