Exercise 6.7 1. Find the following products: (i) (x + 4) (x + 7) (ii) (x – 11) (x + 4) (iii) (x + 7) (x – 5) (iv) (x – 3) (x – 2) (v) (y2 – 4) (y2 – 3) (vii) (3x + 5) (3x + 11) (viii) (2×2 – 3) (2×2 + 5) (ix) (z2 + 2) […]
Tag: RD Sharma Class 8 6th Chapter Solution
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and Identities Exercise 6.6 Solution
Exercise 6.6 1. Write the following squares of binomials as trinomials: We know that, (a + b) 2 =a2 + 2ab + b2 and (a–b)2 = a2–2ab+b2 (i) (x + 2) 2 Solution: (x+2) 2 is in the form of (a + b)2 = a2+2ab+b2 here, a = x , b = 2 ⇒ x2 + 2 × […]
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and Identities Exercise 6.5 Solution
Exercise 6.5 Multiply: 1. (5x + 3) by (7x + 2) Solution: To multiply, we will use distributive law as follows: (5x+3)(7x+2) =5x(7x+2)+3(7x+2) =(5x×7x+5x×2)+(3×7x+3×2) =(35×2+10x)+(21x+6) =35×2+10x+21x+6 =35×2+31x+6 Thus, the answer is 35×2+31x+6 2. (2x + 8) by (x – 3) Solution: To multiply, we will use distributive law as follows: (2x+8) (x–3) = 2x(x–3) […]
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and Identities Exercise 6.4 Solution
Exercise 6.4 Find the following products: 1. 2a3 (3a+5b) Solution: To find the product, we will use distributive law as follows: 2a3 (3a + 5b) = 2a3 × 3a + 2a3 × 5b = (2 x 3)(a3 × a) + (2 × 5)a3b = (2×3)a3+1+ (2×5)a3b = 6a4 +10a3b Thus, the answer is 6a4 +10a3b. 2. -11a (3a + 2b) Solution: To find the product, we will use distributive law as […]
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and Identities Exercise 6.3 Solution
Exercise 6.3 Find each of the following products: 1. 5×2 × 4×3 Solution: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws is subject to their applicability in the given expressions. In the present problem, to perform the multiplication, we can proceed […]
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and identities Exercise 6.2 Solution
Exercise 6.2 1. Add the following algebraic expressions: (i) 3a2b, −4a2b, 9a2b Solution: To add the like terms, we proceed as follows: 3a2b + (-4a2b) + 9a2b = 3a2b – 4a2b + 9a2b (Distributive Law) = 8a2b (iii) 4xy2–7x2y,12x2y–6xy2, −3x2y+5xy2 Solution: To add, we proceed as follows: (4xy2–7x2y)+(12x2y)+(−6xy2)+(−3x2y+5xy2) = 4xy2–7x2y+12x2y–6xy2–3x2y+5xy2 = 4xy2–6xy2+5xy2–7x2y+12x2y–3x2y (Collecting like terms) = 3xy2+2x2y (Combining […]
RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and identities Exercise 6.1 Solution
Exercise 6.1 1.Identify the terms, their coefficients for each of the following expressions: (i) 7x2yz − 5xy (ii) x2 + x + 1 (iii) 3x2y2 − 5x2y2z2 + z2 (iv) 9 – ab + bc – ca (v) a2 + b2 − ab (vi) 0.2x – 0.3xy + 0.5y Solution: Definitions: A term in an algebraic expression can be […]