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
@Hero I don't know what extraneous means o.o....
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
\[\frac{6}{x-3}=\frac{3}{x}\]
Yes
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
solve by rewriting as \[6x=3(x-3)\]
multiply out, get \[6x=3x-9\] then go from there
you good or no?
hold on and yes
Am I suppose to factor or something?
no
it is a linear equation solve \[6x=3x-9\] in two steps a) subtract \(3x\) from both sides b) divide by sides by \(3\)
Oh. Then noooo I'm confused >_<.
Ok hold on I will give you a answer
you factor quadratics, ( degree 2) not linear equations
I got 2...Idk what im doing ):
Wait
\[6x=3x-9\\ -3x~~-3x\\ 3x=-9\]
-9 divided by 3 right?
yes
Ok I got -3
yes that is right
Yay!! is extranous
no no-extraneous
ok, thank you so much @satellite73 I am fanning you
yw
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