# Ex 3.3, 1

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 3.3 ,1 Solve the following pair of linear equations by the substitution method. (i) x + y = 14 x – y = 4 x + y = 14 x – y = 4 From equation (1) x + y = 14 x = 14 – y Substituting x = 14 – y in equation (2) x – y = 4 (14 – y) – y = 4 14 – y – y = 4 14 – 2y = 4 –2y = 4 – 14 –2y = –10 y = (−10)/(−2) y = 5 Putting y = 5 in (2) x – y = 4 x = y + 4 x = 5 + 4 x = 9 Hence, x = 9, y = 5 is the solution of the above equation Ex 3.3 ,1 Solve the following pair of linear equations by the substitution method. (ii) s – t = 3 𝑠/3+𝑡/2=6 s – t = 3 𝑠/3+𝑡/2=6 From (1) s – t = 3 s = 3 + t Substituting s = 3 + t in (2) (3 + 𝑡)/3+𝑡/2=6 ((3 + 𝑡) × 2 + 𝑡 × 3)/(3 × 2)=6 (6 + 2𝑡 + 3𝑡)/6=6 (6 + 5𝑡)/6=6 6 + 5t = 6 × 6 6 + 5t = 36 5t = 36 – 6 5t = 30 t = 30/5 t = 6 Putting t = 6 in (1) s – t = 3 s = 3 + 6 s = 9 Hence, s = 9, t = 6 Ex 3.3 ,1 Solve the following pair of linear equations by the substitution method. (iii) 3x – y = 3 9x – 3y = 9 3x – y = 3 9x – 3y = 9 Solving (1) 3x – y = 3 3x = y + 3 x = (𝑦 + 3)/3 Putting value of x in (2) 9x – 3y = 9 9((𝑦 + 3)/3)−3𝑦=9 3(y + 3) – 3y = 9 3y + 9 – 3y = 9 3y – 3y + 9 = 9 0 + 9 = 9 9 = 9 The statement is true for all values of x Hence, there is no single value of x as the answer for this question Reason :- The 2 equations given in question are 3x – y = 3 9x – 3y = 9 From (2), taking 3 common , we get 3 (3x – y) = 9 3x – y = 9/3 3x – y = 3 Which is equation same as equation (1) Hence both equation actually same So there can be infinite values of x and y So there can be infinite values of x and y For e.g. If y = 1, 3x – 1 = 3 3x = 3 + 1 x = (1 + 3)/3 x = 4/3 And so on Ex 3.3 ,1 (Method 1) Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 0.2x + 0.3y = 1.3 0.4 x + 0.5y = 2.3 From (1) 0.2x + 0.3y = 1.3 0.2x = 1.3 – 0.3y x = (1.3 − 0.3𝑦)/0.2 Substituting the value of x in (2) 0.4 x + 0.5y = 23 0.4 ((1.3−0.3𝑦)/0.2)+0.5𝑦=2.3 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 – 0.1y = 2.3 – 2.6 – 0.1y = – 0.3 y = (−0.3)/(−0.1) y = 3 Putting y = 3 equation in (1) 0.2x + 0.3y = 1.3 0.2x + 0.3 (3) = 1.3 0.2x + 0.9 = 1.3 0.2x = 1.3 – 0.9 0.2x = 0.4 x = 0.4/0.2 x = 2 Hence x = 2 and y = 3 is the solution for the equation Ex 3.3 ,1 (Method 2) Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 From (1) 0.2x + 0.3y = 1.3 Multiplying both side by 10 (0.2x + 0.3y ) ×10=1.3×10 2x + 3y = 13 2x = 13 – 3y x = (13 − 3𝑦)/2 Putting value of x in (2) 0.4x + 0.5y = 2.3 0.4((13−3𝑦)/2) + 0.5y = 2.3 Multiplying both sides by 10 10 × 0.4((13−3𝑦)/2) + 10 × 0.5y = 10 × 2.3 4 ((13−3𝑦)/2)+5𝑦=23 2 ( 13 – 3y) + 5y = 23 26 – 6y + 5y = 23 26 – y = 23 26 – 23 = y 3 = y y = 3 Putting y = 3 equation in (1) 0.2x + 0.3y = 1.3 0.2x + 0.3 (3) = 1.3 0.2x + 0.9 = 1.3 0.2x = 1.3 – 0.9 0.2x = 0.4 x = 0.4/0.2 x = 2 Hence x = 2 and y = 3 is the solution for the equation Ex 3.3 ,1 Solve the following pair of linear equations by the substitution method. (v) √2 𝑥+√3 𝑦=0 √3 𝑥−√8 𝑦=0 √2 𝑥+√3 𝑦=0 √3 𝑥−√8 𝑦=0 From (1) √2 𝑥+√3 𝑦=0 √2 𝑥=−√3 𝑦 x = (−√3 𝑦)/√2 Substituting x in (2) √3 𝑥−√8 𝑦=0 √3 ((−√3 𝑦)/√2)−√8 𝑦 = 0 (√3×(−√3 𝑦))/√2−√8 𝑦=0 – 3𝑦/√2−√8 𝑦=0 – 3y – √8 𝑦×√2= √2×0 – 3y – √(8×2) 𝑦=0 – 3y – √16 𝑦=0 – 3y – √(4×4) 𝑦=0 – 3y – √42 𝑦=0 – 3y – 4y = 0 – 7y = 0 y = 0/(−7) y = 0 Putting the value of y in (1) √2 𝑥+√3 𝑦=0 √2 𝑥+√3×0=0 √2 𝑥+0=0 √2 𝑥=0 x = 0/√2 x = 0 Hence, x = 0, y = 0 Ex 3.3 ,1 Solve the following pair of linear equations by the substitution method. (vi) 3𝑥/2−5𝑦/3=−2 𝑥/3+𝑦/2=13/6 3𝑥/2−5𝑦/3=−2 Since the equation(1) is given in the fraction , we remove them 3𝑥/2−5𝑦/3=−2 (3 × 3𝑥 − 2 × 5𝑦)/6=−2 (9𝑥 − 10𝑦)/6=−2 9x – 10y = − 12 Similarly we remove fractions from 𝑥/3+𝑦/2=13/6 (2𝑥 + 3𝑦)/6=13/6 2x + 3y = (13 × 6)/6 2x + 3y = 13 Hence, our equations are 9x – 10y = −12 …(1) 2x + 3y = 13 …(2) From (1) 9x – 10y = –12 9x = –12 + 10y x = ((−12 + 10𝑦)/9) Substitute x in (2) 2x + 3y = 13 2((−12 + 10𝑦)/9)+3𝑦=13 (2(−12 + 10𝑦) + 9 × 3𝑦)/9=13 2(–12 + 10y ) + 27y = 13×9 – 24 + 20y + 27y = 117 – 24 + 20y + 27y – 117 = 0 47y – 141 = 0 47y = 141 y = 141/47 y = 3 Putting y = 3 in (2) 2x + 3y = 13 2x + 3(3) = 13 2x + 9 = 13 2x = 13 – 9 2x = 4 x = 4/2 x = 2 Hence, x = 2, y = 3 is the solution of equation .

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .