For `x in [1, 5], y= x ^2- 5x + 3` has :

2x + 5y = 1 , 2x + 3y = 3 .

If (2x + y + 2)/(5) = ( 3x - y + 1)/( 3 ) = ( 2 x + 2y + 1 )/(6) then

The system 2x + 3y + z= 5, 3x + y + 5z =7, x + 4y -2z=3 has

If x ^(2) - 2x ^(2) y ^(2) + 5x + y-5 =0 and y (1) then y '' (1)=

If (3x -4y) : (2x -3y) = (5x -6y) : (4x -5y) , find : x : y .

Simplify : (i) (5x - 9y) - (-7x + y) (ii) (x^(2) -x) -(1)/(2)(x - 3 + 3x^(2)) (iii) [7 - 2x + 5y - (x -y)]-(5x + 3y -7) (iv) ((1)/(3)y^(2) - (4)/(7)y + 5) - ((2)/(7)y - (2)/(3)y^(2) + 2) - ((1)/(7)y - 3 + 2y^(2))