- Thread starter
- #1
lol typoWhat happened to the 9 originally on the right side?
Last line should be:Completing the square
$ (2x)^2 + 2(2x)y -6y - 9 = 0 $
$ (2x+y)^2 -y^2 -6y - 9 = 0 $
$(2x+y)^2 - (y+3)^2 = 0 $
$(2x+y)^2 = (y+3)^2 $
$ 2x+y = |y+3| $
what is wrong in my solution ?Last line should be:
\(2x+y = \pm(y+3)\)
CB
Solve for x:
$$4x^2 +4xy - 6y - 9 = 9 $$
Since x is the variable and y is considered to be a constant your solution should readwhat is wrong in my solution ?