Ex 3.3

Chapter 3 Class 10 Pair of Linear Equations in Two Variables
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Ex 3.3, 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 3.3, 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 