Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.2

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.2

Question 1. Fill in the blanks
(i) \( \frac { 18m^{4} }{ 2m^{3}n^{3} } = \_\_\_ mn^{3} \).
Answer:
(i) \( \frac{18 m^{4} n^{8}}{2 m^{3} n^{3}} = 9 mn^{5} \) Here, the exponent of 'n' in the numerator allows for this simplification, combining terms using exponent rules.
In simple words: When you divide algebraic terms, subtract the exponents of the same variables. The result fills the blank.

🎯 Exam Tip: Always remember to apply the rules of exponents correctly for each variable when dividing, \( \frac{a^m}{a^n} = a^{m-n} \).

Question 1. Fill in the blanks
(ii) \( \frac { l^{4}m^{5}n^{7} }{ 2l^{3}m^{3}n^{6} } = \_\_\_ \).
Answer:
(ii) \( \frac{l^{4} m^{5} n^{7}}{2 l^{3} m^{3} n^{6}} = \frac{l^{4-3} m^{5-3} n^{7-6}}{2} = \frac{l^{1} m^{2} n^{1}}{2} = \frac{l m^{2} n}{2} \) Division involves subtracting the exponents of identical variables.
In simple words: To simplify this fraction, divide the numbers and subtract the powers of the same letters.

🎯 Exam Tip: Treat each variable's exponent separately, and remember that a coefficient of 1 is often not written.

Question 1. Fill in the blanks
(iii) \( \frac { 42a^{2}b^{5}(\_\_\_) }{ 6a^{4}b^{2} } = (\_\_\_)b^{3}c^{2} \).
Answer:
(iii) \( \frac{42 a^{4} b^{5}(c^{2})}{6(a)^{4}(b)^{2}} = (7)b^{3}c^{2} \) By comparing the given expression with the result, we can find the missing terms. The 'a' terms cancel out if the numerator has `a^4`.
In simple words: We need to find the missing parts. By dividing the numbers and comparing the powers of the letters in the question and answer, we can find what's missing.

🎯 Exam Tip: Carefully cross-check the coefficients and exponents of each variable on both sides of the equation to find the unknown terms.

Question 2. Say True or False
(i) \( 8x^{3}y \div 4x^{2} = 2xy \)
Answer: True
In simple words: When we divide \( 8x^3y \) by \( 4x^2 \), we get \( 2xy \). So the statement is correct.

🎯 Exam Tip: Perform the division operation yourself to verify if the given equality holds true.

Question 2. Say True or False
(ii) \( 7ab^{3} \div 14ab = 2b^{2} \)
Answer: False
In simple words: Dividing \( 7ab^3 \) by \( 14ab \) gives \( \frac{1}{2}b^2 \), not \( 2b^2 \). So the statement is incorrect.

🎯 Exam Tip: Double-check the numerical coefficients as well as the exponents. \( \frac{7}{14} \) simplifies to \( \frac{1}{2} \), not 2.

Question 3. Divide
(i) \( 27 y^{3} \) by \( 3y \)
Answer:
(i) \( \frac{27 y^{3}}{3 y} = \frac{27}{3} y^{3-1} = 9y^{2} \) We divide the numbers and subtract the exponents of the same variable.
In simple words: Divide the numbers \( 27 \) by \( 3 \) to get \( 9 \). Then, for the letter \( y \), subtract the small numbers (exponents) to get \( y^2 \).

🎯 Exam Tip: Remember that \( y \) can be written as \( y^1 \) when performing exponent subtraction.

Question 3. Divide
(ii) \( x^{3} y^{2} \) by \( x^{2}y \)
Answer:
(ii) \( \frac{x^{3} y^{2}}{x^{2} y} = x^{3-2} y^{2-1} = x^{1} y^{1} = xy \) Here, we apply the exponent rule for division to both variables 'x' and 'y'.
In simple words: For 'x', subtract the powers \( 3-2=1 \). For 'y', subtract the powers \( 2-1=1 \). So the answer is \( xy \).

🎯 Exam Tip: When an exponent is 1, it is usually not written, so \( x^1 \) is just \( x \).

