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1. Solve the following pair of linear equations by the substitution method (i) x + y = 14 x – y = 4 (ii) s – t = 3 (s/3) + (t/2) = 6 (iii) 3x – y = 3 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 (v) √2 x+√3 y = 0 √3 x-√8 y = 0 (vi) (3x/2) – (5y/3) = -2 (x/3) + (y/2) = (13/6)

1. Solve the following pair of linear equations by the substitution method

(i) x + y = 14

x – y = 4

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0

√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2

(x/3) + (y/2) = (13/6)

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8741 Points
3 months ago
Solutions: (i) Given, x + y = 14 and x – y = 4 are the two equations. From 1st equation, we get, x = 14 – y Now, substitute the value of x in second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the exact value of x; ∵ x = 14 – y ∴ x = 14 – 5 Or x = 9 Hence, x = 9 and y = 5. (ii) Given, s – t = 3 and (s/3) + (t/2) = 6 are the two equations. From 1st equation, we get, s = 3 + t ________________(1) Now, substitute the value of s in second equation to get, (3+t)/3 + (t/2) = 6 ⇒ (2(3+t) + 3t )/6 = 6 ⇒ (6+2t+3t)/6 = 6 ⇒ (6+5t) = 36 ⇒5t = 30 ⇒t = 6 Now, substitute the value of t in equation (1) s = 3 + 6 = 9 Therefore, s = 9 and t = 6. (iii) Given, 3x – y = 3 and 9x – 3y = 9 are the two equations. From 1st equation, we get, x = (3+y)/3 Now, substitute the value of x in the given second equation to get, 9(3+y)/3 – 3y = 9 ⇒9 +3y -3y = 9 ⇒ 9 = 9 Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values. (iv) Given, 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations. From 1st equation, we get, x = (1.3- 0.3y)/0.2 _________________(1) Now, substitute the value of x in the given second equation to get, 0.4(1.3-0.3y)/0.2 + 0.5y = 2.3 ⇒2(1.3 – 0.3y) + 0.5y = 2.3 ⇒ 2.6 – 0.6y + 0.5y = 2.3 ⇒ 2.6 – 0.1 y = 2.3 ⇒ 0.1 y = 0.3 ⇒ y = 3 Now, substitute the value of y in equation (1), we get, x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2 Therefore, x = 2 and y = 3. (v) Given, √2 x + √3 y = 0 and √3 x – √8 y = 0 are the two equations. From 1st equation, we get, x = – (√3/√2)y __________________(1) Putting the value of x in the given second equation to get, √3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0 ⇒ y = 0 Now, substitute the value of y in equation (1), we get, x = 0 Therefore, x = 0 and y = 0. (vi) Given, (3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations. From 1st equation, we get, (3/2)x = -2 + (5y/3) ⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1) Putting the value of x in the given second equation to get, ((-12+10y)/9)/3 + y/2 = 13/6 ⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6 Now, substitute the value of y in equation (1), we get, (3x/2) – 5(3)/3 = -2 ⇒ (3x/2) – 5 = -2 ⇒ x = 2 Therefore, x = 2 and y = 3.

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