hard fraction!

**andyboy179** · February 18th 2011, 09:11 AM

hi,
i need to work out this fraction:

3/2x-1 - 4/3x-1

i can do the first part and worked it out and you get a quadratic equation as shown in my picture:
$hard fraction!-quadratic.jpg$

when i type in to my calculator 36-4x6x6 it =-108?

-108 doesn't work!!! why is it coming up with this? have i started my quadratic equation wrong?

help would be much appreciated!!!!

thanks!

**Unknown008** · February 18th 2011, 09:33 AM

If you need to work out, then you only need to simplify. There are no equal signs and hence, you cannot solve for x.

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

$= \dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}$

$= \dfrac{x+1}{6x^2 - 5x + 1}$

That's about all you can do...

**masters** · February 18th 2011, 09:33 AM

Originally Posted by andyboy179

hi,
i need to work out this fraction:

3/2x-1 - 4/3x-1

i can do the first part and worked it out and you get a quadratic equation as shown in my picture:

Click image for larger version.

Name: quadratic.jpg
Views: 20
Size: 34.6 KB
ID: 20861

when i type in to my calculator 36-4x6x6 it =-108?

-108 doesn't work!!! why is it coming up with this? have i started my quadratic equation wrong?

help would be much appreciated!!!!

thanks!

Hi andyboy179,

What do you mean "work out this fraction". What is there to work out? And there is no equation here.

And, please use grouping symbols (parenthesis) to show us what's in the denominators of your fractions.

Restate the problem exactly the way it's stated in your book or worksheet or whatever.

**andyboy179** · February 18th 2011, 09:59 AM

Originally Posted by Unknown008

If you need to work out, then you only need to simplify. There are no equal signs and hence, you cannot solve for x.

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

$= \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}$

$= \dfrac{1-5x}{6x^2 - 5x + 1}$

That's about all you can do...

on this part:

$= \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}$

how did you get 3x-3 wouldn't it be 9x-3

**andyboy179** · February 18th 2011, 10:02 AM

really really sorry, miss typed my original question!!

it is:

$\dfrac{3}{2x - 1} - \dfrac{4}{3x-1} =1$

**harish21** · February 18th 2011, 10:28 AM

Originally Posted by andyboy179

on this part:

$= \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}$

how did you get 3x-3 wouldn't it be 9x-3

I think Unknown008 made a small mistake. The numerator should be $9x - 3 - 8x + 4$

so, you have:

$\dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}=1$

can you finish?

**andyboy179** · February 18th 2011, 10:33 AM

Originally Posted by harish21

I think Unknown008 made a small mistake. The numerator should be $9x - 3 - 8x + 4$

so, you have:

$\dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}=1$

can you finish?

would it go into a quadratic equation and end as 0.6 and 0??

**harish21** · February 18th 2011, 10:43 AM

It does go into a quadratic equation, but only one of the answers you have written is correct! How did you get 0.83? Was that a guess?If not, Please show your work

**andyboy179** · February 18th 2011, 10:46 AM

sorry i forgot to take an x away from 5x! my answers are 0.6 and 0

**Unknown008** · February 18th 2011, 10:46 AM

Sorry for that, I'm really tired and got a lot of things to do >.< Thanks harish

**harish21** · February 18th 2011, 10:52 AM

Originally Posted by andyboy179

sorry i forgot to take an x away from 5x! my answers are 0.6 and 0

NO! Still incorrect. My question is HOW did you get there?! Please show the process!!

**andyboy179** · February 18th 2011, 10:57 AM

$hard fraction!-.jpg$

then quadratic equation with answers as 0.6 and 0

**harish21** · February 18th 2011, 11:01 AM

okay. you subtract x and 1 from both the left hand side and the right hand side, that is,

$6x^2-5x+1=x+1$

$6x^2-5x+1-x-1=x+1-x-1$

$6x^2-6x+0=0$

$6x^2-6x=0$

$x=....$

**andyboy179** · February 18th 2011, 01:24 PM

Originally Posted by harish21

okay. you subtract x and 1 from both the left hand side and the right hand side, that is,

$6x^2-5x+1=x+1$

$6x^2-5x+1-x-1=x+1-x-1$

$6x^2-6x+0=0$

$6x^2-6x=0$

$x=....$

i don't understand that because you have:
X+1=6x^2-5+1
minus x and 1 from each side leaves you with
0=6x^2-4+0

how did you get $6x^2-5x+1=x+1$ ?

**harish21** · February 18th 2011, 01:46 PM

Originally Posted by andyboy179

i don't understand that because you have:
X+1=6x^2-5+1
minus x and 1 from each side leaves you with
0=6x^2-4+0

No it does not. You are subtracting x and 1 from both sides of the equation.

$(x+1)=(6x^2-5x+1)$

subtract 1 from both the sides:

$(x+1)-1=(6x^2-5x+1)-1$

that leaves you with:

$x = 6x^2-5x$

now you subtract x from both the sides. this gives you:

$x-x=(6x^2-5x)-x$

which gives you:

$0=6x^2-5x-x$

therefore $0=6x^2-6x$

************************************************** *************
Note: $\boxed{-5x-x=-6x} \;\;and \;\;\boxed{-5x+x=-4x}$

In the first box, you are subtracting x from -5x. In the second box, you are adding x to -5x. This is where your confusion lies.

If you still dont understand it, please refer to your books/notes/instructor to avoid trouble in the future.

how did you get $6x^2-5x+1=x+1$ ?

Please refer to post Number 2 by Unknown008

your question was $\dfrac{3}{2x - 1} - \dfrac{4}{3x-1}=1$

then Unknown008's method (which you've been told multiple number of times) takes you :

$\dfrac{x+1}{6x^2 - 5x + 1}=1$

multiplying both sides of this equation by $6x^2 - 5x + 1$ .gives you:

$x+1=6x^2 - 5x + 1$ .

Have fun!

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