Ex 3.3, 1 - Solve by substitution method (i) x + y = 14 - Ex 3.3

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3

 

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6

 

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 9 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 10

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 11 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 12 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 13

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 14 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 15 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 16

Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 17 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 18 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 19 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 20

 

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise

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 value of x 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 So, there are infinitely many solutions 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 Example And so on Ex 3.3, 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 0.4x + 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 Removing fractions from both equations 𝟑𝒙/𝟐−𝟓𝒚/𝟑=−2 Multiplying both equations by 6 6 × 3𝑥/2−"6 ×" 5𝑦/3="6 ×"−2 9x – 10y = − 12 𝒙/𝟑+𝒚/𝟐=𝟏𝟑/𝟔 Multiplying both equations by 6 6 × 𝑥/3+"6 ×" 𝑦/2="6 × " 13/6 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) Substituting value of x in (2) 2x + 3y = 13 2((−12 + 10𝑦)/9)+3𝑦=13 Multiplying both sides by 9 9 × 2((−12 + 10𝑦)/9) + 9 × 3y = 9 × 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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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.