## timo86m 3 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?

1. jygebs

also if you divide by a negative number

2. timo86m

what about the reciprocal how does that work?

3. Syderitic

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

4. myininaya

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

5. timo86m

can you give me a simple example involving a variable when you have to use the reciprocal?

6. timo86m

oh i see thank @myininaya

7. timo86m

what if it was -1/x<7

8. timo86m

-x>1/7 x<-1/7 ?

9. myininaya

oh wait @timo86m i made a mistake sorry

10. Syderitic

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

11. myininaya

$\frac{1}{x}<7$ 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

12. myininaya

unless x is always greater 0

13. myininaya

but like i said i didn't say that so the way to solve this one is...

14. timo86m

ah ok

15. myininaya

$\frac{1}{x}-7 <0$ $\frac{1}{x}-\frac{7x}{x} <0$ $\frac{1-7x}{x}<0$ 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

16. myininaya

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

17. myininaya

Like I did before

18. timo86m

we could just assume that and make it easier lol.

19. myininaya

but if is doesn't say that x>0 then you cannot assume it lol but yes it would have made things none easier

20. myininaya

Ok let me give an example where you don't have to worry about being wrong when taking reciprocal ... one sec

21. timo86m

well specify like x>0 and x>n

22. myininaya

$\frac{1}{e^x}<7$ $\frac{1}{e^x}-7<0$ $\frac{1}{e^x}-\frac{7 e^x}{e^x}<0$ $\frac{1-7e^x}{e^x}<0$ 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

23. timo86m

thanks :) Sorry i had to go suddenly family emergency came up :)