solve (x-5/x )-1=x+3/x
This is all about order of operations. You need to get x alone. Do you know how you would go about that?
not really
OK. If you look at it, everything on the right has an x. But the left is mixed up. So why not move things over.
could you add one to both sides
Yes, but that won't get all the xes together.
\(\left(x-\frac{5}{x} \right)-1=x+\frac{3}{x}\) \(\left(x-\frac{5}{x} \right)-\left(x-\frac{5}{x} \right)-1=x+\frac{3}{x}-\left(x-\frac{5}{x} \right)\) \(-1=x+\frac{3}{x}-\left(x-\frac{5}{x} \right)\)
See, that gets all the Xes on one side. Now, it still needs to be simplified.
-2/x on the right side
Yes! \(-1=-\frac{2}{x}\) So, what do you think will be good to do next?
could you multiply both sides by -2
Well, that would still leave the x on the bottom of a fraction. The goal here is to get rid of the fraction next.
divide by-2
Still leaves a fraction... so not the best choice. Great you are thinking about it, but there is a better choice.
i dont know
Ever heard about multiplying by the inverse to get rid of a fraction?
oh
We don't need the whole inverse this time. Mostly we just need the x.
In fact... you could finish this in one step if you multiply by a negative x. That would take care of the - on the 1 and the x under the fraction all at the same time! =)
so it would be x=2
Yep! \(-1\cdot (-x)=-\frac{2}{x}\cdot (-x)\implies x=2\)
thanks
wait thats not an answer choice
np. This is all about figuring out how to get the x alone and out of any fractions. That is all about using the order of operation rules. If you work with those more, these should be a lot easier for you. =)
Oops! I missed one mistake!
8 0 −1 −8
Look at this again: \(-1=x+\frac{3}{x}-\left(x-\frac{5}{x} \right)\) The negative distribution! \(-1=x+\frac{3}{x}-x+\frac{5}{x} \)
\(-1=\frac{8}{x}\)
isnt it x-5/x
The - outdide the parentisis distributes to both. |dw:1372093683746:dw|
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