Question 3. Divide
(iii) \( 45x^{3} y^{2} z^{4} \) by \( (-15 xyz) \)
Answer:
(iii) \( \frac{45 x^{3} y^{2} z^{4}}{-15 x y z} = \frac{45}{-15} x^{3-1} y^{2-1} z^{4-1} = -3x^{2} y z^{3} \) Dividing a positive number by a negative number results in a negative number.
In simple words: Divide \( 45 \) by \( -15 \) to get \( -3 \). Then subtract the powers for each letter: \( x \) becomes \( x^2 \), \( y \) becomes \( y^1 \), and \( z \) becomes \( z^3 \).

🎯 Exam Tip: Pay close attention to the signs when dividing (positive divided by negative is negative).

Question 3. Divide
(iv) \( (3xy)^{2} \) by \( 9xy \)
Answer:
(iv) \( \frac{(3 x y)^{2}}{9 x y} = \frac{9 x^{2} y^{2}}{9 x y} = \frac{9}{9} x^{2-1} y^{2-1} = 1xy = xy \) First, expand the square term, then divide the coefficients and subtract the exponents.
In simple words: First, \( (3xy)^2 \) means \( 3xy \) multiplied by itself, which is \( 9x^2y^2 \). Then divide this by \( 9xy \) to get \( xy \).

🎯 Exam Tip: Remember to apply the exponent to both the coefficient and the variables inside the parenthesis, \( (ab)^n = a^n b^n \).

Question 4. Simplify
(i) \( \frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}} \)
Answer:
(i) \( \frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}} = 3m^{2-1} + 2m^{4-3} = 3m^{1} + 2m^{1} = 3m + 2m = (3 + 2)m = 5m \) We simplify each fraction by subtracting exponents, then combine like terms.
In simple words: Simplify each part by subtracting the powers of 'm'. Then, add the two parts together since they both have 'm' raised to the power of one.

🎯 Exam Tip: Combine like terms only after simplifying each fraction completely. Remember \( m^1 = m \).

Question 4. Simplify
(ii) \( \frac{14 p^{5} q^{3}}{2 p^{2} q} - \frac{12 p^{3} q^{4}}{3 q^{2}} \)
Answer:
(ii) \( \frac{14 p^{5} q^{3}}{2 p^{2} q} - \frac{12 p^{3} q^{4}}{3 q^{2}} = \frac{14}{2} p^{5-2} q^{3-1} - \frac{12}{3} p^{3} q^{4-2} = 7p^{3}q^{2} - 4p^{3}q^{2} = 3p^{3}q^{2} \) We simplify each term by dividing coefficients and subtracting exponents, then subtract the resulting like terms. This problem involves both division and subtraction of algebraic expressions.
In simple words: First, divide the numbers and subtract the powers in the first part. Do the same for the second part. Then, subtract the second result from the first one.

🎯 Exam Tip: Ensure that you correctly identify the operation between the terms (in this case, subtraction) and combine only like terms at the end.

Question 5. Divide:
(i) \( 32y^{2} – 8yz \) by \( 2y \)
Answer:
(i) \( \frac{32y^{2} - 8yz}{2y} = \frac{32y^{2}}{2y} - \frac{8yz}{2y} = 16y^{2-1} - 4z = 16y - 4z \) Each term in the numerator is divided by the denominator separately.
In simple words: Divide each part of the top expression by \( 2y \). This means dividing the numbers and subtracting the powers of \( y \) for each term.

🎯 Exam Tip: When dividing a polynomial by a monomial, divide each term of the polynomial by the monomial individually.

Question 5. Divide:
(ii) \( (4m^{2}n^{3} + 16m^{2} n^{2} – mn) \) by \( 2mn \)
Answer:
(ii) \( \frac{4m^{2}n^{3} + 16m^{2} n^{2} - mn}{2mn} = \frac{4m^{2}n^{3}}{2mn} + \frac{16m^{2} n^{2}}{2mn} - \frac{mn}{2mn} \)
\( \implies 2m^{2-1}n^{3-1} + 8m^{2-1}n^{2-1} - \frac{1}{2}m^{1-1}n^{1-1} \)
\( \implies 2mn^{2} + 8mn - \frac{1}{2}m^{0}n^{0} \)
\( \implies 2mn^{2} + 8mn - \frac{1}{2} \) Here, we distribute the division to each term and simplify using exponent rules. Any variable raised to the power of zero equals 1.
In simple words: Take each part of the top expression and divide it by \( 2mn \). For each part, divide the numbers and subtract the powers of 'm' and 'n'. Remember that anything divided by itself is 1.

