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Last updated at Dec. 8, 2016 by Teachoo

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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 /3+ /2=6 s t = 3 /3+ /2=6 From (1) s t = 3 s = 3 + t Substituting s = 3 + t in (2) (3 + )/3+ /2=6 ((3 + ) 2 + 3)/(3 2)=6 (6 + 2 + 3 )/6=6 (6 + 5 )/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 = ( + 3)/3 Putting value of x in (2) 9x 3y = 9 9(( + 3)/3) 3 =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 0.3 )/0.2 Substituting the value of x in (2) 0.4 x + 0.5y = 23 0.4 ((1.3 0.3 )/0.2)+0.5 =2.3 0.4 ((1.3 0.3 )/0.2)+0.5 =2.3 2(1.3 0.3 )+0.5 =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 3 )/2 Putting value of x in (2) 0.4x + 0.5y = 2.3 0.4((13 3 )/2) + 0.5y = 2.3 Multiplying both sides by 10 10 0.4((13 3 )/2) + 10 0.5y = 10 2.3 4 ((13 3 )/2)+5 =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 + 3 =0 3 8 =0 2 + 3 =0 3 8 =0 From (1) 2 + 3 =0 2 = 3 x = ( 3 )/ 2 Substituting x in (2) 3 8 =0 3 (( 3 )/ 2) 8 = 0 ( 3 ( 3 ))/ 2 8 =0 3 / 2 8 =0 3y 8 2= 2 0 3y (8 2) =0 3y 16 =0 3y (4 4) =0 3y 42 =0 3y 4y = 0 7y = 0 y = 0/( 7) y = 0 Putting the value of y in (1) 2 + 3 =0 2 + 3 0=0 2 +0=0 2 =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) 3 /2 5 /3= 2 /3+ /2=13/6 3 /2 5 /3= 2 Since the equation(1) is given in the fraction , we remove them 3 /2 5 /3= 2 (3 3 2 5 )/6= 2 (9 10 )/6= 2 9x 10y = 12 Similarly we remove fractions from /3+ /2=13/6 (2 + 3 )/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 + 10 )/9) Substitute x in (2) 2x + 3y = 13 2(( 12 + 10 )/9)+3 =13 (2( 12 + 10 ) + 9 3 )/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 .

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.