Question

3/√x+4/√y=2, 5/√x+7/√y=41/12, x>0,y>0 ,Solve the given pairs of linear equation.

Answer 1

Solution :

3/√x + 4/√y = 2 --( 1 )

5/√x + 7/√y = 41/12 ---( 2 )

Let 1/√x = a , 1/√y = b ,

3a + 4b = 2 ---( 3 )

5a + 7b = 41/12---( 4 )

[ 5 × ( 3 ) - 3 × ( 4 ) ]

15a + 20b = 10 --- ( 5 )

15a + 21b = 41/4 --- ( 6 )
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••••••• - b = 10 - 41/4

=> - b = ( 40 - 41)/4

=> b = ( -1 )/( -4 )

=> b = 1/4

Put b = 1/4 in equation ( 3 ), we get

3a + 4 × ( 1/4 ) = 2

=> 3a = 2 - 1

=> a = 1/3

Therefore ,

1/√x = a = 1/3

=> √x = 3

=> x = 9

1/√y = b = 1/4

=> √y = 4

=> y = 16

( x , y ) = ( 9 , 16 )

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