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Mathematics 26 Online
OpenStudy (gracygirl):

Double check my answers? (2 questions) 1. 3y + 2 ≥ 5y + 8 3y + 2 - 5y ≥ 5y + 8 - 5y -2y + 2 ≥ 8 -2y + 2 - 2 ≥ 8 - 2 -2y ≥ 6 -2y/-2 ≥ 6/-2 y ≥ 4? 2. 7y - 3 ≤ 10y + 9 7y - 3 - 10y ≤ 10y + 9 - 10y -3y - 3 ≤ 9 -3y - 3 + 3 ≤ 9 + 3 -3y ≤ 12 -3y/-3 ≤ 12/-3 y ≤ 4? I'm not sure I did these right?

OpenStudy (gracygirl):

@AloneS @wwhitlock @welshfella

alones (alones):

#1 seems wrong

alones (alones):

Waitt I'm confused both of them are wrong?

OpenStudy (gracygirl):

I don't think either of them are right but I don't know where I messed up or how to fix it.

alones (alones):

Oh lol

alones (alones):

Well than let's go over it

OpenStudy (gracygirl):

Mkay

OpenStudy (ninja4535):

I think in problem 1 your division is wrong. 6 / -2

OpenStudy (gracygirl):

OH wait it should be -3 because a negative + negative = positive, right?

OpenStudy (gracygirl):

For #1

alones (alones):

So this is how i'll do, and you check \(\ 3y+2−5y≥5y+8−5y\) \(\ −2y+2≥8\) subtract 2 from both sides \(\ −2y+2−2≥8−2\) \(\ −2y≥6\) and then divide both sides by -2. \[\frac{ -2y }{ -2 }=\frac{ 6 }{ -2 }\] y≤−3 is what i got for ma answer

OpenStudy (ninja4535):

That's what I got too

OpenStudy (gracygirl):

Yeah, that's it.

OpenStudy (gracygirl):

That means for the other problem it should be -4...

alones (alones):

Yes for #2 it should be y≥−4

OpenStudy (ninja4535):

Yep

OpenStudy (gracygirl):

Thank you both!

alones (alones):

(:<

OpenStudy (ninja4535):

No prob :)

OpenStudy (wwhitlock):

This looks like math.

OpenStudy (gracygirl):

Well observed XD

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