CBSE Class 8 Maths Notes Chapter 2: Chapter 2 of CBSE Class 8 Maths, Linear Equations in One Variable, focuses on solving equations of the form ax+b=0 , where x is the variable. The chapter teaches students how to isolate the variable using operations like addition, subtraction, multiplication, and division.
It also covers equations with variables on both sides, fractional and decimal coefficients, and negative values. Students are introduced to word problems, translating real-life situations into linear equations. The chapter emphasizes the transposition method, preparing students for more complex algebraic equations and real-world applications.CBSE Class 8 Maths Notes Chapter 2 PDF
Question:
The sum of two numbers is 74. If one of the numbers is 10 more than the other number, find the two numbers.Solution:
Let one of the numbers be x. Then the other number is x + 10. Given that the sum of the two numbers is 74. So, x+(x+10)=74 ⇒2x+10=74 ⇒2x=74−10=64 x = 64/2 x = 32 Therefore, one number is 32. As given, the one of the numbers is 10 more than the other, we get x + 10 = 32 + 10 = 42. Therefore, the two numbers are 32 and 42.Question:
(x+1)/(2x + 3) = ⅜Solution:
Given equation: (x+1)/(2x + 3) = ⅜ Now, cross multiply the equation, we get 8(x + 1) = 3(2x + 3) 8x + 8 = 6x + 9 Now, bring the variables on one side and constants on the other side, 8x – 6x = 9 – 8 On simplifying the above equation, 2x = 1 x = ½. Now, this can be written as: x – (½) = 0Question:
Simplify the equation: [(6x + 1)/3] + 1 = (x – 3)/6. Also, justify your answer.Solution:
Given Equation: [(6x + 1)/3] + 1 = (x – 3)/6 To simplify the given equation, take the LCM of 3 and 6, which is 6. Now, multiply both sides of the equation by 6, we get [6(6x + 1)/3] + 1(6) = 6(x – 3)/6 2(6x+1) + 6 = x – 3 12x + 2 + 6 = x – 3 Now, bring all the terms with variables on one sides and constants on the other side. So, we get 12x + 8 = x – 3 12x – x = -3 – 8 11x = -11 x = -11/11 x = -1 Hence, the simplification of [(6x + 1)/3] + 1 = (x – 3)/6 is x = -1, which is the required solution.Justification:
Now, substitute x = -1 in the given equation. LHS: [(6x + 1)/3] + 1 LHS = [(6(-1) + 1)/3] + 1 LHS = [(-6 + 1)/3] + 1 LHS = (-5/3) + 1 LHS = (-5 + 3) /3 LHS = -⅔ Now, take the expression on the right side of the given equation. RHS = (x – 3)/6 RHS = (-1 – 3)/6 RHS = -4/6 RHS = -⅔ Therefore, LHS = RHS. Hence, verified.