How do I solve for y and write a solution set for 3|y-9|

Question

How do I solve for y and write a solution set for 3|y-9|<27 and graph this?

tkhunny · Answer

You must utilize this idea:

|x| = x for $x\ge 0$
|x| = -x for $x\lt 0$

Let's do some examples of each of these:

|x| = x for $x\ge 0$

|3| = 3
|12| = 12

|x| = -x for $x\lt 0$

|-2| = -(-2) = 2
|-5| = -(-5) = 5

You must apply this idea to your |y-9|.

Let's see what you get.

anonymous · Answer

When I solved it. 3|y-9|<27 I got One solution as y>18 and another I solved as 3|y-9|<27 and I got y>0

anonymous · Answer

3|y-9|<-27 Sorry Thats how I got my second solution

tkhunny · Answer

You seem to be on the right path, but that doesn't quite make enough sense.

You should have two answers differentiated by $y-9\ge 0\;and\;y-9\lt 0$

anonymous · Answer

I took the 3 and I distributed to the y-9

tkhunny · Answer

I see.  Don't do that.  The difficulty with these problem types is the absolute value.  It is not helpful to make the value inside the absolute value more complicated.  Make things easier if you can, not harder.

3|y-9|<27

Divide by 3

|y-9| < 9

For $y \ge 9$, this is equivalent to (y-9) < 9, which leads to y < 18

For $y \lt 9$, this is equivalent to (9-y) < 9, which leads to y > 0

One more thing to point out before you think you are done.  Do you know what it is?

anonymous · Answer

Ohhhh ! I see. So I need to divide first

anonymous · Answer

Im going to re-do the problems. and see what I get again

anonymous · Answer

Ok. So I ended up getting y<18 and y<0 why did the signs change on y<0 for you?

tkhunny · Answer

Two things:

1) As far as your solutions, they are incomplete.  $y < 18$ is valid ONLY when $y \ge 9$.

2) I did not reverse the inequality.  I merely solved for the given variable.

For y < 9

|y-9| = -(y-9) = 9-y < 9

Solve for y.

anonymous · Answer

So I see what you're saying. As far a solution set what would that be?

tkhunny · Answer

$y < 18$ for $y \ge 9$, this makes $y\in [9,18)$

You do the other piece.

anonymous · Answer

-9 & -18 ?

tkhunny · Answer

This is an inequality.  You are not likelt to get a single point as the answer set.  you should get ranges.

$y<18\;for\; y≥9$ , this makes $y∈[9,18)$

$y>0\;for\; y<9$ , this makes $y∈(0,9)$

Can you combine those two ranges to a single range?  They seem to come pretty close to each other around y = 9.