anonymous
  • anonymous
Solve the inequality: 3y-2<5/y
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the answer comes outs to 3y^2-2y-5=0 is this correct?
anonymous
  • anonymous
\[ 3y^2-2y-5<0 \]Now you want to find to roots.
anonymous
  • anonymous
Or rather, you'd find the roots to help factor. What you really want to do here is factor.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
answers come out to 5/3 and -1?
anonymous
  • anonymous
Then \( (y-r_1)(y-r_2)<0 \) then \(y-r_1 < 0\) and \(y-r_2 > 0\) or the reverse.
anonymous
  • anonymous
Since it has to be negative, one of them has to be negative, but not both and not none.
anonymous
  • anonymous
so is it not 5/3 and -1
anonymous
  • anonymous
Hold on, first of all we have: \[ \left(y-\frac 5 3\right)(y+1) < 0 \]Assuming your roots are true.
anonymous
  • anonymous
We know that \[ \left(y-\frac 5 3\right) < (y+1) \]So that means if either one is going to be negative, then it will be \(y-5/3\)
anonymous
  • anonymous
This gives us: \[ \left(y-\frac 5 3\right)<0 \]And \[ (y+1)>0 \]
anonymous
  • anonymous
Or, simply put: \[ y < \frac 5 3, \,y>-1 \]
anonymous
  • anonymous
Both must be true for the equality to be true.
anonymous
  • anonymous
ok thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.