Question

Solve the inequality:
3y-2<5/y

Answer 1

the answer comes outs to 3y^2-2y-5=0 is this correct?

Answer 2

\[
3y^2-2y-5<0
\]Now you want to find to roots.

Answer 3

Or rather, you'd find the roots to help factor. What you really want to do here is factor.

Answer 4

answers come out to 5/3 and -1?

Answer 5

Then \(
(y-r_1)(y-r_2)<0
\) then \(y-r_1 < 0\) and \(y-r_2 > 0\) or the reverse.

Answer 6

Since it has to be negative, one of them has to be negative, but not both and not none.

Answer 7

so is it not 5/3 and -1

Answer 8

Hold on, first of all we have: \[
\left(y-\frac 5 3\right)(y+1) < 0
\]Assuming your roots are true.

Answer 9

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 \(y-5/3\)

Answer 10

This gives us: \[
\left(y-\frac 5 3\right)<0
\]And \[
(y+1)>0
\]

Answer 11

Or, simply put: \[
y < \frac 5 3, \,y>-1
\]

Answer 12

Both must be true for the equality to be true.

Answer 13

ok thanks