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Mathematics 14 Online
OpenStudy (anonymous):

Solve: 2x-1 = 3 x-1 x+3

OpenStudy (anonymous):

hmmm, I can't really make out the equation(s) here. what are the x-1 and x+3 for?

OpenStudy (anonymous):

i think its soppused to be like 2x-1 over x -1 and 3 over x+3

OpenStudy (anonymous):

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) (2x^2-x)/(x^2+2x-3) I'll catch my breath now

razor99 (razor99):

@brook aint that simulataneous equation

OpenStudy (anonymous):

oups, made a mistake here

OpenStudy (anonymous):

it's ((2x^2+5x-3) - (3x-3))/(x^2+2x-3) then (2x^2+2x)/(x^2+2x-3)

OpenStudy (anonymous):

I get the feeling I did something wrong here... @razor99 got an idea?

OpenStudy (anonymous):

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 ;-)

razor99 (razor99):

no guyz 3 would be the answer

OpenStudy (anonymous):

2*3/2 doesn't equal 3/(3+3), unless I'm missing something?

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