Would someone mind verifying this for me?

Thanks

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- Nov 13th 2009, 01:36 PMcrashdummy[SOLVED] Something doesn't look quite right
Would someone mind verifying this for me?

Thanks - Nov 13th 2009, 01:48 PMskeeter
you're right ... it's not quite right.

$\displaystyle \frac{3x-5}{x} > 0$

there are two critical values ... $\displaystyle x = 0$ , where the expression is undefined, and $\displaystyle x = \frac{5}{3}$ , where the expression equals 0.

these two "critical" values break up the values for x into three sections ...

$\displaystyle x > \frac{5}{3}

$

$\displaystyle 0 < x < \frac{5}{3}$

and $\displaystyle x < 0$

pick any value for x in each section and test it in the original inequality to see if it makes the inequality true ... if the inequality is true for that value of x, then all values of x in that interval will make the inequality true.

if false, then all values of x in that interval make the inequality false. - Nov 13th 2009, 06:56 PMcrashdummy
OK so I took your suggestions and played around with a few numbers for x.

(3x-5) / 3 > 0, when:

x > 5/3 and;

x < 0.

Now how would I write this in interval notation?

Would I need two notations? ie. (infinity, 0) & (5/3, infinity)