Ex 3.4, 1 (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)

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Ex 3.4, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) 𝑥/2+2𝑦/3=−1 𝑎𝑛𝑑 𝑥−𝑦/3=3 Given x/2+2y/3=−1 (3(x) + 2(2y))/(2 × 3)=−1 (3𝑥 + 4y)/6=−1 3x + 4y = −1 × 6 3x + 4y = −6 x – y/3=3 (3𝑥 − 𝑦 )/3=3 3x – y = 3(3) 3x – y = 9 We use elimination method with equation (1) & (2) 5y = – 15 y=(−15)/( 5) y = −3 Putting y = −3 in equation (2) 3x – y = 9 3x – (−3) = 9 3x + 3 = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = −3 is the solution of the given equations Ex 3.4, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) 𝑥/2+2𝑦/3=−1 𝑎𝑛𝑑 𝑥−𝑦/3=3 Given x/2+2y/3=−1 (3(x) + 2(2y))/(2 × 3)=−1 (3𝑥 + 4y)/6=−1 3x + 4y = −1 × 6 3x + 4y = −6 x – y/3=3 (3𝑥 − 𝑦 )/3=3 3x – y = 3(3) 3x – y = 9 From (1) 3x + 4y = –6 3x = –6 – 4y Putting value of 3x in (2) 3x – y = 9 (–6 – 4y) – y = 9 – 4y – y = 9 + 6 –5y = 15 y = (−15)/( 5) y = –3 Putting y = −3 in equation (2) 3x – y = 9 3x – (−3) = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = −3 is the solution of the given equations