Question

How do you solve `5/(x+3) ≥ 3/x`?

Answer 1

Solution: `-3 < x < 0 or x >= 4.5`. In interval notation: `(-3,0) uu[4.5,oo)`

Explanation:

`5/(x+3) >= 3/x or 5/(x+3) - 3/x >= 0 or (5x - 3(x+3))/(x(x+3)) >=0` or
`(2x-9)/( x(x+3)) >=0` . Critical points are `x=-3 , x =0 ,x =4.5`
`x != -3 , x != 0` (since denominator should not be zero.)

Check for `>=0`
1) `x < -3 ; (2x-9)/( x(x+3)) (-)/(-*-) i.e< 0`

2) `-3 < x < 0` ; `(2x-9) / ( x(x+3)) (-)/(-*+) i.e> 0`

3) `0 < x < 4.5 ; (2x-9)/( x(x+3)) (-)/(+*+) i.e<0`

4) `x > 4.5 ; (2x-9)/ ( x(x+3)) >=0 (+) / (+*+) i.e >0`

Solution: `-3 < x < 0 or x >= 4.5`. In interval notation: `(-3,0) uu[4.5,oo)` [Ans]