RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions In Text Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Algebraic Expressions |

Exercise |
In Text Exercise |

Number of Questions |
9 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions In Text Exercise

**Page No: 103**

Question 1.

Give five different examples of numeric and algebraic expression. Then(RBSESolutions.com)categorize them into monomials, binomials and trinomials.

Solution

5 Numeral expressions are

4, 100, – 17, 0, \(\frac { 2 }{ 3 }\)

5 algebric expressions are—

2y², 3x² – 5, 13 – y + y²,

4p²q – 3pq² + 5, xy + 4

Monomial Expressions—2y²

Binomial Expressions—3x² – 5, xy + 4

Trinomial Expressions—13 – y + y², 4p²q – 3pq² + 5

**Page No: 104**

Question 2.

From the following tick the essential condition for like terms.

(i) Same signs

(ii) Same coefficient

(iii) Same exponents

(iv) Same number of variable.

Solution

For an expression to be like the(RBSESolutions.com)necessary conditions are same exponents and same number of variables.

**Page No: 104**

Question 3.

Find the like terms from the following—

ax²y, 2n, 5y² – 7x², – 3n, 7xy, 25y²

Solution

2n, – 3n are like terms.

Question 4.

Write three like terms for the expression 7xy².

Solution

Like terms similar to 7xy² are – 7xy², 3xy², – 4xy2²

**Page No: 105**

Question 5.

Fill in the blanks by adding the following like terms

4n + (- 3n) =__ – 5x²y + ( – 3x²y) = ___

5pq + 12pq =__ 2ab² + 11ab² = ___

Solution

4n + (-3 n) = n

5pq + 12pq = 17pq

– 5x²y + (-3x²y) = – 8x²y

2ab² + 11ab² = 13ab²

Question 6.

Sheela says that(RBSESolutions.com)the sum of 2pq and 4pq is 8p²q². Is she right?

Solution

2pq + 4pq = 6pq ≠ 8p²q²

Hence she is not right.

Question 7.

Raees adds 4p and 7q and gets 11pq as its answer. Do you agree with his answer?

Solution

No.

**Page No: 106**

Question 8.

If A = 2y² + 3x – x² and B = 3x² – y² then, find A + B and A – B?

Solution

A + B

= 2y² + 3x – x² + (3x² – y²)

= 2y² + 3x – x² + 3x² – y²

= 2y² – y² – x² + 3x² + 3x

= y² + 2x² + 3x

A – B = 2y² + 3x – x² – (3x² – y²)

= 2y² + 3x – x² – 3x² + y²

= 2y² + y² – x² – 3x² + 3x

= 3y² – 4x² + 3x

**Page No: 112**

Question 9

Put – b in place of b in identity (I). Do you get identity (II)?

Solution

Identity (I) is

(a + b)² = a² + 2ab + b²

We put – b in place of b

{a + (- b)}² = a² + 2a (- b) + (- b)² = (a – b)² = a² – 2ab + b²

Which is(RBSESolutions.com)identity (II). Yes, we get identity (II).

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