How to simplify this my answer to 2x+2y-13=0

My answer is :-

$\displaystyle \frac{4x-2y-15}{2\sqrt{5}} $= $\displaystyle \frac{2x-y-1}{\sqrt{5}} $

How to simplify my answer to get 2x+2y-13=0

Thank you

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- Jun 1st 2009, 07:33 AMmj.alawamiHow to simplify this answer?
How to simplify this my answer to 2x+2y-13=0

My answer is :-

$\displaystyle \frac{4x-2y-15}{2\sqrt{5}} $= $\displaystyle \frac{2x-y-1}{\sqrt{5}} $

How to simplify my answer to get 2x+2y-13=0

Thank you - Jun 1st 2009, 07:45 AMzenjenn
http://www.mathhelpforum.com/math-he...13f56835-1.gif= http://www.mathhelpforum.com/math-he...97b7d9f0-1.gif

Hmm... It looks like you tried to factor a 2 out of the numerator and then cancel 2 from both the numerator and denominator but..

Factor 2 out of -15 and get... -1?

I'm not sure what your original problem was, but you may want to re-check your arithmetic over all. - Jun 1st 2009, 07:46 AMyeongil
Can you post the original problem?

01 - Jun 1st 2009, 08:13 AMVonNemo19

$\displaystyle \frac{4x-2y-15}{2\sqrt{5}} $= $\displaystyle \frac{2x-y-1}{\sqrt{5}} $

multiplying both sides by $\displaystyle 2\sqrt{5}$ Yields

$\displaystyle 4x-2y-15=2(2x-y-1)\Longleftrightarrow{4x-2y-15}=4x-2y-2\Longleftrightarrow-15=-2$??(Wondering)

The logic has gone bad.............(Crying)

It appears that something has gone way way wrong here. The only way to simplify your answer to the given equation would be to develop a new branch of Mathematics(Giggle)

Sorry buddy. but if you post the original problem, I'm here for you. - Jun 1st 2009, 08:14 AMSoroban
Hello, mj.alawami!

You made an error somewhere.

Your answer is quite impossible.

Quote:

My answer is: .$\displaystyle \frac{4x-2y-15}{2\sqrt{5}} \:=\: \frac{2x-y-1}{\sqrt{5}} $ . . . . wrong!

How to simplify my answer to get: .$\displaystyle 2x+2y-13\:=\:0$

You have: .$\displaystyle \frac{4x-2y - 15}{2\sqrt{5}} \;=\;\frac{2x-y-1}{\sqrt{5}} $

Multiply by $\displaystyle 2\sqrt{5}\!:\quad 4x - 2y - 15 \;=\;2(2x-y-1) \quad\Rightarrow\quad 4x - 2y - 15 \:=\:4x - 2y - 2$

. . and we have: .$\displaystyle -15 \:=\:-2 \quad\Rightarrow\quad 0 \:=\:13 \;\;??$

I believe you need to have: .$\displaystyle \frac{4x-2y - 15}{2\sqrt{5}} \;=\;\frac{{\color{red}x - 2y} - 1}{\sqrt{5}} $