🎯 Exam Tip: Be careful with the last term \( \frac{mn}{2mn} \). The variables cancel out completely, leaving only the numerical fraction \( \frac{1}{2} \).

Question 5. Divide:
(iii) \( 5xy^{2} – 18x^{2}y^{3} + 6xy \) by \( 6xy \)
Answer:
(iii) \( \frac{5xy^{2} - 18x^{2}y^{3} + 6xy}{6xy} = \frac{5xy^{2}}{6xy} - \frac{18x^{2}y^{3}}{6xy} + \frac{6xy}{6xy} \)
\( \implies \frac{5}{6}y^{2-1} - \frac{18}{6}x^{2-1}y^{3-1} + 1 \)
\( \implies \frac{5}{6}y - 3xy^{2} + 1 \) We divide each term in the polynomial by the monomial \( 6xy \).
In simple words: Divide each section of the long expression by \( 6xy \). Simplify the numbers and subtract the small numbers (powers) for each letter. The last term becomes 1 because it's divided by itself.

🎯 Exam Tip: Ensure that when a term is divided by itself, the result is 1, not 0. Also, remember to simplify numerical fractions.

Question 5. Divide:
(iv) \( 81(p^{4} q^{2} r^{3} + 2p^{3}q^{3} r^{2} – 5p^{2}q^{2}r^{2}) \) by \( (3pqr)^{2} \)
Answer:
(iv) \( \frac{81(p^{4} q^{2} r^{3} + 2p^{3}q^{3} r^{2} - 5p^{2}q^{2}r^{2})}{(3pqr)^{2}} \)
\( \implies \frac{81 p^{2} q^{2} r^{2} (p^{2}r + 2pq - 5)}{9 p^{2} q^{2} r^{2}} \)
\( \implies \frac{81}{9} (p^{2} q^{2} r^{2})^{1-1} (p^{2}r + 2pq - 5) \)
\( \implies 9 (p^{2}r + 2pq - 5) \)
\( \implies 9 p^{2}r + 18pq - 45 \) First, expand the denominator and factor out common terms from the numerator. Then, simplify the resulting expression. The term \( (p^{2} q^{2} r^{2})^{1-1} \) simplifies to 1.
In simple words: First, square \( 3pqr \) to get \( 9p^2q^2r^2 \). Then, take out \( p^2q^2r^2 \) from each part of the top expression. Now you can divide the numbers and cancel out the common letters. Finally, multiply the result by 9.

🎯 Exam Tip: Factoring out common terms from the numerator before division can significantly simplify the process.

Question 6. Identify the errors and correct them.
(i) \( 7y^{2} – y^{2} + 3y^{2} = 10y^{2} \)
Answer:
(i) \( 7y^{2} - y^{2} + 3y^{2} = (7 - 1 + 3)y^{2} = (6 + 3)y^{2} = 9y^{2} \) The given statement \( 7y^{2} – y^{2} + 3y^{2} = 10y^{2} \) is incorrect. The correct sum is \( 9y^2 \).
In simple words: Add and subtract the numbers in front of \( y^2 \) (the coefficients). \( 7-1+3 \) gives \( 9 \), so the answer should be \( 9y^2 \), not \( 10y^2 \).

🎯 Exam Tip: When combining like terms, only add or subtract the numerical coefficients; the variable and its exponent remain unchanged.

Question 6. Identify the errors and correct them.
(ii) \( 6xy + 3xy = 9x^{2}y^{2} \)
Answer:
(ii) \( 6xy + 3xy = (6 + 3)xy = 9xy \) The given statement \( 6xy + 3xy = 9x^{2}y^{2} \) is incorrect. When adding like terms, only the coefficients are added, not the exponents of the variables. The correct sum is \( 9xy \).
In simple words: When you add terms like \( xy \), you only add the numbers in front. \( 6+3=9 \), so the answer should be \( 9xy \), not \( 9x^2y^2 \). The powers of letters do not change when adding.

