help algebra problem!

[elositocandela]

thanks a bunch guys one of the other reasons i love cb7

[blizzard335]

he's right

Its been 3 years since i've done that shit.

Thanks blizz

anytime

this is easy.

try integration or vectors or line integrals. now thats stirring ur noodles.

[d112crzy]

1 + (2/x) = (3/x^2)

x^2[1+(2/x)]=(3/x^2)x^2 multiply both sides by x^2

x^2+(2x^2/x^3)=3 Cancel out x's, and you're left with;

x^2+2x=3 Bring the 3 to the other side

x^2+2x-3=0 Then simplify it down

(x-1)(x+3) Solve for x

x=-3, 1

[blizzard335]

so x =1 i plug it in and it works it cant be -3 since it doesn't work

both will work

depends on what u need man.

if its like distance or something that needs a positive number, rule out the negative.

other than that, both will work fine.

x= -3 --> 9-6-3=0

x= 1 --> 1+2-3=0

[elositocandela]

both will work

depends on what u need man.

if its like distance or something that needs a positive number, rule out the negative.

other than that, both will work fine.

x= -3 --> 9-6-3=0

x= 1 --> 1+2-3=0

but dont i have to plug in, into the original?

[d112crzy]

yes, and it will work either way.

All those steps are the same exact equation, just simplified. The answer will be the same to all of them.

That's algebra.

[avendi]

awesome, I was partially right.
Thank god. I was worried I had completely forgotten my algebra.

[Straight Success]

ok so here is another one ...

1 + (2/x) = (3/x^2)

any suggestions?

1 + (2/x) = (3/x^2)

(x+3)(x-1)=0
x = -3 or x= +1