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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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jygebsBest ResponseYou've already chosen the best response.0
also if you divide by a negative number
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
what about the reciprocal how does that work?
 one year ago

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

myininayaBest ResponseYou've already chosen the best response.2
5>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
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
can you give me a simple example involving a variable when you have to use the reciprocal?
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
oh i see thank @myininaya
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
oh wait @timo86m i made a mistake sorry
 one year ago

SyderiticBest ResponseYou've already chosen the best response.1
okay, 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
 one year ago

myininayaBest 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
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
unless x is always greater 0
 one year ago

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

myininayaBest 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
 one year ago

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

timo86mBest ResponseYou've already chosen the best response.0
we could just assume that and make it easier lol.
 one year ago

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

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

timo86mBest ResponseYou've already chosen the best response.0
well specify like x>0 and x>n
 one year ago

myininayaBest 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
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
thanks :) Sorry i had to go suddenly family emergency came up :)
 one year ago
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