Question

Solve 2 + x > 15/x

Answer 1

The first thing you do is put your variables on the same side.

Answer 2

I've actually done the problem I just wanted someone else to do it too to see if our answers matched!

Answer 3

Oh okay i got it!

Answer 4

Could anyone double check me? I got an answer of x<-5 and x>3.
Could you show your work if you got different?

Answer 5

\[2+x >15\div x\]

Answer 6

The problem to solve is:
2 + x ›15/x

First, let's work on the left hand side of your inequality, the 2 + x
This means, for instance, to see if it can be simplified at all.
2+x evaluates to 2+x
So, all-in-all, the left hand side of your inequality can be written as: 2+x

Now, let's work on the right hand side of your inequality, the 15/x
To divide 1 by x

Answer 7

Basically is equals -5

Answer 8

So would you say my answer is incorrect?

Answer 9

\[2+x>\frac{15}{x}\] ??

Answer 10

Is it only greater then or is it equal to? because the way you solved it is equal to...

Answer 11

Im doubting -5 as well....

Answer 12

don't even think about multiplying both sides by \(x\) because you have no idea if \(x\) is positive or negative

Answer 13

you must start with the 2.

Answer 14

\[x+2\geq\frac{15}{x}\]
\[2+x-\frac{15}{x}\leq 0\]
\[\frac{x^2+2x-15}{x}\geq 0\]
\[\frac{(x+5)(x-3)}{x}\geq0\]

Answer 15

Or that way!

Answer 16

you have to put a zero on one side of the equal sign, so you can determine if it is positive or negative

Answer 17

\[\frac{(x+5)(x-3)}{x}\] changes sign at \(x=-5, x=0\) and \(x=3\)

Answer 18

Because its nonlinear...should have known that.

Answer 19

you will have to divide the line in two four intervals
\[(-\infty, -5), (-5,0), (0,3),(3,\infty)\] and check to see over what intervals it is positive

Answer 20

* in TO four intervals

Answer 21

@OpenSessame i don't think your answer is correct, because of the missing 0
i think it is positive on \([-5,0)\) and \([3,\infty)\)

Answer 22

i said -5 but not 3...

Answer 23

But i think your correct

Answer 24

@satellite73, what would your final answer be then?

Answer 25

@OpenSessame I am a bit confused would you mind filling me in?

Answer 26

When you have a nonlinear equation you must set it to zero and then factor it. Then you have to plot it on a graph. SO in the end you were originally correct.

Answer 27

-5,3:)

Answer 28

Thanks for the clarification! x>-5 & x>3

Answer 29

@satellite73 do you agree with my above answers?

Answer 30

\[(-5,0)∪(3,\infty)\]

Answer 31

what? haha