Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10    1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Ex 3.3

Transcript

Ex 3.3 ,3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball. Let cost of one bat = Rs. x & Cost of one ball = Rs. y Given that Coach buys 7 bats and 6 balls for Rs. 3800 So, 7 × (Cost of one bat ) + 6 × (Cost of one ball) = 3800 7x + 6y = 3800 Also, Coach buys 3 bats and 5 balls for Rs 1750 So, 3 × (Cost of one bat ) + 5 × (Cost of one ball) = 1750 3x + 5y = 1750 Thus, our equations are 7x + 6y = 3800 …(1) 3x + 5y = 1750 …(2) From (1) 7x + 6y = 3800 7x = 3800 – 6y x = ((3800−6𝑦)/7) Putting x in (2) 3x + 5y = 1750 3((3800−6𝑦)/7)+5𝑦=1750 Multiplying by 7 both sides 7×(3(3800−6𝑦)/7)+5𝑦×7=1750×7 3( 3800 – 6y ) + 35y = 1750×7 3×3800−6𝑦×3+35𝑦=1750×7 11400 – 18y + 35y = 12250 –18y + 35y = 12250 – 11400 17y = 850 y = 850/17 y = 50 Putting y = 50 in (1) 7x + 6y = 3800 7x + 6(50) = 3800 7x + 300 = 3800 7x = 3800 – 300 7x = 3500 x = 3500/7 x = 500 Thus, x = 500, y = 50 Hence , Cost of one bat = x = Rs. 500 Cost of one ball = y = Rs. 50

Ex 3.3 