katherinesmith
  • katherinesmith
inequality question - equation inside; PLEASE HELP!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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katherinesmith
  • katherinesmith
\[\frac{ x + 5 }{ 4x } > 3\]
katherinesmith
  • katherinesmith
solve for x
phi
  • phi
these are tricky. start with \[ \frac{ x + 5 }{ 4x } -3 > 0\]

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katherinesmith
  • katherinesmith
okay... then what D:
phi
  • phi
use a common denominator of 4x to change 3 to 12x/4x and subtract the two fractions. Can you do that ?
katherinesmith
  • katherinesmith
you lost me.
katherinesmith
  • katherinesmith
wait.... i got it
katherinesmith
  • katherinesmith
nevermind. i'm lost.
phi
  • 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
  • katherinesmith
so then you get -11x + 5...all that over 4x?
katherinesmith
  • katherinesmith
i don't think anything can confuse me more than inequalities and stupid fractions.
phi
  • 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
  • katherinesmith
well A isn't positive
phi
  • 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
  • katherinesmith
i don't freaking know, i got x < 5/11 and x < 0
katherinesmith
  • katherinesmith
i'm angry at myself not you. i'm so frustrated with not understanding this.
phi
  • 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
  • katherinesmith
but you said x > 0, 0 < x is this opposite of that isn't it?
phi
  • 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
  • phi
Second Case: both are negative -11x+5< 0 and (at the same time) 4x < 0 what do you get for this case ?
katherinesmith
  • 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
  • phi
can you finish this one?
phi
  • 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
  • katherinesmith
so if i get x < 3/5 and x < 0, would my final answer be 0 < x < 3/5 ?
phi
  • 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
  • katherinesmith
yes it is a different question. so would my answer be correct? because x < 0
phi
  • 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
  • katherinesmith
so what the heck am i supposed to put. x < 0 < 3/5 ?
katherinesmith
  • katherinesmith
or 3/5 > 0 > x ?
phi
  • 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
  • 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
  • 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
  • katherinesmith
\[\frac{ x + 3 }{ 3x } > 2\] there's the problem. go for it ;)
phi
  • phi
I get -5x + 3 > 0 --> 3/5 > x and 3x > 0 -------> x > 0 that is different from what you posted.
katherinesmith
  • 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
  • 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
  • katherinesmith
i understand now. so my final answer is x > 3/5 > 0 ?
phi
  • phi
x > 3/5 is not the same as x < 3/5
katherinesmith
  • katherinesmith
oh my
phi
  • 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
  • katherinesmith
yeah
phi
  • 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
  • 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
  • katherinesmith
thank you for your help

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