1. Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) x/2+ 2y/3 = -1 and x-y/3 = 3

Pawan Prajapati
2 years ago
Solutions: (i) x + y = 5 and 2x – 3y = 4 By the method of elimination. x + y = 5 ……………………………….. (i) 2x – 3y = 4 ……………………………..(ii) When the equation (i) is multiplied by 2, we get 2x + 2y = 10 ……………………………(iii) When the equation (ii) is subtracted from (iii) we get, 5y = 6 y = 6/5 ………………………………………(iv) Substituting the value of y in eq. (i) we get, x=5−6/5 = 19/5 ∴x = 19/5 , y = 6/5 By the method of substitution. From the equation (i), we get: x = 5 – y………………………………….. (v) When the value is put in equation (ii) we get, 2(5 – y) – 3y = 4 -5y = -6 y = 6/5 When the values are substituted in equation (v), we get: x =5− 6/5 = 19/5 ∴x = 19/5 ,y = 6/5 (ii) 3x + 4y = 10 and 2x – 2y = 2 By the method of elimination. 3x + 4y = 10……………………….(i) 2x – 2y = 2 ………………………. (ii) When the equation (i) and (ii) is multiplied by 2, we get: 4x – 4y = 4 ………………………..(iii) When the Equation (i) and (iii) are added, we get: 7x = 14 x = 2 ……………………………….(iv) Substituting equation (iv) in (i) we get, 6 + 4y = 10 4y = 4 y = 1 Hence, x = 2 and y = 1 By the method of Substitution From equation (ii) we get, x = 1 + y……………………………… (v) Substituting equation (v) in equation (i) we get, 3(1 + y) + 4y = 10 7y = 7 y = 1 When y = 1 is substituted in equation (v) we get, A = 1 + 1 = 2 Therefore, A = 2 and B = 1 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 By the method of elimination: 3x – 5y – 4 = 0 ………………………………… (i) 9x = 2y + 7 9x – 2y – 7 = 0 …………………………………(ii) When the equation (i) and (iii) is multiplied we get, 9x – 15y – 12 = 0 ………………………………(iii) When the equation (iii) is subtracted from equation (ii) we get, 13y = -5 y = -5/13 ………………………………………….(iv) When equation (iv) is substituted in equation (i) we get, 3x +25/13 −4=0 3x = 27/13 x =9/13 ∴x = 9/13 and y = -5/13 By the method of Substitution: From the equation (i) we get, x = (5y+4)/3 …………………………………………… (v) Putting the value (v) in equation (ii) we get, 9(5y+4)/3 −2y −7=0 13y = -5 y = -5/13 Substituting this value in equation (v) we get, x = (5(-5/13)+4)/3 x = 9/13 ∴x = 9/13, y = -5/13 (iv) x/2 + 2y/3 = -1 and x-y/3 = 3 By the method of Elimination. 3x + 4y = -6 …………………………. (i) x-y/3 = 3 3x – y = 9 ……………………………. (ii) When the equation (ii) is subtracted from equation (i) we get, -5y = -15 y = 3 ………………………………….(iii) When the equation (iii) is substituted in (i) we get, 3x – 12 = -6 3x = 6 x = 2 Hence, x = 2 , y = -3 By the method of Substitution: From the equation (ii) we get, x = (y+9)/3…………………………………(v) Putting the value obtained from equation (v) in equation (i) we get, 3(y+9)/3 +4y =−6 5y = -15 y = -3 When y = -3 is substituted in equation (v) we get, x = (-3+9)/3 = 2 Therefore, x = 2 and y = -3