The coach of a cricket team buys 3 bats and 6 balls for 3900. Later, she buys another bat and 3
more balls of the same kind for 1300. Represent this situation algebraically and geometrically.
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➨let rate of bat = xRs.
➨ and rate of ball = yRs.
➨3x+6y=3900 (i)
➨and x+3y=1300 (ii)
➨or x=(1300−3y)
now equation (i) can be written as
➨ 3(1300−3y)+6y=3900
➨ 3900−9y+6y=3900
➨y=0
Ans:- x=1300
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