` 6x + 5y = 7x + 3y + 1 = 2 ( x + 6y - 1 )`

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

The orthocentre of the triangle formed by the lines x - 7y + 6 = 0, 2x - 5y - 6 = 0 and 7x + y - 8 = 0 is

Solve for x and y : (1)/( 7x ) + ( 1) /( 6y) = 3, (1)/( 2x) - (1)/( 3y ) = 5 (x ne 0, y ne 0)

Add 2x - 3y + z , 5y - x + 7z and 3x - y - 6z.

2x + 3y-5z = 7, x + y + z = 6,3x-4y + 2z = 1, then x =

Subtract 5x^2 - 4y^2 + 6y - 3 from 7x^2 - 4xy + 8y^2 + 5x - 3y .

Subtract 5x^2 – 4y^2 + 6y – 3 from 7x^2 – 4xy + 8y^2 + 5x – 3y .