3x - 5 > 0
x
I have one interval notation as (5/3, infinity).
but x is also less than 0 to negative infinity.
help please.
$\displaystyle \frac{3x-5}{x}>0$
- 1st step (the denominator not equal 0 (zero))
since, the denominator is x so
$\displaystyle x (!=) 0$ Note: (!=) not equal
- 2nd step (the numerator)
$\displaystyle 3x-5=0$
so, we have $\displaystyle x=\frac{5}{3}$
- 3rd step (create the number line)
create the number line with those two point we already have
first point is 0 and second point is $\displaystyle \frac{5}{3}$
$\displaystyle -\infty$-----0----------5/3-------$\displaystyle \infty$
the line number devided the line into 3 intervals
($\displaystyle -\infty$,0); (0,5/3] and [5/3,$\displaystyle \infty$)
- 4th step (check the point)
from the number line, you check the value of the point between the intervals and choose the positive value ($\displaystyle >0$ from the question).
......... hope it help...........