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Ex 3.5, 3 - Solve by substitution and cross multiplication

Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3

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Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5

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Transcript

Ex 3.5, 3 (Substitution method) Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9 3x + 2y = 4 8x + 5y = 9 3x + 2y = 4 From (1) 8x + 5y = 9 8x = 9 – 5y x = ((9 − 5𝑦))/8 Putting value of x in (2) 3x + 2y = 4 3 (((9 − 5𝑦))/8) + 2y = 4 (3(9 − 5𝑦))/8 + 2y = 4 (3(9 − 5𝑦) + 8(2𝑦) )/8 = 4 3(9 – 5y) + 8(2y) = 4 × 8 27 – 15y + 16y = 32 27 + y = 32 y = 32 – 27 y = 5 Putting y = 5 in (2) 3x + 2y = 4 3x + 2(5) = 4 3x + 10 = 4 3x = 4 – 10 3x = –6 x = (−6)/3 x = –2 Therefore, x = – 2 & y = 5 is the solution of the given pair of linear equations Ex 3.5, 3 (Cross – multiplication method) Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9 3x + 2y = 4 For cross-multiplication 8x + 5y – 9 = 0 3x + 2y – 4 = 0 𝑥/(5 ×(−4) − 2 ×(−9) ) = 𝑦/((−9) × 3 − (−4) × 8 ) = 1/(8 × 2 − 3 × 5 ) 𝑥/((−20) + 18 ) = 𝑦/((−27) + 32 ) = 1/(16 − 15) 𝑥/(−2 ) = 𝑦/(5 ) = 1/1 Now, Hence, x = – 2, y = 5 is the solution

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.