9 taken away from the sum of `x\ a n d\ y` is (a) `x+y-9` (b) `9-(x+y)` (c) `(x+y)/9` (d) `9/(x+y)`

9 taken away from the sum of x and y is x+y-9( b) 9-(x+y)(x+y)/(9) (d) (9)/(x+y)

3x-y=3 , find the value of x and y 9x-3y=9

If the mean of 6,\ 7,\ x ,\ 8,\ y ,\ 14 is 9, then (a) x+y=21 (b) x+y=19 (c) x-y=19 (d) x-y=21

If the mean of 6,\ 7,\ x ,\ 8,\ y ,\ 14 is 9, then (a) x+y=21 (b) x+y=19 (c) x-y=19 (d) x-y=21

Find the value of x and y when 3x+y=3 9x-3y=9

Add the following: (i) 3x\ a n d\ 7x (ii) -5x y\ a n d\ 9x y

If x+y=k is normal to y^2=12 x , then k is (a)3 (b) 9 (c) -9 (d) -3

If x+y=k is normal to y^2=12 x , then k is (a)3 (b) 9 (c) -9 (d) -3

The value of x and y from the equation 3x-y=3 and 9x-3y=9 are

The equations of tangents to the circle x^2+y^2-6x-6y+9=0 drawn from the origin in (a) x=0 (b) x=y (c) y=0 (d) x+y=0