Working out an inequality?

y^2<9 or,y^2-9<0 or,(y)^2-(3)^2<0 or,(y+3)(y-3)<0
We know, if a*b<0 then a & b are of opposite signs.
Case- 1: y+3>0 & y-3<0 i.e. y>-3 & y<3 i.e -3<y<3 is the solution.
Case- 2: y+3<0 & y-3>0 i.e. y<-3 & y>3. This has no solution.
Thus the required solution is -3<y<3. [Answer]

Unfortunately, we can't simply take the square root of both sides, even if we flip the inequality sign. We need to think about this question. Here are three ways you can think about this:

1) When we solve y^2 = 9, we get two solutions: y = +3 or -3. Thus -3, 3 are "critical points", and we can test on either side of them to determine whether the inequality holds there or not. We have three sections:

y < -3
-3 < y < 3
y > 3

If you try a number in each, e.g. y = -4, 0, 4, then we see that only y = 0 causes the inequality to hold, and so -3 < y < 3 is our answer.

2) You can also change the inequality to this:

y^2 - 9 < 0

or even better:

x^2 - 9 < 0

You can then graph the function y = x^2 - 9, and determine where it's less than 0. You should get the same answer.

3) You can also factorise:

(y - 3)(y + 3) < 0

We have two numbers multiplied to be less than 0, which means that one is positive and one is negative. Of course, y + 3 is always bigger than y - 3, so y + 3 must be positive and y - 3 must be negative. So we get two inequalities:

y + 3 > 0 AND
y - 3 < 0

y > -3 AND
y < 3

-3 < y < 3

- 3 > y < 3

There is really no working.

y^2 < 9
y < √9 -- or -- y > -√9
y < 3 -- or -- y > -3

∴ -3 < y < 3

_____________________________________

   y² < 9

0 ≤ y² < 9         ← any number squared is never negative

√y² < √9            ← since y² is ≥ to zero, you can take the square root of
                                            both sides without changing the order.

│y│< 3            ← because, in general:   √x² = │x│        and        √9 = 3


-3 < y < 3            ← because, in general:   If │x│< a,  then  -a < x < a



                      ANSWER
                      -3 < y < 3
                               OR
           open interval (-3,3)
                               OR
         y is between -3 and 3

_____________________________________

Ysquared<9
then yousquare root both sides so it look like this
√ysquared<√9
the sqare and square rootcross out and you get
y<3

y^2 < 9

Y^2 -9 < 0
y^2 - 3^2 < 0
(y + 3)(y-3) < 0

For it to be true both factors must be of opposite sign

Case 1; when y + 3 > 0 then y - 3 < 0
=> y > -3 and y < 3

Case 2: when y + 3 < 0 and y - 3 > 0
=> y < -3 and y > 3
not possibe as no value of y satisfies this.

y^2<9
y<3

y^2<9
⇔ -3<y<3

y^2 < 9 (square root both sides)
-3 < y < 3 <===ANSWER