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Mathematics 14 Online
ThisGirlPretty:

Solve 6 over x minus 3 equals 3 over x for x and determine if the solution is extraneous or not. x = −2, extraneous x = −2, non-extraneous x = −3, non-extraneous x = −3, extraneous

ThisGirlPretty:

@Hero I don't know what extraneous means o.o....

satellite73:

it means "not really a solution" sort of silly because either you have a solution or you do not but maybe we can do it by example

satellite73:

\[\frac{6}{x-3}=\frac{3}{x}\]

ThisGirlPretty:

Yes

satellite73:

ok here is the deal: if you solve and get \(x=3\) as the answer, it will be "extraneous" because you cannot really replace \(x\) by \(3\) that would make the denominator equal to zero

satellite73:

solve by rewriting as \[6x=3(x-3)\]

satellite73:

multiply out, get \[6x=3x-9\] then go from there

satellite73:

you good or no?

ThisGirlPretty:

hold on and yes

ThisGirlPretty:

Am I suppose to factor or something?

satellite73:

no

satellite73:

it is a linear equation solve \[6x=3x-9\] in two steps a) subtract \(3x\) from both sides b) divide by sides by \(3\)

ThisGirlPretty:

Oh. Then noooo I'm confused >_<.

ThisGirlPretty:

Ok hold on I will give you a answer

satellite73:

you factor quadratics, ( degree 2) not linear equations

ThisGirlPretty:

I got 2...Idk what im doing ):

ThisGirlPretty:

Wait

satellite73:

\[6x=3x-9\\ -3x~~-3x\\ 3x=-9\]

ThisGirlPretty:

-9 divided by 3 right?

satellite73:

yes

ThisGirlPretty:

Ok I got -3

satellite73:

yes that is right

ThisGirlPretty:

Yay!! is extranous

satellite73:

no no-extraneous

ThisGirlPretty:

ok, thank you so much @satellite73 I am fanning you

satellite73:

yw

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