inequality question - equation inside; PLEASE HELP!

- katherinesmith

inequality question - equation inside; PLEASE HELP!

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- katherinesmith

\[\frac{ x + 5 }{ 4x } > 3\]

- katherinesmith

solve for x

- phi

these are tricky.
start with
\[ \frac{ x + 5 }{ 4x } -3 > 0\]

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## More answers

- katherinesmith

okay... then what D:

- phi

use a common denominator of 4x to change 3 to 12x/4x
and subtract the two fractions. Can you do that ?

- katherinesmith

you lost me.

- katherinesmith

wait.... i got it

- katherinesmith

nevermind. i'm lost.

- phi

\[ \frac{ x + 5 }{ 4x } -3 > 0 \\ \frac{ x + 5 }{ 4x } +\frac{-3\cdot 4x}{4x} > 0 \]
when you have two fractions with the same denominator, you can add or subtract their tops, and put that result over the common denominator.

- katherinesmith

so then you get -11x + 5...all that over 4x?

- katherinesmith

i don't think anything can confuse me more than inequalities and stupid fractions.

- phi

yes,
\[ \frac{-11x + 5}{4x} > 0\]
now you need to use this idea: a fraction A/B > 0 means A/B is a positive number. A/B will be positive if A is positive and B is positive OR both are negative.

- katherinesmith

well A isn't positive

- phi

First Case: both top and bottom are positive. for that to be true we need
-11x+5>0
and (at the same time)
4x > 0
can you solve for x ?

- katherinesmith

i don't freaking know, i got x < 5/11 and x < 0

- katherinesmith

i'm angry at myself not you. i'm so frustrated with not understanding this.

- phi

let's first do 4x > 0
we can divide both sides by 4
x > 0/4 or x> 0
in other words 4x is positive as long as x is positive.... i.e. x>0
now -11x+5>0
it is *always safer* to add to both sides . add +11x to both sides
-11x + 11x +5 > 11x
5 > 11x
5/11 > x
or x < 5/11
we have
x>0 and x<5/11
which we can write as
0 < x < 5/11

- katherinesmith

but you said x > 0, 0 < x is this opposite of that isn't it?

- phi

the "big side" of > is next to the bigger number
x > 0 (x bigger than zero) is the same thing as 0 < x (0 is less than x)
if x is bigger than 0, then 0 is smaller than x

- phi

Second Case: both are negative
-11x+5< 0
and (at the same time)
4x < 0
what do you get for this case ?

- katherinesmith

okay. thank you. i have two more problems like this so i'm gonna try both of them in the same way this one was done

- phi

can you finish this one?

- phi

You should get
-11x+5< 0 --> 5/11 < x
4x < 0 -----> x<0
you need both of these to be true at the same time (so we get a negative up top and a negative in the bottom).
but x cannot be both less than 0 and at the same time greater than 5/11
so this case cannot happen.
the final answer is
0 < x < 5/11 (from the first case)

- katherinesmith

so if i get x < 3/5 and x < 0, would my final answer be 0 < x < 3/5 ?

- phi

I assume this is a different question.
But if you need both conditions to be true *at the same time*
x < 3/5 and x < 0
then you need x<0 (if x < 0 then both conditions are satisfied)

- katherinesmith

yes it is a different question. so would my answer be correct? because x < 0

- phi

0 < x < 3/5 is short for x bigger than 0 and smaller than 3/5
(in other words, x is between 0 and 3/5 )
that is not the same as x<0 and x< 3/5

- katherinesmith

so what the heck am i supposed to put. x < 0 < 3/5 ?

- katherinesmith

or 3/5 > 0 > x ?

- phi

remember, the pointy end (smaller, narrow end) of < is next to the smaller number
***so what the heck am i supposed to put. x < 0 < 3/5***
it would be nice to see the original problem, but what you have found is you need one expression to be x<0 and the other expression to be x<3/5
(to get the correct signs)
you can think of it like this: x<0 (for example x=-1, "works" for -1<3/5)
both expressions will have the correct sign.
but even though x=1/5 works for 1/5< 3/5, it does not work for 1/5 < 0 (no, 0 is smaller than 1/5)

- katherinesmith

one of my expressions was x < 0 and the other was x < 3/5. it works. my question is how do i write the final answer.
do you want me to post the whole problem? i think i did it right but if you would like to check my work and do it yourself i don't mind posting it.

- phi

one of my expressions was x < 0 and the other was x < 3/5
The main idea is that both must be true at the same time.
if we plot them, like this
|dw:1377562259382:dw|

- katherinesmith

\[\frac{ x + 3 }{ 3x } > 2\]
there's the problem. go for it ;)

- phi

I get
-5x + 3 > 0 --> 3/5 > x
and
3x > 0 -------> x > 0
that is different from what you posted.

- katherinesmith

so i got the same thing as you except my inequality signs were flipped... don't you flip the signs when you divide?

- phi

you "flip" the relation if you divide by a negative number
so if you did
-5x + 3 > 0
-5x > -3
now divide by -5 and "flip"
x < 3/5
It is safer to only divide by positive numbers. we could do this
-5x + 3 > 0
add 5x to both sides
3 > 5x
3/5 > x (which is the same as x < 3/5 )

- katherinesmith

i understand now. so my final answer is x > 3/5 > 0 ?

- phi

x > 3/5 is not the same as x < 3/5

- katherinesmith

oh my

- phi

If I were you, I would try to sort out the confusion
notice these are true:
1<2
2 > 1 (same thing written two different ways)

- katherinesmith

yeah

- phi

if you are unsure, try a test case with numbers:
4 > 2
divide both sides by 2:
2 > 1
still ok
-4 < -2
divide by -2:
2 < 1 (not true). dividing by a negative flips the relation

- phi

-4 < -2
add +4 to both sides
0 < +2
still true
add -2 to both sides
-2 < 0
still true.
adding or subtracting is "safe". the relation stays the same

- katherinesmith

thank you for your help

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