How do you solve and graph #0 < 2y + 8<12# OR #3> 2 (5-y) + 3 > -17#?
1 Answer
I'm going to start off with the first one,
0<2y+8<12
This is simple and easy to learn.
so to figure this out, you can start by subtracting 8 on the middle area so the 8 and the -8 cancel eachother out, and whatever you do to one side you have to do to the other, so subtract 8 from the right and left sides. This should get you to -8<2y<4. Now you have to divide 2 from all sides so /2 and 2y cancel eachother out to make 2 and so on...
You should have -4<y<2 and your simplified!! (btw you cannot add four to all area to cancel the right side out if you were wondering you cant leave the right side hanging and it goes against graphing rules!
The same with your second problem. Which your second problem =5>y>6
(BTW IMPORTANT graphing teqnique... When solving for y, if you divide by a negitive number all the greater than/less than signs change to opposites)
To CHECK your work you can go to desmos.com and mathway.com but, please, for my sake, since I helped ya maybe, please don't cheat! If you have any questions feel free to comment below.
