1. ## Re:Help with Algebra??

It did not work. I will try again.

3=16x^2

2. ## Re:Help with Algebra??

x = 0.5, try again.

3. ## Re:Help with Algebra??

It&#039;s not working. I either get 3=16x or 3=16x^2

4. ## Re:Help with Algebra??

First the bonus points:

1 2
--- + --- = 2
2x 4x

notice that this is the same as the above

1 1
--- + --- = 2
2x 2x

then:

2
--- = 2
2x

4x = 2

x = 0.5

that was for the bonus points:

Here goes the universial method:

4x 4x
--- + --- = 2
8x² 8x²

8x
---- = 2
8x²

16x² = 8x

16x = 8

x = 0.5
---

Easy!
I&#039;m begining to think you are jerking my chain.

5. ## Re:Help with Algebra??

I don&#039;t know how to do that. That was not easy it is very hard. I did not understand anything on your last post. I have to tell you something that i never told you before. I&#039;m a slow learner and i have a learning disability. How did i make it through algebra 1, 2 years ago it beats me.

6. ## Re:Help with Algebra??

Get tutoring, that&#039;s all I can say - it&#039;s MUCH easier to understand if you get tutoring in person.

7. ## Re:Help with Algebra??

Yo hold up. I got something better. Do you use yahoo messenger?? There is something called doodle it&#039;s and IMVironment where you could draw out your problems and answers. I use yahoo. I know i&#039;m going to need a tutor when it comes down to Analytic Geometry &amp; Calculus 1, 2 and 3. Also Linear algebra. Please reply back. Thanks.

23

9. ## Re:Help with Algebra??

&quot;crossed over&quot;

1 * 2x 4*(x-1)
-------- + ----------- = 10
(x -1)*2x 2x*(x-1)

Now we do the math - nothing changes but appearences

2x 4x - 1
--------- + ----------- = 10
2x² - 2x 2x² - 2x
[/quote]
Not to piss anyone off or anything, but this action should give a result of:
2x 4x - 4
--------- + ----------- = 10
2x² - 2x 2x² - 2x

So I guess the result in that one, isn&#039;t what you ended up with.

-edit-
And in the end you&#039;ll end up with solving the roots of a second degree equation, like 0=ax²+bx+c
Which either can have only one solution, two solutions or no solution at all.
The way to find these is by using the formular:
-b +- sqrt(b²-4ac)
x = ------------------------
2a
If (b²-4ac) gives a negative value, it has no solution, atleast not within the number group R, but if we use Z, then it has.

10. ## Re:Help with Algebra??

WTF... did I miscalculated also... oh I&#039;m so embarrased now.

oh it&#039;s a minor typo I see, I should have catched that as well.

I think the Z domain is a tad out of scope for the project at hand, but true it&#039;s an awesome tool.

But you are of course correct about that root solving thing, I&#039;m guessing I&#039;m chalk this up to picking a bad example, it was maybe a tad to advanced (even for me it seems - I glad stumbled into the common trap, forgetting prior knowledge)

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