How to simplify this answer?

• Jun 1st 2009, 08:33 AM
mj.alawami
How to simplify this answer?
How to simplify this my answer to 2x+2y-13=0

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

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

Thank you
• Jun 1st 2009, 08:45 AM
zenjenn
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, 08:46 AM
yeongil
Can you post the original problem?

01
• Jun 1st 2009, 09:13 AM
VonNemo19
Quote:

Originally Posted by mj.alawami
How to simplify this my answer to 2x+2y-13=0

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

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

Thank you

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

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

$4x-2y-15=2(2x-y-1)\Longleftrightarrow{4x-2y-15}=4x-2y-2\Longleftrightarrow-15=-2$??(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, 09:14 AM
Soroban
Hello, mj.alawami!

You made an error somewhere.

Quote:

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

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

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

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

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

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