Class 10 Maths Chapter 3 Exercise 3.3 Question 1 (iii, iv) | Elimination Method | NCERT Solution

In this video, we have solved Exercise 3.3 Question 1 (Parts iii and iv) of Chapter 3 (Pair of Linear Equations in Two Variables) for CBSE and State Board Class 10 Maths in Hindi. Topic: Elimination Method (विलोपन विधि) Question Covered: Solve the following pair of linear equations by the elimination method and the substitution method: (निम्न समीकरणों के युग्म को विलोपन विधि तथा प्रतिस्थापन विधि से हल कीजिए:) (iii) 3x - 5y - 4 = 0 and 9x = 2y + 7 (iv) (x/2) + (2y/3) = -1 and x - (y/3) = 3 Topics and Steps Explained: Part (iii): • Given equations: 3x - 5y = 4 ---(1) 9x - 2y = 7 ---(2) • Multiplying equation (1) by 3: 9x - 15y = 12 ---(3) • Subtracting equation (2) from equation (3): (9x - 15y) - (9x - 2y) = 12 - 7 -13y = 5 y = -5/13 • Putting value of y in equation (1): 3x - 5(-5/13) = 4 3x + 25/13 = 4 3x = 4 - 25/13 3x = (52 - 25) / 13 3x = 27/13 x = 9/13 • Final Answer: x = 9/13, y = -5/13 Part (iv): • Given equations: (x/2) + (2y/3) = -1 ---(1) x - (y/3) = 3 ---(2) • Simplifying equation (1) by taking LCM (6): (3x + 4y) / 6 = -1 3x + 4y = -6 ---(3) • Simplifying equation (2) by taking LCM (3): (3x - y) / 3 = 3 3x - y = 9 ---(4) • Subtracting equation (4) from equation (3): (3x + 4y) - (3x - y) = -6 - 9 5y = -15 y = -15 / 5 y = -3 • Putting value of y in equation (4): 3x - (-3) = 9 3x + 3 = 9 3x = 9 - 3 3x = 6 x = 2 • Final Answer: x = 2, y = -3 • Easy step-by-step handwriting method for CBSE and All State Boards ----------------------------------------------- Subscribe to The Maths Paper for complete Class 10 Maths NCERT solutions with easy handwriting step-by-step methods! Video pasand aaye toh Like aur apne dosto ke sath Share zaroor karein. ----------------------------------------------- #Class10Maths #LinearEquationsInTwoVariables #EliminationMethod #VilopanVidhi #Class10MathsChapter3 #Exercise3_3 #NCERTClass10Maths #TheMathsPaper #CBSEClass10 #Class10BoardExam