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 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?
 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?

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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 :)
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