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the coach buy 3 bats and 6 balls at price. is 3900
then buy a bat and 2 balls at price is 1300
algebraically
1 ×3 = 3 bat and 2×3 =6 balls at price is 1300×3 = 3900
we need to form linear equation for the situation.
let cost of one ball = ₹x
let cost of one bat = ₹y
In first case,three bats and 6 balls cost him 3900₹ therefore our equation becomes
= 3x +6y =3900_________(1)
dividing equation by 3 both side
=x +2y =1300
x =1300 - 2y
for plotting the equation of graph, take different value of x from equation or you can do vice versaAt y =0= = x= 1300 -2(0) = x =1300 now finding the value at x = 0
0 = 1300-2y
2y = 1300
y = 650
in second case he buys one bat and 3 balls for 1300 therefore,
= x +2y = 1300_________(2)
x=1300 -2y
At y =0
x=1300 -2(0)
x= 1300
x =0
0=1300-2y
2y=1300
y=650