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timo86m

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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  1. jygebs
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    also if you divide by a negative number

    • one year ago
  2. timo86m
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    what about the reciprocal how does that work?

    • one year ago
  3. Syderitic
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    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
  4. myininaya
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    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
  5. timo86m
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    can you give me a simple example involving a variable when you have to use the reciprocal?

    • one year ago
  6. timo86m
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    oh i see thank @myininaya

    • one year ago
  7. timo86m
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    what if it was -1/x<7

    • one year ago
  8. timo86m
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    -x>1/7 x<-1/7 ?

    • one year ago
  9. myininaya
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    oh wait @timo86m i made a mistake sorry

    • one year ago
  10. Syderitic
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    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
  11. myininaya
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    \[\frac{1}{x}<7\] 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
  12. myininaya
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    unless x is always greater 0

    • one year ago
  13. myininaya
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    but like i said i didn't say that so the way to solve this one is...

    • one year ago
  14. timo86m
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    ah ok

    • one year ago
  15. myininaya
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    \[\frac{1}{x}-7 <0\] \[\frac{1}{x}-\frac{7x}{x} <0\] \[\frac{1-7x}{x}<0\] The expression (1-7x)/x is zero when 1-7x=0 => x=1/7 The expression (1-7x)/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 \[\frac{1-7x}{x}\] For -1, we get (1-7(-1))/(-1)=8/(-1)=-8 But -8<0 This satisfies the inequality For 1/10, we get (1-7(1/10))/(1/10)=10(1-7(1/10))=10-7=3>0 But 3>0 This does not satisfy the inequality For 25, we get (1-7(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
  16. myininaya
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    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
  17. myininaya
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    Like I did before

    • one year ago
  18. timo86m
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    we could just assume that and make it easier lol.

    • one year ago
  19. myininaya
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    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
  20. myininaya
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    Ok let me give an example where you don't have to worry about being wrong when taking reciprocal ... one sec

    • one year ago
  21. timo86m
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    well specify like x>0 and x>n

    • one year ago
  22. myininaya
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    \[\frac{1}{e^x}<7 \] \[\frac{1}{e^x}-7<0\] \[\frac{1}{e^x}-\frac{7 e^x}{e^x}<0\] \[\frac{1-7e^x}{e^x}<0\] The expression on bottom is always greater than 0 so this expression is defined everywhere But 1-7e^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 1-7=-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
  23. timo86m
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    thanks :) Sorry i had to go suddenly family emergency came up :)

    • one year ago
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