Question

1-1/4(2-(3-x)/(5)) ≤ (8-x)/(3)

Answer 1

o.o wow I have no clue im sorry wish I could help

Answer 2

@L-Lawliet-L, Can you write it out using the draw feature?

Answer 3

\[1 - \frac{1}{4\left(2-\frac{3-x}{5}\right)} \le \frac{8-x}{3}\]

Answer 4

That's what I think you posted.

Answer 5

No it was :\[1-\frac{ 1 }{ 4 }(2-\frac{ 3-x }{ 5 })\le \frac{ 8-x }{ 3 }\]

Answer 6

I tried to get you to post something sooner.

Answer 7

I am really sorry.

Answer 8

\[1-\frac{ 1 }{ 4 }\left(2-\frac{ 3-x }{ 5 }\right)\le \frac{ 8-x }{ 3 }\]
\[1-\frac{ 1 }{ 4 }\left(\frac{10}{5}-\frac{ 3-x }{ 5 }\right)\le \frac{ 8-x }{ 3 }\]
\[1-\frac{ 1 }{ 4 }\left(\frac{10-(3-x)}{5}\right)\le \frac{ 8-x }{ 3 }\]
\[1-\frac{ 1 }{ 4 }\left(\frac{10-3+x}{5}\right)\le \frac{ 8-x }{ 3 }\]
\[1-\frac{ 1 }{ 4 }\left(\frac{7+x}{5}\right)\le \frac{ 8-x }{ 3 }\]
\[1-\frac{7+x}{20}\le \frac{ 8-x }{ 3 }\]
\[\frac{20}{20}-\frac{7+x}{20}\le \frac{ 8-x }{ 3 }\]
\[\frac{20-(7+x)}{20}\le \frac{ 8-x }{ 3 }\]
\[\frac{13-x}{20}\le \frac{ 8-x }{ 3 }\]

Answer 9

Thank you.

Answer 10

I'm sure you know you're not done

Answer 11

But if i have to find x ?

Answer 12

Yes, you have to continue to isolate until you have isolated x

Answer 13

The good thing is, we don't have to worry about any denominator restrictions

Answer 14

O.K.
Thank you again.

Answer 15

I have a question.

Answer 16

@Hero ?

Answer 17

What's the question

Answer 18

If I substract the right side does it becomes :
\[\frac{ 13-x }{ 20 }- \frac{ 8-x }{ 3 }\le0\]
?

Answer 19

Yes, you could do that, but you don't have to

Answer 20

I do not know if i am going to find x by this way.

Answer 21

The x dissapears.
I think.

Answer 22

I take that back. There IS a solution. Continue solving for x.

Answer 23

Just try solving for x

Answer 24

I have to substract ?

Answer 25

You could have simply cross multiplied

Answer 26

\[\frac{13-x}{20}\le \frac{ 8-x }{ 3 }\]

At this point you could have done this:

\[3(13 - x) \le 20(8 - x)\]

Answer 27

Are you solving for x or are you stuck?

Answer 28

If you used my approach you would distribute to get

\[39 - 3x = 160 - 20x\]

Then place like terms on the same side:
\[20x - 3x = 160 - 39\]
\[(20 - 3)x = 121\]
\[17x = 121\]

Answer 29

Oops, the equals should be \(\le\)

Answer 30

I WAS.
Thanks to you.
Thank you again @Hero.

Answer 31

I had not seen that you had cross multiplied.
Perhaps if i had seen i would not been have stucked.