A community for students.
Here's the question you clicked on:
 0 viewing
timo86m
 2 years ago
It's been a while but what are the rules for sign flipping when solving inequalities
For example I know when you multiply by a negative number you have to flip the sign as in
2x>4
x<2 Notice that it flipped. What are the other rules for sign flipping?
timo86m
 2 years ago
It's been a while but what are the rules for sign flipping when solving inequalities For example I know when you multiply by a negative number you have to flip the sign as in 2x>4 x<2 Notice that it flipped. What are the other rules for sign flipping?

This Question is Closed

jygebs
 2 years ago
Best ResponseYou've already chosen the best response.0also if you divide by a negative number

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0what about the reciprocal how does that work?

Syderitic
 2 years ago
Best ResponseYou've already chosen the best response.1hum, it flips every time you multiply both sides of the equation by 1 ex 2x>4 <=> 4>2x ,rearrange and you get 2x<4

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.25>4 But 1/5<1/4 So flip when doing reciprocal of both sides  Also 5>4 But 5<4 So flip also when multiplying(or dividing) both sides by a negative

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0can you give me a simple example involving a variable when you have to use the reciprocal?

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0oh i see thank @myininaya

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2oh wait @timo86m i made a mistake sorry

Syderitic
 2 years ago
Best ResponseYou've already chosen the best response.1okay, 4/x>1/3 =>(mult by x) 4x/x>x/3 => 4>x/3 (mult by 3) 3*4>3x/3 =>12>x , since x switched, you switch the equation getting x<12

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{x}<7\] I made the assumption earlier that x>0 but I didn't tell you that so we can't just simply flip .... i will show you how to solve this

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2unless x is always greater 0

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2but like i said i didn't say that so the way to solve this one is...

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{x}7 <0\] \[\frac{1}{x}\frac{7x}{x} <0\] \[\frac{17x}{x}<0\] The expression (17x)/x is zero when 17x=0 => x=1/7 The expression (17x)/x is undefined when x=0 So we do a number line and test the intervals around 0 and 1/7  0 1/7 Choose number in each of the three intervals 1 1/10 25 Ok So plug into expression \[\frac{17x}{x}\] For 1, we get (17(1))/(1)=8/(1)=8 But 8<0 This satisfies the inequality For 1/10, we get (17(1/10))/(1/10)=10(17(1/10))=107=3>0 But 3>0 This does not satisfy the inequality For 25, we get (17(25))/25=some negative number Which also satisfies the inequality So The intervals (inf,0) U (1/7,inf) are the solution to the inequality

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2Ok but @timo86m if x was always greater than 0 then we could have simply took reciprocal of both sides and flip the inequality sign

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0we could just assume that and make it easier lol.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2but if is doesn't say that x>0 then you cannot assume it lol but yes it would have made things none easier

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2Ok let me give an example where you don't have to worry about being wrong when taking reciprocal ... one sec

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0well specify like x>0 and x>n

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{e^x}<7 \] \[\frac{1}{e^x}7<0\] \[\frac{1}{e^x}\frac{7 e^x}{e^x}<0\] \[\frac{17e^x}{e^x}<0\] The expression on bottom is always greater than 0 so this expression is defined everywhere But 17e^x=0 when 1=7e^x => 1/7=e^x => ln(1/7)=x  ln(1/7) Test both intervals 2 0 So plug in 2 we get some positive number Plug in 0 we get 17=6 <0 which satisfies the inequality So the solution is (ln(1/7),inf) But we could have did \[\frac{1}{e^x} <7\] \[e^x>\frac{1}{7}\] \[x>\ln(\frac{1}{7})\] much eaiser ! :) We can do this because e^x is always greater than 0

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0thanks :) Sorry i had to go suddenly family emergency came up :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.