Ex 3.5, 3 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)

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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 = 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