1. ## Fractions again!

$\displaystyle \frac{-4x}{x-3}=\frac{x-1}{2}$

When I input this into wolfram and click on step by step solution, the next step looks like this:

$\displaystyle -8x(x-3)(x-1)$

so I assume the step they did was to multiple the numerator of the first fraction by the denominator of the second giving $\displaystyle -4x\times 2$ and then joining the $\displaystyle x-3$ and $\displaystyle x-1$

If this is true then does this apply to every fraction of this type?

So my next move would be to multiple $\displaystyle -8x$ by each term in the brackets? I don't get what to do next. Fractions really suck

2. ## Re: Fractions again!

Originally Posted by uperkurk
$\displaystyle \frac{-4x}{x-3}=\frac{x-1}{2}$

When I input this into wolfram and click on step by step solution, the next step looks like this:

$\displaystyle -8x(x-3)(x-1)$
That makes no sense at all! What happened to the "= "?

so I assume the step they did was to multiple the numerator of the first fraction by the denominator of the second giving $\displaystyle -4x\times 2$ and then joining the $\displaystyle x-3$ and $\displaystyle x-1$

If this is true then does this apply to every fraction of this type?

So my next move would be to multiple $\displaystyle -8x$ by each term in the brackets? I don't get what to do next. Fractions really suck
You can always "get rid" of fractions by multiplying both sides of the equation by all the denominators.
What do you get when you multiply
$\displaystyle \frac{-4x}{x- 3}(x- 3)(2)= \frac{x- 1}{2}(2)(x- 3)$?

(It's not the fractions that suck.)

3. ## Re: Fractions again!

$\displaystyle -8x = x^2-4x+3$
$\displaystyle -x^2=8x-4x+3$
$\displaystyle -x^2=4x+3$
$\displaystyle (-x+3)(x+1)$

$\displaystyle x=[-3,-1]$

Looks correct...

But what if I have something like:

$\displaystyle \frac{-4x}{x-3}(x-3)(x-2)=\frac{x-1}{x-2}(x-2)(x-3)$

$\displaystyle (-4x)(x-2)=(x-1)(x-3)$

and then start factoring them out?

any ideas?

5. ## Re: Fractions again!

Originally Posted by uperkurk
$\displaystyle \frac{-4x}{x-3}=\frac{x-1}{2}$

$\displaystyle \frac{-4x}{x-3}=\frac{x-1}{2}\\-8x=x^2-4x+3\\x^2+4x+3=0$

Finish?

6. ## Re: Fractions again!

Originally Posted by Plato
$\displaystyle \frac{-4x}{x-3}=\frac{x-1}{2}\\-8x=x^2-4x+3\\x^2+4x+3=0$

Finish?
huh? I thought I did complete it? Besides when you rearrange

$\displaystyle -8x=x^2-4x+3$ does it not become $\displaystyle -x^2=8x-4x+3=0$ because I move the postitive x^2 over the other side of the equals sign making it negative?

Sorry I'm confused now I'm not sure what you're asking me for.

7. ## Re: Fractions again!

Still trying to get an answer for this

8. ## Re: Fractions again!

Trying to get an answer to what? That was your fifth post in this thread and you still haven't said what you want to do with the equation you posted.

You give an equation $\displaystyle \frac{-4x}{x-3}= \frac{x- 1}{2}$. You then multiplied both sides by 2(x- 3) to get $\displaystyle -4x(2)= (x- 1)(x- 3)$. Plato then multiplied it out for you getting $\displaystyle -8x= x^2- 4x+ 3$, rewriting that as $\displaystyle x^2+ 4x+ 3= 0$ and suggested you finish it. You responded "I thought I did complete it".

So, what are you trying to do that you had thought you had completed? I thought at first that you wanted to solve for x. If so, then you want to solve the quadratic equation $\displaystyle x^2+ 4x+ 3= 0$. And that means being able to write "x= " specific numbers. Do you know how to solve quadratic equations? The simplest way is to factor to two linear terms. Then use the fact that if ab= 0, either a= 0 or b= 0.

9. ## Re: Fractions again!

Originally Posted by uperkurk

$\displaystyle -8x = x^2-4x+3$
$\displaystyle -x^2=8x-4x+3$
$\displaystyle -x^2=4x+3$
$\displaystyle (-x+3)(x+1)$

$\displaystyle x=[-3,-1]$

Looks correct...

But what if I have something like:

$\displaystyle \frac{-4x}{x-3}(x-3)(x-2)=\frac{x-1}{x-2}(x-2)(x-3)$

$\displaystyle (-4x)(x-2)=(x-1)(x-3)$

and then start factoring them out?
I think this post isn't being shown? This is now what I am asking.

10. ## Re: Fractions again!

Originally Posted by uperkurk
I think this post isn't being shown? This is now what I am asking.

No, that post does indeed show.
The problem is that the sort of help you need is so very much beyond anything we can offer.
Because you are so very challenged by basic arithmetic and pre-algebra there is little we can do.
This is not a tutorial service. You need a live tutor.

11. ## Re: Fractions again!

Originally Posted by uperkurk

$\displaystyle -8x = x^2-4x+3$
$\displaystyle -x^2=8x-4x+3$
$\displaystyle -x^2=4x+3$
$\displaystyle (-x+3)(x+1)$

$\displaystyle x=[-3,-1]$
In standard notation this would be wrong- it means "x between -3 and -1". You want $\displaystyle x\in {-3, -1}$.

Looks correct...

But what if I have something like:

$\displaystyle \frac{-4x}{x-3}(x-3)(x-2)=\frac{x-1}{x-2}(x-2)(x-3)$

$\displaystyle (-4x)(x-2)=(x-1)(x-3)$

and then start factoring them out?
No, they are already factored! You want to multiply them. That gives you
$\displaystyle -4x^2+ 8x= x^2- 4x+ 3$
$\displaystyle 5x^2- 12x+ 3= 0$.

If your intention is to solve the original equation (you still haven't said) then you could "complete the square" or use the "quadratic formula". That equation will not factor easily since its roots are not rational numbers.

12. ## Re: Fractions again!

Thank you HallsofIvy, that was all I ever wanted from about the 3rd post.

@Plato, no I am not challenged by basic arithmetic and I do not need a live tutor. If you read post number 3 it was very clear indeed what I was asking for help with. I'm not sure if someimes you like to patronise people who do not have the same understanding as you but it's very rude.

Should me havng to ask how you could not understand what I was asking for in post number 3 lead me to question your basic literacy skills?