How to simplify this answer?

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June 1st 2009, 07:33 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
June 1st 2009, 07:45 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.
June 1st 2009, 07:46 AM
yeongil

Can you post the original problem?

01
June 1st 2009, 08:13 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.
June 1st 2009, 08:14 AM
Soroban

Hello, mj.alawami!

You made an error somewhere.
Your answer is quite impossible.

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