Ex 3.5

Ex 3.5, 1 (i)
Deleted for CBSE Board 2022 Exams

Ex 3.5, 1 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 3.5, 1 (iii) Deleted for CBSE Board 2022 Exams

Ex 3.5, 1 (iv) Important Deleted for CBSE Board 2022 Exams

Ex 3.5, 2 (i) Important

Ex 3.5, 2 (ii)

Ex 3.5, 3 Deleted for CBSE Board 2022 Exams You are here

Ex 3.5, 4 (i)

Ex 3.5, 4 (ii)

Ex 3.5, 4 (iii)

Ex 3.5, 4 (iv) Important

Ex 3.5, 4 (v) Important

Last updated at July 24, 2021 by Teachoo

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