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hmmm, I can't really make out the equation(s) here. what are the x-1 and x+3 for?
i think its soppused to be like 2x-1 over x -1 and 3 over x+3
oh, ok, gotcha
so you need a common denominator. let's put everything on the same side
(2x-1)/(x-1) - 3/(x+3)=0
to get the common dnominator,
((2x-1)*(x+3))/((x-1)*(x+3)) - 3*(x-1)/((x+3)*(x-1)
((2x^2+2x-3) - (3x-3))/(x^2+2x-3)
I'll catch my breath now
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@brook aint that simulataneous equation
oups, made a mistake here
it's ((2x^2+5x-3) - (3x-3))/(x^2+2x-3)
I get the feeling I did something wrong here... @razor99 got an idea?
oh, no, ok, got it...
since that equation equals 0, you can scratch the denominator, leaving you with :
2x^2+2x = 0
x^2+x = 0
x^2 = -x
x = -1
when you plug it back in, it works, so that's fine by me. May have been unnecessarely long, but I got there ;-)