## anonymous one year ago help me solve this equation please? will medal

1. anonymous

$\left| x+1 \right|-3<4$

2. anonymous

3. jim_thompson5910

|dw:1441587101373:dw|

4. anonymous

I'm sorry, it was x<7

5. jim_thompson5910

This is not an equation. It's an inequality The first step is to add 3 to both sides |dw:1441587163774:dw|

6. jim_thompson5910

|dw:1441587178859:dw|

7. jim_thompson5910

so we're left with |x+1| < 7

8. jim_thompson5910

now you'll use the rule If $$\Large |x| < k$$ then $$\Large -k < x < k$$ where k is a positive number

9. anonymous

okay

10. jim_thompson5910

so in this case, that rule makes |x+1| < 7 turn into -7 < x+1 < 7

11. anonymous

whats our next step?

12. jim_thompson5910

|dw:1441587371735:dw|

13. jim_thompson5910

next we subtract 1 from all sides to fully isolate x |dw:1441587398846:dw|

14. jim_thompson5910

the '1's in the middle cancel since they add to 0 |dw:1441587429310:dw|

15. anonymous

-6<x<6?

16. jim_thompson5910

-7 - 1 is not equal to -6

17. anonymous

on a number line would that look like this?

18. anonymous

The draw function isnt working for me v.v

19. anonymous

there would be an open circle at six and the arrow would go to the left continuously?

20. jim_thompson5910

-7 - 1 = -8 instead think of it as "you're seven dollars in debt. You buy something worth 1 dollar. So you're now 8 dollars in debt"

21. anonymous

Or would it stop at 8?

22. jim_thompson5910

-7 < x+1 < 7 becomes -8 < x < 6

23. jim_thompson5910

|dw:1441587719085:dw|

24. jim_thompson5910

here's what -8 < x < 6 looks like |dw:1441587737649:dw| open circle at -8 open circle at 6 shading in between the open circles (do NOT fill in the open circles)

25. anonymous

Thank you!

26. jim_thompson5910

you're welcome