# hard fraction!

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• Feb 18th 2011, 02:00 PM
andyboy179
Quote:

Originally Posted by harish21
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.

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!

okay i think i get it now, so would 0=6x^2 -6x be the answer?
• Feb 18th 2011, 02:03 PM
harish21
Does your question tell you to find the value of x?
• Feb 18th 2011, 02:11 PM
andyboy179
usually with questions like these i usually just find X via quadratic equation or by dividing a number by the amount of X's.

how would i work out x with this question?
• Feb 18th 2011, 02:17 PM
harish21
You have:

$0=6x^2-6x$

Find x from here..
• Feb 18th 2011, 02:22 PM
andyboy179
would i square root 6x^2=6
0=6-6

plus 1 to each side to get:
6=6
6/6=1
x=1?
• Feb 18th 2011, 04:34 PM
Quote:

Originally Posted by andyboy179
would i square root 6x^2=6
0=6-6

plus 1 to each side to get:
6=6
6/6=1
x=1?

No, to find all answers, write as a multiplication of factors...

$6x^2-6x=0$

$6xx-6x=0$

$(x)6x+(-1)6x=0$

6x is common to both, so factor...

$6x(x-1)=0$

6 is not 0, so $x(x-1)=0$

Those factors now tell you the answers because any multiple of zero is zero.
• Feb 18th 2011, 09:18 PM
Unknown008
Yes, x = 1 but you forgot 0

If you so much want to use the formula, you can make this:

$0 = ax^2 + bx + c$

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, you have $0 = 6x^2 - 6x$

So, a = 6, b = -6, c = 0
• Feb 19th 2011, 01:21 AM
andyboy179
So by doing the quadratic equation I would end with the answers of 1 and 0 for x??
• Feb 19th 2011, 01:27 AM
Unknown008
Did you get something else than 1 and 0? (Wondering)
• Feb 19th 2011, 01:51 AM
andyboy179
No I did the quadratic and get 1 and 0.
To get it I did:
X=6+ (square root)36-4x6x0/12=1
X=6- (square root)36-4x6x0/12=0
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