Form the pair of linear equations for the following problem and find their solution by substitution method.

The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Let cost of each bat be ₹ x and each ball be ₹ y. Given, cost of 7 bats and 6 balls ₹3800.

⇒ 7x + 6y = 3800 …………(1)

Given,

Cost of 3 bats and 5 balls for ₹1750.

⇒ 3x + 5y = 1750 …………(2)

Solving equation (1), we get :

⇒ 7x + 6y = 3800

⇒ 6y = 3800 - 7x

⇒ y = $\dfrac{3800 - 7x}{6}$ …………(3)

Substituting value of y from equation (3) in equation (2), we get :

$\Rightarrow 3x + 5 \times \dfrac{3800 - 7x}{6} = 1750 \\[1em] \Rightarrow 3x + \dfrac{19000 - 35x}{6} = 1750 \\[1em] \Rightarrow \dfrac{18x + 19000 - 35x}{6} = 1750 \\[1em] \Rightarrow \dfrac{19000 - 17x}{6} = 1750 \\[1em] \Rightarrow 19000 - 17x = 10500 \\[1em] \Rightarrow 17x = 19000 - 10500 \\[1em] \Rightarrow 17x = 8500 \\[1em] \Rightarrow x = \dfrac{8500}{17} = ₹500.$

Substituting value of x in equation (3), we get :

$\Rightarrow y = \dfrac{3800 - 7x}{6} \\[1em] \Rightarrow y = \dfrac{3800 - 7 \times 500}{6} \\[1em] \Rightarrow y = \dfrac{3800 - 3500}{6} \\[1em] \Rightarrow y = \dfrac{300}{6} \\[1em] \Rightarrow y = ₹50.$

Hence, pair of linear equations are 7x + 6y = 3800, 3x + 5y = 1750, where x and y are the costs (in ₹) of one bat and one ball respectively; x = 500, y = 50.