Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Last updated at Dec. 20, 2020 by Teachoo

Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Transcript

Ex 3.4, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4 x + y = 5 2x – 3y = 4 Multiplying equation (1) by 2 2(x + y) = 2 × 5 2x + 2y = 10 Solving (3) and (2) by Elimination –5y = –6 5y = 6 y = 𝟔/𝟓 Putting y = 6/5 in (1) x + y = 5 x + 6/5 = 5 x = 5 – 6/5 x = (5 × 5 − 6)/5 x = (25 − 6)/5 x = 𝟏𝟗/𝟓 Hence, x = 19/5,𝑦=6/5 Ex 3.4, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4 x + y = 5 2x – 3y = 4 From (1) x + y = 5 x = 5 – y Substituting x in (2) 2x – 3y = 4 2 (5 – y) – 3y = 4 10 – 2y – 3y = 4 10 – 5y = 4 –5y = 4 – 10 –5y = −6 y = (−6)/(−5) y = 𝟔/𝟓 Putting y = 6/5 in equation (1) x + y = 5 x + 6/5 = 5 x = 5 – 6/5 x = (5 × 5 − 6)/5 x = (25 − 6)/5 x = 𝟏𝟗/𝟓 Hence, x = 19/5,y=6/5 is the solution of the equations Ex 3.4 , 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x – 2y = 2 3x + 4y = 10 2x – 2y = 2 We multiply equation (2) by 2 2(2x – 2y) = 2 × 2 4x – 4y = 4 Using elimination with equations (3) & (1) 7x = 14 x = 14/7 x = 2 Putting x = 2 in (2) 2x – 2y = 2 2(2) – 2y = 2 4 – 2y = 2 –2y = 2 – 4 –2y = –2 y = (−2)/(−2) y = 1 Thus, x = 2, y = 1 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 : (ii) 3x + 4y = 10 and 2x – 2y = 2 3x + 4y = 10 2x – 2y = 2 From (1) 3x + 4y = 10 3x = 10 – 4y x = ((𝟏𝟎 − 𝟒𝒚)/𝟑) Putting value of x in (2) 2x – 2y = 2 2((10 − 4𝑦)/3)−2𝑦=2 (2(10 − 4𝑦))/3−2𝑦=2 (2(10 − 4𝑦) − 2𝑦 × 3)/3 =2 2(10 – 4y) – 6y = 2 × 3 20 – 8y – 6y = 6 –8y – 6y = 6 – 20 –14y = –14 y = (−14)/(−14) y = 1 Putting y = 1 in (2) 2x – 2y = 2 2x – 2(1) = 2 2x – 2 = 2 2x = 2 + 2 2x = 4 x = 4/2 x = 2 Therefore, x = 2, y = 1 are the solution of the given equations. Ex 3.4, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 Given 3x – 5y – 4 = 0 3x – 5y = 4 Also given that 9x = 2y + 7 9x – 2y = 7 Now, we multiply first equation by 3 3(3x – 5y) = 3 × 4 9x – 15y = 12 We use elimination method with equations (3) & (2) 13y = –5 y = (−𝟓)/𝟏𝟑 Putting y = (−5)/13 in equation (2) 9x – 2y = 7 9x −" 2"×((−5)/13)=7 9x + 10/13=7 9x = 7 −10/13 9x = (7 × 13 − 10)/13 9x = (91 − 10)/13 9x = 81/13 x = 81/13×1/9 x = 𝟗/𝟏𝟑 Therefore, x = 9/13 & y = −5/13 are the solutions of the given equations Ex 3.4, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 3x – 5y – 4 = 0 9x = 2y + 7 From (1) 3x – 5y – 4 = 0 3x = 4 + 5y x = (𝟒 + 𝟓𝒚)/𝟑 Putting x in (2) 9x = 2y + 7 9 ((4 + 5𝑦)/3)=2𝑦+7 3 (4 + 5y) = 2y + 7 12 + 15y = 2y + 7 15y – 29 = 7 – 12 13y = −5 y = (−𝟓)/𝟏𝟑 Putting y = (−5)/13 in (1) 3x – 5y – 4 = 0 3x – 5((−5)/13) – 4 = 0 3x + 25/13 – 4 = 0 3x = 4 – 25/13 3x = (4(13) − 25)/13 3x = (52 − 25)/13 3x = 27/13 x = 27/(13 × 3) x = 𝟗/𝟏𝟑 Hence, x = 9/13 & y = (−5)/13 are the solution of the given equations 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.