Results 1 to 4 of 4

- Nov 5th 2012, 10:10 AM #1

- Joined
- Nov 2012
- From
- sweden
- Posts
- 2

## Not sure I'm doing this the right way

I need to solve this in terms of x

3y(3y-x) = 2z(x+2z)

9y^{2}- 3yx = 2zx + 4z^{2}

9y^{2}- 4z^{2}= 3yx + 2zx

(9y - 4z)(9y + 4z) = x(2z + 3y)

x = (9y - 4z)(9y + 4z) / (2z + 3y)

don't know if I can cancel to give this

x = (3y - 2z)(3y + 2z)

x = 9y^{2}+ 3y2z - 3y2z - 4b^{2}

x = 9y^{2}- 4b^{2}

Really don't know if I've done this properly. Can anyone help?

Thanks.

- Nov 5th 2012, 12:44 PM #2

- Nov 5th 2012, 03:00 PM #3

- Joined
- Nov 2012
- From
- sweden
- Posts
- 2

- Nov 5th 2012, 03:17 PM #4