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

[ivaralink]

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?

 

[MathLover1]

(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)

 

[nycmathguy]

(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.

 

[ivaralink]

(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)

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?

 

[MathLover1]

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

 

[nycmathguy]

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.

 

[ivaralink]

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

 

[MathLover1]

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

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

 

[nycmathguy]

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.