
Re:Help with Algebra??
[quote author=dsantamassino link=board=14;threadid=7355;start=0#msg67863 date=1057957160]
I understand that just a little. Well anyway i also know 8

6
can be reduce to
1 and 1

3
[/quote]
While true it's fairly irrelevant in this context  but it's none the less correct.

Re:Help with Algebra??
Why don't you give me an algebra problem so you could tell me how much you could help me with?? Let's start off from the beginning then to advance. Please reply back.

Re:Help with Algebra??
The simple one:
2x + 3 = 2
The slightly more advanced one
1 1
 +  = 3
2x 5x

Re:Help with Algebra??
Tell me if i'm correct.
The simple one:
x=  1

2
The advance one:
x= 3

7
Is that right?? Please reply back. If you want to know how i got the answer i will show you my steps. Thanks.

Re:Help with Algebra??
Okay here goes
1)
2x = 1
1
x= 
2
So that one is correct.
2)
5x 2x
 +  = 3
10x² 10x²
7x
 = 2
10x²
20x² = 7x
20x = 7
7
x = 
20
So that one isn't correct.
Let's try again
1 4
 +  = 10
(x1) 2x
Now remember these are off the top of my head so as much as I would like it they don't always end out with a nice even result, but hey... it's it jerryrigged math.
Give me all your steps for that one btw.

Re:Help with Algebra??
I did it another way and i got a different answer.
1 4
 + 
(x1) 2x
ok here we go. First we want to get rid of those fractions. We multiply each side by x1. That leaves us with 1+40x4. Once again we have to get rid of that other fraction. We multiply each side by 2x so 2x could cancel. That leaves us with 2x+40x4.
42x4. Is that right?? Please reply back. Thanks.

Re:Help with Algebra??
okay:
1 4
 +  = 10
(x1) 2x
first we multiply both fractions by 1  we do this by 2x/2x for first one and (x1)/(x1) for the other, since multipling any number by 1 does not alter it's value we are allowed to do this.
We pick these specific numbers from the divisor ( the lower part of the fraction) as you can see we do this "crossed over"
1 * 2x 4*(x1)
 +  = 10
(x 1)*2x 2x*(x1)
Now we do the math  nothing changes but appearences
2x 4x  1
 +  = 10
2x²  2x 2x²  2x
Now we see that we can make these to fractions into a common fraction since they have the same divisor. This is also perfectly allowable since it does not alter the value of equation.
2x + 4x  1
 = 10
2x²  2x
 this is the end of the part you have problems with understanding if I'm correct 
Now we multiply by 2x² 2x on both sides of the equation
2x + 4x 1 = 20x²  2x
substract 2x from each side
4x 1 = 20x²
divide by x
4  1/x = 20x
We can now realise that this has no finite solution within the given scope  thus we can only express x as a function of x.
thus
1 1
x =  + 
5 20x
So this was a tricky one.

Let's try the same on a different equation
1 2
 +  = 2
2x 4x
Now this one has a "shortcut" builtin  so bonus point for that one if you spot it, but try mindless to follow the method I showed you anyways, because the method is never wrong.
One way to check if you are correct in your calculations is to insert the value for x you found in the equation you were given  if it matches up you have found a usable solution.

Re:Help with Algebra??
First can you show me how to use my keyboard to raise a power of an expoent?? If not then i'm going to use ^ and then the number i want as a power. Please reply back. Thanks.

Re:Help with Algebra??
^2 will give you ² if typed in one go, no spaces etc.

Re:Help with Algebra??
The answer is 3=16x^2. Is that correct?? Please reply back. Thanks.
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