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lol typoWhat happened to the 9 originally on the right side?

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- #6

- Jan 26, 2012

- 890

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| $

\(2x+y = \pm(y+3)\)

CB

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- #8

what is wrong in my solution ?Last line should be:

\(2x+y = \pm(y+3)\)

CB

- Jan 30, 2012

- 74

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 ?

$$|2x+y| = y+3$$

as CB has pointed out.

- Oct 11, 2012

- 1

Or you can use POLYSMLT if you have a TI83