🎯 Exam Tip: Never change the exponents of variables when adding or subtracting like terms; they only change during multiplication or division.

Question 6. Identify the errors and correct them.
(iii) \( m(4m – 3) = 4m^{2} – 3 \)
Answer:
(iii) \( m(4m – 3) = m(4m) + m(-3) = 4m^{2} - 3m \) The given statement \( m(4m – 3) = 4m^{2} – 3 \) is incorrect. According to the distributive property, 'm' must be multiplied with both terms inside the parenthesis. The correct expansion is \( 4m^2 - 3m \).
In simple words: When multiplying 'm' by \( (4m-3) \), you must multiply 'm' by \( 4m \) AND by \( -3 \). This gives \( 4m^2 \) and \( -3m \). The error was not multiplying 'm' by the second term.

🎯 Exam Tip: Always apply the distributive property to every term inside the parenthesis when multiplying a monomial by a polynomial.

Question 6. Identify the errors and correct them.
(v) \( (x – 2)(x + 3) = x^{2} – 6 \)
Answer:
(v) \( (x – 2)(x + 3) = x(x + 3) – 2(x + 3) = x(x) + x(3) + (-2)(x) + (-2)(3) = x^{2} + 3x - 2x – 6 = x^{2} + x – 6 \) The given statement \( (x – 2)(x + 3) = x^{2} – 6 \) is incorrect. This multiplication requires using the FOIL method (First, Outer, Inner, Last) or distributive property twice. The correct expansion is \( x^2 + x - 6 \).
In simple words: When you multiply two brackets like \( (x-2)(x+3) \), you need to multiply each part of the first bracket by each part of the second. This gives four terms, which then combine to \( x^2 + x - 6 \). The error was missing the middle terms.

🎯 Exam Tip: When multiplying two binomials, remember to include and combine the "Outer" and "Inner" terms from the FOIL method, not just the "First" and "Last".

Question 6. Identify the errors and correct them.
(vi) \( -3p^{2} + 4p – 7 = -(3p^{2} + 4p – 7) \)
Answer:
(vi) \( -3p^{2} + 4p - 7 = -(3p^{2} - 4p + 7) \) The given statement \( -3p^{2} + 4p – 7 = -(3p^{2} + 4p – 7) \) is incorrect. When factoring out a negative sign, the sign of every term inside the parenthesis must be flipped. The correct factored form is \( -(3p^2 - 4p + 7) \).
In simple words: If you take a minus sign out of the front, you must change the sign of ALL the numbers inside the bracket. So \( +4p \) becomes \( -4p \), and \( -7 \) becomes \( +7 \).

🎯 Exam Tip: Always remember that distributing a negative sign or factoring it out changes the sign of every term it affects.

Question 7. Statement A: If \( 24p^{2}q \) is divided by \( 3pq \), then the quotient is \( 8p \). Statement B: Simplification of \( \frac{(5 x+5)}{5} \) is \( 5x \).
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer: (ii) A is true but B is false
In simple words: Statement A is true because dividing \( 24p^2q \) by \( 3pq \) correctly gives \( 8p \). Statement B is false because simplifying \( (5x+5)/5 \) gives \( x+1 \), not \( 5x \).

🎯 Exam Tip: For statement A, perform the division accurately. For statement B, remember to factor out the common term in the numerator before simplifying the fraction.

Question 8. Statement A: \( 4x^{2} + 3x – 2 = 2(2x^{2} + \frac{3 x}{2} – 1) \) Statement B: \( (2m – 5) – (5 – 2m) = (2m – 5) + (2m – 5) \).
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer: (i) Both A and B are true
In simple words: Statement A is true because if you multiply \( 2 \) by each part inside the bracket on the right side, you get the left side. Statement B is true because subtracting \( (5-2m) \) is the same as adding \( (2m-5) \).

🎯 Exam Tip: For Statement A, use the distributive property on the right side. For Statement B, remember that \( -(a-b) = b-a \) or \( b+(-a) \).