anonymous
  • anonymous
1-1/4(2-(3-x)/(5)) ≤ (8-x)/(3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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demonickittenlover
  • demonickittenlover
o.o wow I have no clue im sorry wish I could help
Hero
  • Hero
@L-Lawliet-L, Can you write it out using the draw feature?
Hero
  • Hero
\[1 - \frac{1}{4\left(2-\frac{3-x}{5}\right)} \le \frac{8-x}{3}\]

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More answers

Hero
  • Hero
That's what I think you posted.
anonymous
  • anonymous
No it was :\[1-\frac{ 1 }{ 4 }(2-\frac{ 3-x }{ 5 })\le \frac{ 8-x }{ 3 }\]
Hero
  • Hero
I tried to get you to post something sooner.
anonymous
  • anonymous
I am really sorry.
Hero
  • Hero
\[1-\frac{ 1 }{ 4 }\left(2-\frac{ 3-x }{ 5 }\right)\le \frac{ 8-x }{ 3 }\] \[1-\frac{ 1 }{ 4 }\left(\frac{10}{5}-\frac{ 3-x }{ 5 }\right)\le \frac{ 8-x }{ 3 }\] \[1-\frac{ 1 }{ 4 }\left(\frac{10-(3-x)}{5}\right)\le \frac{ 8-x }{ 3 }\] \[1-\frac{ 1 }{ 4 }\left(\frac{10-3+x}{5}\right)\le \frac{ 8-x }{ 3 }\] \[1-\frac{ 1 }{ 4 }\left(\frac{7+x}{5}\right)\le \frac{ 8-x }{ 3 }\] \[1-\frac{7+x}{20}\le \frac{ 8-x }{ 3 }\] \[\frac{20}{20}-\frac{7+x}{20}\le \frac{ 8-x }{ 3 }\] \[\frac{20-(7+x)}{20}\le \frac{ 8-x }{ 3 }\] \[\frac{13-x}{20}\le \frac{ 8-x }{ 3 }\]
anonymous
  • anonymous
Thank you.
Hero
  • Hero
I'm sure you know you're not done
anonymous
  • anonymous
But if i have to find x ?
Hero
  • Hero
Yes, you have to continue to isolate until you have isolated x
Hero
  • Hero
The good thing is, we don't have to worry about any denominator restrictions
anonymous
  • anonymous
O.K. Thank you again.
anonymous
  • anonymous
I have a question.
anonymous
  • anonymous
@Hero ?
Hero
  • Hero
What's the question
anonymous
  • anonymous
If I substract the right side does it becomes : \[\frac{ 13-x }{ 20 }- \frac{ 8-x }{ 3 }\le0\] ?
Hero
  • Hero
Yes, you could do that, but you don't have to
anonymous
  • anonymous
I do not know if i am going to find x by this way.
anonymous
  • anonymous
The x dissapears. I think.
Hero
  • Hero
I take that back. There IS a solution. Continue solving for x.
Hero
  • Hero
Just try solving for x
anonymous
  • anonymous
I have to substract ?
Hero
  • Hero
You could have simply cross multiplied
Hero
  • Hero
\[\frac{13-x}{20}\le \frac{ 8-x }{ 3 }\] At this point you could have done this: \[3(13 - x) \le 20(8 - x)\]
Hero
  • Hero
Are you solving for x or are you stuck?
Hero
  • Hero
If you used my approach you would distribute to get \[39 - 3x = 160 - 20x\] Then place like terms on the same side: \[20x - 3x = 160 - 39\] \[(20 - 3)x = 121\] \[17x = 121\]
Hero
  • Hero
Oops, the equals should be \(\le\)
anonymous
  • anonymous
I WAS. Thanks to you. Thank you again @Hero.
anonymous
  • anonymous
I had not seen that you had cross multiplied. Perhaps if i had seen i would not been have stucked.

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