Solving the inequality (y+2)(y-3) ≤ 0

[Thiru07]

Homework Statement

Given : (y+2)(y-3) <= 0

Homework Equations

The Attempt at a Solution

Now, I have y-3 <= 0 or y+2 <= 0
Hence, y <= 3 or y <= -2
But how

-2 <= y <= 3

is correct?
I think

-2 <= y <= 3

is wrong because y <= -2.
Can someone please clarify?

 

[PeroK]

Homework Statement

Given : (y+2)(y-3) <= 0

Homework Equations

The Attempt at a Solution

Now, I have y-3 <= 0 or y+2 <= 0
Hence, y <= 3 or y <= -2
But how is correct?
I think is wrong because y <= -2.
Can someone please clarify?

You could draw a graph of $(y+2)(y-3)$ to see what's happening.

What happens if both $y+2 < 0$ and $y-3 < 0$?

 

[Thiru07]

You could draw a graph of $(y+2)(y-3)$ to see what's happening.

What happens if both $y+2 < 0$ and $y-3 < 0$?

In that case , we will keep y < -2 and ignore y < 3.
I think I got it. We have to ignore y+2 <= 0 and y-3 >=0 as

y <= -2 and y >= 3

is not possible and keep

y >= -2 and y <= 3

Thanks PeroK :)

 

[Mark44]

Given : (y+2)(y-3) <= 0

Now, I have y-3 <= 0 or y+2 <= 0

No, the above doesn't follow from the inequality. I think you might already have solved this inequality, but showing where you went wrong is worthwhile.
For the product of two expressions to be negative, one of them has to be negative, and the other has to be positive. Since this can happen in either of two ways, you need to examine two different cases.

Hence, y <= 3 or y <= -2