## AonZ 2 years ago Represent the solution sets on the number line for |x+3|>1 and |2x+5| < 3

1. AonZ

@whpalmer4

2. AonZ

|dw:1362125049826:dw|

3. AonZ

|dw:1362125068647:dw|

4. AonZ

ok now idk how to graph it

5. amorfide

to graph it take each side to be y= you have |x+3|>1 draw line y=x+3 then you know that because of the | | signs, your line is reflected so you only have a graph above the x axes. now draw the line y=1 then you want the region which y=|x+3| is greater than y=1

6. whpalmer4

No, we're supposed to show the result on the number line, no y involved here

7. Stebie

|dw:1362125192026:dw|

8. amorfide

he said graph it I instantly thought of the actual graph my bad

9. Stebie

|dw:1362125277366:dw|

10. whpalmer4

Ooh, @stebie that's not correct. Try out x = -2 in the original inequality, does it satisfy it?

11. whpalmer4

I just find the endpoints of the segments, then test a point to see whether the area between two endpoints, or between an endpoint and infinity is part of the solution set. I almost always use x=0 as a test point because the arithmetic is usually pretty trivial

12. AonZ

but this has 4 pair of answers

13. whpalmer4

@AonZ do you feel comfortable with this material now?

14. AonZ

umm not with this question still

15. whpalmer4

isn't this two questions?

16. AonZ

no these 2 question should be solved together i think

17. whpalmer4

it does say "solution sets"

18. whpalmer4

Well, if you think they are combined, plot all of the endpoints (solutions to the equalities), then try test points in each segment, plus between the endpoints and infinity. If the test point satisfies both inequalities, then that portion of the number line should be shaded.

19. AonZ

ok thanks