Question

I'm having issues with figuring out set-builder notation, I have a general idea of how it works, but this is giving me problems. 4y + 7 (the symbol is greater than or equal to) >_ 23

Answer 1

well first do you know how to solve the inequality
4y+7>=23?

Answer 2

I've been watching some tutorials on it, but its kind of going over my head.... math is not my best subject.... I'm pretty sure its starting with {y|(its this bit that's giving me issues) }

Answer 3

well to get to that part
you need to first isolate y in the inequality
4y+7>=23

Answer 4

I would start by undoing that addition by 7 there

Answer 5

to undo addition you subtract
whatever you do to one side you do to the other
so
4y+7>=23

so try subtracting 7 on both sides as a first step

Answer 6

what do you after subtracting 7 on both sides?

Answer 7

So its going something like: 4y + 7 - 7 > 23 - 7?

Answer 8

yes?

Answer 9

right

Answer 10

and the whole reason we did that is so we could isolate the 4y

Answer 11

\[4y \ge 16 \]

Answer 12

now to undo the multiplication by 4 you need to divide by 4

Answer 13

whatever you do to one side do to the other

Answer 14

the inequality sign only changes direction when you multiply or divide both sides by a negative number
so the inequality keeps its direction is is already in

Answer 15

so try to divide both sides by 4
and let me know what you have as a result

Answer 16

\[\left\{ y: y \ge 4 \right\}\]

Answer 17

you were right I thought

Answer 18

Ah, that make so much more sense now. Ty

Answer 19

Thanks for explaining it!

Answer 20

np

Answer 21

were you trying to skip the solving part
and go straight to building your set?

Answer 22

No, I needed an understanding of how it works. I like to see an example of how to solve, it gives me an idea for how to do other problems.

Answer 23

Now I can actually move on in progress on my assignment. That was an example problem form my book. The book explains in it very techinical terms that are harder for me to understand

Answer 24

*from

Answer 25

here is another example
and I wanted to include this one just in case you end up multiplying and dividing both sides by a negative
\[-4x-1 \le 6 \\ \text{ add one \to both sides } \\ -4x \le 6+1 \\ -4x \le 7 \\ \text{ divide both sides by -4 } \\ \text{ remember when dividing both sides } \\ \text{ or multiplying both sides by a \neg flip the inequality direction } \\ x \ge \frac{7}{-4} \\ \left\{ x: x \ge \frac{-7}{4} \right\}\]

Answer 26

So the < flips when the problem has division? And > flips when there is multiplication?

Answer 27

when you have division or multiplication of a negative number on both sides
for example
we know that 3>2
but if you multiply both sides by -1
then you -3<-2

Answer 28

notice -3>-2 isn't true at all

Answer 29

but -3<-2 is true

Answer 30

so if you have -x>1
then x<-1

Answer 31

So, its changing the < or > to make the equation true

Answer 32

yes

Answer 33

only when dividing and/or multiplying both sides by a NEGATIVE number

Answer 34

Ok, that helps a lot, ty

Answer 35

like if you have 5x>10 then x>2

Answer 36

but if -5x>10 then x<-2

Answer 37

of if -5x>-10 then x<2 (<--one more example)

Answer 38

Awesome, that makes sense to me now, thank you very very much, you have been a great help! =)