Posted by **maddie** on Sunday, May 24, 2009 at 11:36am.

I'm quite clueless as to how to solve this. I know the quadratic equation but am not sure how to get this equation into the format to plug in to the quadratic...

Solve the quadratic equation using the formula [1/(1+x)]-[1/(3-x)]=(6/35)

Show your work

- algebra II -
**bobpursley**, Sunday, May 24, 2009 at 11:46am
multiply both sides by 35(x+1)(x-3)

then

35(x-3)+35(x+1)=6(x-3)(x+1)

and you should be able to assemble it from there.

- algebra II -
**maddie**, Sunday, May 24, 2009 at 11:50am
so should i just multiply that all out and then plug it in?

- algebra II -
**drwls**, Sunday, May 24, 2009 at 11:51am
Write the two terms on the left with a common denomimnator. Then convert to standard quadratic form.

[(3-x) - (1+x)]/[(1+x)(3-x)] = 6/35

(2-2x)/[(1+x)(3-x)] = 6/35

(1-x) = (3/35)[(1+x)(3-x)]

35 -35x = 3[-x^2 +2x +3)

3x^2 -41x +26 = 0

(3x -2)(x-13)

Check my work. No guarantees

- algebra II -
**drwls**, Sunday, May 24, 2009 at 11:52am
We were lucky that one factored fairly easily. x = 2/3 or 13

- algebra II -
**maddie**, Sunday, May 24, 2009 at 12:26pm
i got the same answer but used the quadratic :) thank you for all your help everyone!

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