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

In summary, Nelo's teacher told him not to look at the numerator because it has no affect on the result, but to look at what makes the denominator = 0. So if he were to solve the equation, he would have to use -3/5. Case 1 would be x < -3/5 which is -0.6, so if he plugged in a number less like -1, the value would become negetive and the inequality would flip. Case 2 would be x > 2, and the sign would be positive, so the equation would not be affected. Case 1 is the stronger condition and the final statement is -1/3 < x < 2.
  • #1
Nelo
215
0

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
 
Physics news on Phys.org
  • #2
Consider the possible signs of both the numerator and the denominator. When is the fraction positive?

ehild
 
  • #3
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?
 
  • #4
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:
  • #5
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 ?
 
  • #6
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.
 
  • #7
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?
 
  • #8
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]
 
  • #9
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!
 

What is a rational inequality question?

A rational inequality question is a type of mathematical question that involves fractions or rational expressions. It usually requires solving for an unknown variable while also considering restrictions on the variable to maintain the inequality.

How do I solve a rational inequality question?

To solve a rational inequality question, start by finding the critical values of the variable, where the inequality may change. Then, test each interval between the critical values to determine which intervals satisfy the inequality. The solution will be the union of all the intervals that satisfy the inequality.

What are the common mistakes when solving rational inequality questions?

The most common mistakes when solving rational inequality questions include forgetting to find the critical values, incorrectly identifying the inequality direction, and not considering restrictions on the variable. It is also essential to simplify the rational expression before solving the inequality.

Can I graph a rational inequality question?

Yes, you can graph a rational inequality question by first solving for the variable and then plotting the critical values and testing points in each interval. The solution will be the shaded region between the critical values that satisfy the inequality.

How can I check if my solution to a rational inequality question is correct?

You can check if your solution to a rational inequality question is correct by plugging in values from each interval into the original inequality and seeing if it satisfies the inequality. Additionally, you can graph the solution and compare it to the original inequality to verify the solution.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
838
  • Precalculus Mathematics Homework Help
Replies
11
Views
968
  • Precalculus Mathematics Homework Help
Replies
7
Views
700
  • Precalculus Mathematics Homework Help
Replies
5
Views
721
  • Precalculus Mathematics Homework Help
Replies
9
Views
707
  • Precalculus Mathematics Homework Help
Replies
3
Views
738
  • Precalculus Mathematics Homework Help
Replies
12
Views
939
Back
Top