## anonymous 5 years ago 7w-3 ------ +3<5 2 how can I find the solution set?

1. anonymous

$\left|7w-3 \right|$

2. anonymous

The 7w-3 is an absoulte value

3. anonymous

Well, start by moving the 3, to the other side and clearing the fractional part.

4. anonymous

3+7w-3=5 right?

5. anonymous

3+7w-3<5 I mean

6. anonymous

No. $\frac{|7w-3|}{2} + 3 < 5$ $\implies \frac{|7w-3|}{2} < 5 -3$ $\implies |7w-3|< 2(5 + 3)$ Right?

7. anonymous

Ack!

8. anonymous

Should be $|7w -3| < 2(5-3)$

9. anonymous

7w-3<22

10. anonymous

Did you get 7w<22/3?

11. anonymous

No.

12. anonymous

2*2 = 4.

13. anonymous

That means that 7w is less than 4 units away from 3.

14. anonymous

The part of the equation where you have 7w-3<2(5+3) =7w-3<10+6

15. anonymous

I corrected that. It should be 2(5-3)

16. anonymous

Not 2(5+3)

17. anonymous

So that means -4 < 7w - 3 < 4

18. anonymous

Your right. And from there 7w<4-3 =7w<1 =w<1/7?

19. anonymous

No.

20. anonymous

$|7w - 3| < 4$$\implies -4 < 7w -3 < 4$

21. anonymous

Then you add 3 to each part of the relation.

22. anonymous

$-4 + 3 < 7w - 3 + 3 < 4 + 3$

23. anonymous

Wow thats alot

24. anonymous

Think about it. If I tell you that | x | < 1, That means that x is somewhere between 1 and -1.

25. anonymous

Right?

26. anonymous

right

27. anonymous

So if I tell you that |x + 3| < 4, that means that x+3 is somewhere between -4 and 4

28. anonymous

So now we simplify and we have: $-1 < 7w < 7$

29. anonymous

ok, so I thought when we were at the 7w-3<4, we were almost finished. But we have to bring the 4 to the other side also and do it again.

30. anonymous

well you can work the two relations at the same time.

31. anonymous

Dividing the whole thing by 7 we get...?

32. anonymous

1/7<w

33. anonymous

-1/7 is part of it. what's the other part?

34. anonymous

-1/7<w<7/7

35. anonymous

Right.

36. anonymous

7/7=1

37. anonymous

-1/7<w<1 So would it be accurate that the solution set would become (-1/7, 1)?

38. anonymous

So w can be anything between -1/7 and 1 and the original relation $\frac{|7w -3|}{2}+3 < 5$ Will be true.

39. anonymous

Yep. (-1/7,1) is the solution set.

40. anonymous

ok, great. I understand now. Thanks for your help.

41. anonymous

Just remember when you take off the absolute values that you have 2 possible solutions. $|x| = a \implies x \in \{a,-a\}$ $|x| < a \implies -a < x < a$ $|x| \le a \implies -a \le x \le a$ $|x| \ge a \implies x \le -a \ or\ a\le x$

42. anonymous

etc.

43. anonymous

Would the technique be the same if the absolute value contained a fraction? -2$-2\left| 7-v/8 \right|+3-11?$ First step move the 8 over?

44. anonymous

$2\left| 7-v/8 \right|$

45. anonymous

$-2\left| 7-v/8 \right|-3$

46. anonymous

hello?

47. anonymous

Sorry, was cleaning

48. anonymous

Yes, it'd be the same technique move everything that's not inside the absolute value to the other side, then evaluate the absolute value by splitting it up into parts.