RBSE Solutions Class 8 Maths Chapter 9 Algebraic Expressions More Ques

Question 1. Give five different examples of numeric and algebraic expression. Then categorize them into monomials, binomials and trinomials.
Answer:
Here are five numeric expressions: \( 4, 100, -17, 0, \frac{2}{3} \). These are simply numbers or fractions.
Here are five algebraic expressions:
\( 2y^2 \)
\( 3x^2 - 5 \)
\( 13 - y + y^2 \)
\( 4p^2q - 3pq^2 + 5 \)
\( xy + 4 \)

Now, let's categorize them based on the number of terms:
Monomial Expressions (one term): \( 2y^2 \)
Binomial Expressions (two terms): \( 3x^2 - 5 \), \( xy + 4 \)
Trinomial Expressions (three terms): \( 13 - y + y^2 \), \( 4p^2q - 3pq^2 + 5 \)
In simple words: Numeric expressions are just numbers. Algebraic expressions have letters and numbers. We sort them by how many parts they have: one part is a monomial, two parts is a binomial, and three parts is a trinomial.

🎯 Exam Tip: Remember that terms in an algebraic expression are separated by plus (+) or minus (-) signs. This helps count them correctly for categorization.

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.
Answer: For terms to be considered 'like terms', they must have both of these conditions met:
(iii) Same exponents (meaning the powers of the variables must be the same)
(iv) Same number of variables (meaning they must contain the exact same variables)
For example, \( 3x^2y \) and \( -5x^2y \) are like terms because they both have \( x^2y \). These conditions allow us to add or subtract them directly.
In simple words: Like terms must have the same letters with the same small numbers (exponents) on them. If they do, you can add or subtract them easily.

🎯 Exam Tip: Coefficients (the numbers in front of the variables) and signs (+ or -) can be different for like terms; only the variable part with its exponents must be identical.

Question 4. Write three like terms for the expression \( 7xy^2 \).
Answer: Three like terms similar to \( 7xy^2 \) could be:
\( -7xy^2 \)
\( 3xy^2 \)
\( -4xy^2 \)
These terms all have the same variables (\( x \) and \( y \)) with the same exponents (\( x^1y^2 \)), only their numerical part (coefficient) is different. This allows them to be combined through addition or subtraction.
In simple words: Like terms have the same letters and the same small power numbers. Only the big number in front can be different.

🎯 Exam Tip: To find a like term, just change the numerical coefficient; do not change the variables or their powers.

Question 5. Fill in the blanks by adding the following like terms
\( 4n + (- 3n) = \_ \)
\( 5x^2y + (- 3x^2y) = \_ \)
\( 5pq + 12pq = \_ \)
\( 2ab^2 + 11ab^2 = \_ \)
Answer: Let's fill in the blanks by adding the given like terms:
\( 4n + (-3n) = 4n - 3n = n \)
\( 5x^2y + (-3x^2y) = 5x^2y - 3x^2y = 2x^2y \)
\( 5pq + 12pq = 17pq \)
\( 2ab^2 + 11ab^2 = 13ab^2 \)
When adding like terms, we simply add their coefficients (the numbers in front) and keep the variable part exactly the same.
In simple words: To add terms that are alike, just add their numbers and keep the letters with their powers the same.

🎯 Exam Tip: Always make sure the terms are 'like terms' before adding or subtracting them. If they are not alike, you cannot combine them.

Question 6. Sheela says that the sum of \( 2pq \) and \( 4pq \) is \( 8p^2q^2 \). Is she right?
Answer: No, Sheela is not right.
When we add like terms like \( 2pq \) and \( 4pq \), we only add the numerical coefficients and keep the variable part the same.
So, \( 2pq + 4pq = (2+4)pq = 6pq \).
Sheela's answer, \( 8p^2q^2 \), would come from multiplying the terms, not adding them. When adding, the exponents of the variables do not change.
In simple words: Sheela is wrong. When you add \( 2pq \) and \( 4pq \), you get \( 6pq \). You only add the numbers in front, not change the letters' powers.

🎯 Exam Tip: Remember the difference between adding and multiplying algebraic terms. Addition combines coefficients of like terms, while multiplication involves multiplying both coefficients and variable parts (adding exponents).

Question 7. Raees adds \( 4p \) and \( 7q \) and gets \( 11pq \) as its answer. Do you agree with his answer?
Answer: No, I do not agree with Raees's answer.
The terms \( 4p \) and \( 7q \) are not like terms because they have different variables (\( p \) and \( q \)). We can only add or subtract terms if they are like terms.
Since \( 4p \) and \( 7q \) are unlike terms, they cannot be combined into a single term like \( 11pq \). The expression remains as \( 4p + 7q \). Combining them like Raees did is a common mistake when dealing with algebraic expressions.
In simple words: Raees is wrong. You cannot add \( 4p \) and \( 7q \) to get \( 11pq \) because \( p \) and \( q \) are different letters. They must stay separate.

🎯 Exam Tip: Always check if terms are 'like' before attempting to combine them through addition or subtraction. If the variable parts are different, the terms cannot be simplified further.

Question 9. Put \( -b \) in place of \( b \) in identity (I). Do you get identity (II)?
Answer: Yes, by substituting \( -b \) in place of \( b \) in Identity (I), we can derive Identity (II).
Identity (I) is: \( (a + b)^2 = a^2 + 2ab + b^2 \)
Now, we replace every instance of \( b \) with \( (-b) \):
\( \{a + (-b)\}^2 = a^2 + 2a(-b) + (-b)^2 \)
\( (a - b)^2 = a^2 - 2ab + b^2 \)
This result is indeed Identity (II). This shows how one algebraic identity can lead to another through simple substitution.
In simple words: If you take the first math rule (identity) and change every \( b \) to \( -b \), you will end up with the second math rule. So, yes, you get identity (II).

🎯 Exam Tip: When substituting a negative value into an identity, always use parentheses, like \( (-b) \), to ensure the signs are handled correctly, especially when squaring or multiplying.