Solving Rational Inequalities: How to Determine the Interval of Solutions?

[Nelo]

Homework Statement

2x-1
_____ > 0
5x+3

Homework Equations

The Attempt at a Solution

Just wondering, my teacher taught us that youre only supposed to look at what makes the denominator = 0, and don't look at the numerator because it has no affect on anything.

So, if i were to solve that id have to use -3/5

so, case 1 would be x < -3/5 which is -0.6, so if i plug in a number less like -1 into the bottom id get a negetive value, however am i plugging it in the top as well...? she said just look @ bot


or, suppose i have something with two vertical asymptotes like..
_(x+5)_______ < 0
(x+3) (x-1)

Im just looking at the values that make it negetive right? so -3, and 1. if i were to set a interval chart up and say x < -3 , i plug like -4 into (x+3) and into (x-1) and id get a negetive by a negetive which makes it positive. so the sign wouldn't flip on case 1

 

[ehild]

Consider the possible signs of both the numerator and the denominator. When is the fraction positive?

ehild

 

[Nelo]

This is how

2x-1
_____ > 0
5x+3
was solved. She said DONT look @ the numerator, just at what the denominator = 0. so

case 1: x < 2 ( since its negetive at the bottom the sign flips

so 3x +1 < 0

x < -1/3

case 2: x > 2 , sign will pos so no affect on ineq.

so 3x+1 > 0

x > -1/3.

x>2 is the stronger condition so the final statement was -1/3 < x < 2
So do i only look @ denominator when there's a x^2 on top and bottom to?

 

[willem2]

I don't think the advice of "don't look at the numerator" is very useful.

The expression will change sign when the denominator is zero, but also when the numerator is zero.

I don't understand all what you're trying to compute. x = 2, and x = -1/3 are NOT values where either the numerator or the denominator changes sign.

Are you sure your solution is of the same problem as the one in your first post?

 

[Nelo]

Oh, musta been a dif example. k , so.

2x-1
_____ > 0
5x+3


since we only look @ denominator you plug in -3/5 in the bottom and get two cases

case 1 will be x < -3/5 which is -0.6, so if i plug in -2 . 5(-2) +3 the value is negetive so inequality will flip


c1 : x < -3/5

2x-1
_____ > 0
5x+3

2x-1 < 0
2x < 1

x < 1/2


c2: x > -3/5

2x-1
_____ > 0
5x+3

2x-1 > 0
x > 1/2

so i know those are the proper answers, but i don't know how to really set up the therefore statement. Like , if you look at it on a number line all those values seem to intersect at some point

ie) x < -3/5 , x < 1/2 . this conforms, they go the same direction, but how do i know which one is stronger? Dont we have more information if we know x < 1/2, or do we have more info if we know x < -3/5 ? same for the second case.. How do i figure out what case is predominant?

-0.6<-----------------0.5 c1 ( on a # line, roughly) (these intersect)

c2: -0.6 0.5 ----------------> (is 1/2 stronger?)

so final statement would be -0.6<x<0.5 ?

 

[willem2]

This is terriby confusing, and you've managed to flip a sign somewhere, because for x=1, the whole expression is 1/8 which is bigger than 0, so x=1 should be included in the answer.

just answer these questions.

1. when is the denominator >0
2. when is the numerator >0

and then

3. when do the denominator and the numerator have the same sign?
(there 2 cases here, both positive and both negative)

draw the intervals on 2 numberlines above each other if you have to.

 

[Nelo]

Thats the way we learned it, through cases. Yes, when the denom = 0 then inequality sign flips, that's why it flipped. Still right tho?

 

[willem2]

Thats the way we learned it, through cases. Yes, when the denom = 0 then inequality sign flips, that's why it flipped. Still right tho?

the answer you calculated is the solution of

\frac {2x-1} { 5x + 3} &lt; 0

 

[symbolipoint]

The basic advice you need is in post #5 of https://www.physicsforums.com/showthread.php?t=538060 . The current inequality,

\frac {2x-1} { 5x + 3} &lt; 0

Is simpler and easier to solve. It has maybe two critical values to use. Check all three intervals.

 

[berkeman]

bump

Nelo should be back in about 10 days. Thank you to everybody helping him in this thread.

 

[HallsofIvy]

Thanks to you for putting him out of our misery!