Make y the subject of the formula x = SQRT((y+1)/(y-2))

Square both sidesMultiply by the RHS denominator (x^2(y-2)) = y+1Expand your brackets y(x^2) - 2x^2 = y + 1Begin re-arranging towards making y the subject so moving y terms to the left would get you y(x^2 - 1) = 2x^2 +1isolate y, making it the subject, by dividing the co-efficient on both sides (div by (x^2 - 1)) y(x^2 - 1) = 2x^2 +1 (div by (x^2 - 1))y = (2x^2 +1)/(x^2 - 1)

YE

Related Further Mathematics GCSE answers

All answers ▸

The equation of the line L1 is y = 3x – 2 The equation of the line L2 is 3y – 9x + 5 = 0 Show that these two lines are parallel.


Find the coordinates of the minimum/maximum of the curve: Y = 8X - 2X^2 - 9, and determine whether it is a maximum or a minimum.


How do I know I can multiply two matrices and if so, how do I do it?


Solve the following simultanious equations: zy=28 and 2z-3y=13