# help with rearrange equation

• June 23rd 2011, 07:42 AM
yuenj01
help with rearrange equation
just want to ask which is correct way of doing ax + by + x = 0
example is y = 2x + 8

y -y = 2x -y + 8 (basically take away y both sides)
0 = 2x -y + 8

or

y - 2x -8 = 2x + 8 - 2x - 8 (instead of taking y away, take 2x + 8 away both sides)
-2x + y - 8 = 0

which one is correct?
• June 23rd 2011, 07:46 AM
Quacky
Re: help with rearrange equation
Quote:

Originally Posted by yuenj01
just want to ask which is correct way of doing ax + by + x = 0
example is y = 2x + 8

y -y = 2x -y + 8 (basically take away y both sides)
0 = 2x -y + 8

or

y - 2x -8 = 2x + 8 - 2x - 8 (instead of taking y away, take 2x + 8 away both sides)
-2x + y - 8 = 0

which one is correct?

The first option was surely far more straightforward? Although either method is correct.
• June 24th 2011, 04:35 AM
yuenj01
Re: help with rearrange equation
Quote:

Originally Posted by Quacky
The first option was surely far more straightforward? Although either method is correct.

that is what i was thinking. in da exams, is it acceptable?
• June 24th 2011, 04:37 AM
Glitch
Re: help with rearrange equation
If you multiply either solution by -1 on both sides, you'll find that they're exactly the same.
• June 24th 2011, 04:38 AM
e^(i*pi)
Re: help with rearrange equation
Quote:

Originally Posted by yuenj01
just want to ask which is correct way of doing ax + by + x = 0

Do you mean $ax+by+c=0$?

Quote:

Originally Posted by yuenj01
example is y = 2x + 8

y -y = 2x -y + 8 (basically take away y both sides)
0 = 2x -y + 8

or

y - 2x -8 = 2x + 8 - 2x - 8 (instead of taking y away, take 2x + 8 away both sides)
-2x + y - 8 = 0

which one is correct?