Question

The coach of a cricket teams buys 3 bats and 6 balls for 3900.Later,she buys another bat and 3more balls of the same kind for 1300.Represent this situation algebraically and geometrically.
CLASS 10 LINEAR EQUATONS IN 2 VARIABLES​

Answer 1

Answer:

Solution: Let cost of one bat  = Rs x

Cost of one ball      = Rs y

3 bats and 6 balls for Rs 3900 So that

3 x + 6y = 3900                     … (1)

Divide complete equation by 3 we get

X+ 2y = 1300                      

Subtract 2y both side we get

X         = 1300 – 2y  

Plug y = - 1300, 0 and 1300  we get

X = 1300 – 2 ( - 1300 )         = 1300 + 2600           =  3900

X= 1300 -2(0)                        = 1300 -0                   = 1300

X= 1300 – 2(1300)               = 1300 – 2600           = - 1300  

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. Given that  she buys another bat and 2 more balls of the same kind for Rs 1300

So we get

x  + 2 y = 1300                      … (2)

Subtract 2y both side we get

X         = 1300 – 2y

Plug y = - 1300, 0 and 1300  we get

X         = 1300 – 2 ( - 1300 )             = 1300  +  2600         =  3900

X         = 1300 – 2 ( 0 )                      = 1300 - 0                  = 1300

X         = 1300 – 2(1300)                  = 1300 – 2600           = -1300

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Algebraic representation

3 x + 6 y = 3900                    … (1)

x + 2 y    = 1300                     … (2)

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Answer 2

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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