3 / (x-1) < -2 / x

The text with this example is: "We would like to multiply by x * (x- 1) to clear the inequality of fractions,
but this would require considering three cases separately. (What are they?)"

if i follow those steps:
(3 / (x-1) < -2 / x) * (x(x-1))
= 3x(x-1) / (x-1) + 2x(x-1) < 0
= 3x + 2x -2 < 0
= 5x < 2
= x < 2/5

Im clearly doing something wrong, how do i get to three cases?

 

(3 / (x-1) < -2 / x) * (x(x-1))
= 3x(x-1) / (x-1) + 2x(x-1) < 0

you can't put equal sign there, you are already given sign <


....cross multiply

........expand right side

..................add 2x to both sides





.................one of your your answers

then,
...........you see x is in both denominators, and they could not be equal to zero

so, x-1<0 => x<1
and x<0

interval solution is :
(-∞, 0)
(2/5, 1)

 

Very informative.

 

Thank you, though i had already solved it. I was particurely interested in was multiplying by x * (x- 1): "We would like to multiply by x * (x- 1) to clear the inequality of fractions, but this would require considering three cases separately. (What are they?)"

What would be the three cases that are reffered to?

 

to first clear all the fractions, you must multiply every term by the LCD (least common denominator)

3 / (x-1) < -2 / x ...........in this case LCD=x (x-1)

(3*x (x-1)) / (x-1) < -(2*x (x-1)) / x.............now simplify

3x < -2 (x-1) ............here we came to the point same as cross multiplying

3x < -2 x+2
3x+2x < 2
5x<2
x<2/5

 

It's like solving a fractional equation. Of course, using the rules for inequalities is crucial here.

 

And what would be the three cases that are reffered to?

 

x<2/5
x<1
and x<0

interval solution is :
(-∞, 0) -> one solution
[0,2/5]-> no solution
(2/5, 1)-> second solution

 

Very cool.