Question

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?

Answer 1

also if you divide by a negative number

Answer 2

what about the reciprocal how does that work?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

oh i see thank @myininaya

Answer 7

what if it was -1/x<7

Answer 8

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

Answer 9

oh wait @timo86m i made a mistake sorry

Answer 10

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

Answer 11

\[\frac{1}{x}<7\]
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

Answer 12

unless x is always greater 0

Answer 13

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

Answer 14

ah ok

Answer 15

\[\frac{1}{x}-7 <0\]
\[\frac{1}{x}-\frac{7x}{x} <0\]
\[\frac{1-7x}{x}<0\]

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

Answer 16

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

Answer 17

Like I did before

Answer 18

we could just assume that and make it easier lol.

Answer 19

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

Answer 20

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

Answer 21

well specify like x>0 and x>n

Answer 22

\[\frac{1}{e^x}<7 \]

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

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

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

Answer 23

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