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Ex 3.5, 3 - Solve by substitution and cross multiplication - Cross Multiplication Method

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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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 = 27 – 32 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 4x + 2y – 4 = 0 Now, 𝑥/(−2 ) = 1/(1 ) x = -2 × 1 x = - 2 Hence, x = – 2 , y = 4 is the solution

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