factorisation of a quadratic equation that is a fraction :/

Hey guys im stuck on an equation and just cant seem to get the right answer. I know there must be something wrong with my working. Could someone please show me how you multiply out the denominator to solve this equation.

** 1+ 6/x-1 = 1/x+1**

Thank you in advance for any help.

Re: factorisation of a quadratic equation that is a fraction :/

Hi , multiply equation by $\displaystyle (x-1)(x+1)$

Re: factorisation of a quadratic equation that is a fraction :/

Quote:

Originally Posted by

**princeps** Hi , multiply equation by $\displaystyle (x-1)(x+1)$

Thank you for your quick reply. I tried that and still cant seem to get the correct answer. I know the answer because I have checked and I cant even work backwards to get it. Ill show my working out and maybe someone will know where im going wrong..

1+ 6/x-1 = 1/x+1

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

1+6x+6-x-1 = 0

5x+6 = 0

)

(

)

But the proper answer is= X^{2}+5x+6=0

x=-3 & -2

I cant seem to find where x^{2} came from.

Re: factorisation of a quadratic equation that is a fraction :/

You have to multiply $\displaystyle 1$ by $\displaystyle (x-1)(x+1)$ also .

Re: factorisation of a quadratic equation that is a fraction :/

In such cases go by the following algorithm.

Step-1: Multiply the equation be the LCM of the denominator which in this case is (x-1)(x=1)

.(x-1)(x+1)[ 1 + 6/(x-1) = 1/(x+1) OR x^2 -1 + 6(x+1) = (x-1)

Step 2: Simplify the equation

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

x^2 -1 + 6x+6 -x+1= 0

x^2 + 5x+6 = 0

(x+3)(x+2) = 0

etc.

Re: factorisation of a quadratic equation that is a fraction :/

Ahhh I think I get it now! Thanks heaps for all the help.

Re: factorisation of a quadratic equation that is a fraction :/

Quote:

Originally Posted by

**jewelmaule** Hey guys im stuck on an equation and just cant seem to get the right answer. I know there must be something wrong with my working. Could someone please show me how you multiply out the denominator to solve this equation.

** 1+ 6/x-1 = 1/x+1**

Thank you in advance for any help.

I assume that your equation is $\displaystyle \displaystyle 1 + \frac{6}{x-1} = \frac{1}{x + 1}$. Some brackets where they're needed would be nice... Anyway...

$\displaystyle \displaystyle \begin{align*} 1 + \frac{6}{x - 1} &= \frac{1}{x + 1} \\ \frac{x-1}{x-1} + \frac{6}{x-1} &= \frac{1}{x+1} \\ \frac{x+5}{x-1} &= \frac{1}{x+1} \\ \frac{(x+5)(x+1)}{(x-1)(x+1)} &= \frac{x-1}{(x-1)(x+1)} \\ (x+5)(x+1) &= x-1 \\ x^2 + 6x + 5 &= x-1 \\ x^2 +5x + 6 &= 0 \\ (x + 3)(x+2) &= 0 \\ x+ 3 = 0\textrm{ or } x+2 &= 0 \\ x = -3 \textrm{ or } x &= -2 \end{align*}$

Re: factorisation of a quadratic equation that is a fraction :/

Quote:

Originally Posted by

**Prove It** I assume that your equation is $\displaystyle \displaystyle 1 + \frac{6}{x-1} = \frac{1}{x + 1}$. Some brackets where they're needed would be nice... Anyway...

Thank you for your help. Yeah I have been trying to figure out how everyone gets their equations in that format on this forum.