RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions In Text Exercise

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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