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

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Homework Help Overview

The discussion revolves around solving rational inequalities, specifically focusing on the inequality involving the expression (2x-1)/(5x+3) > 0. Participants are exploring the implications of the numerator and denominator on the sign of the overall expression.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to understand when to consider the numerator versus the denominator in determining the sign of the rational expression. There are questions about the validity of only focusing on the denominator and whether the numerator's zero points also affect the solution.

Discussion Status

There is an ongoing exploration of different cases and conditions under which the inequality holds. Some participants are questioning the advice given about ignoring the numerator, while others are trying to clarify the relationship between the critical points derived from both the numerator and denominator.

Contextual Notes

Participants are discussing the implications of specific values that make the numerator and denominator zero, as well as the need for a number line to visualize the intervals of solutions. There is a mention of confusion regarding the intersection of intervals and which conditions are stronger in the context of the inequality.

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

ehild
 
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?
 
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?
 
Last edited:
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 ?
 
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.
 
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?
 
Nelo said:
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

[tex]\frac {2x-1} { 5x + 3} < 0[/tex]
 
The basic advice you need is in post #5 of https://www.physicsforums.com/showthread.php?t=538060 . The current inequality,

[tex]\frac {2x-1} { 5x + 3} < 0[/tex]

Is simpler and easier to solve. It has maybe two critical values to use. Check all three intervals.
 
  • #10
Nelo said:
bump

Nelo should be back in about 10 days. Thank you to everybody helping him in this thread.
 
  • #11
Thanks to you for putting him out of our misery!
 

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