## cbm Group Title Someone please check. q+12-2(q-22)>0 q+10q-220>0 11q(-220+220)>0 11q>220 q>20 one year ago one year ago

1. NickR Group Title

where do you get 10q from?

2. cbm Group Title

12-2, then I distributed by parenthesis

3. cbm Group Title

Is that wrong NickR?

4. etemplin Group Title

you need to distribute first

5. theEric Group Title

$q+12-2(q-22)>0$

6. theEric Group Title

That's the start, right?

7. theEric Group Title

@cbm (just making a notification pop up :P )

8. jayz657 Group Title

i think you made a mistake there q+12-2(q-22)>0 q+12-2q+44 >0 -q +56 >0 -q > -56 q>56

9. jayz657 Group Title

sorry q<56

10. theEric Group Title

I agree with jayz657's work. Any questions, @cbm ?

11. cbm Group Title

Thank ya'll so much! I was reworking it. Yep, once I distributed I got the same answer. The sign at the end changed right? From > to <

12. cbm Group Title

I wish you could give more than one best response

13. theEric Group Title

Yep! Because you multiplied or divided by a negative!

14. cbm Group Title

@theEric thanks!

15. theEric Group Title

My pleasure. I'm glad you got it worked out :) Good luck with any more you have to do!

16. cbm Group Title

Thanks, I'm struggling with one more now trying to set up an inequality.

17. theEric Group Title

You can ask it here or in another question. I have time if you'd like a hand. Algebra like this can be hard at first.

18. cbm Group Title

Ok, it will take me a minute to type it. Thanks for your help. Yes, I'm too old to learn this stuff.

19. theEric Group Title

Haha, no such thing. It might just require some thinking and rethinking, right? Maybe thinking in a different light! I started thinking about counting cookies/blocks again before so the numbers meant something.

20. theEric Group Title

And take your time!

21. cbm Group Title

Your making a rectangular patio that is 10ft wide and 20ft long, using square pavers that have an area of 0.81 sq ft, what is the # of pavers u will need? Write and solve an inequality to find the answer.

22. cbm Group Title

Lol, I'm trying. I thought I had equations but not the fractions

23. cbm Group Title

would it be like (10x20).81=p

24. cbm Group Title

idn

25. theEric Group Title

The first step is always understanding the problem. Then we move on.. So I have a quick question... Each "paver" is .81 square feet, and we need to know how many of the will fit in our 10 foot by 20 foot area?

26. cbm Group Title

Yep, thats the question.

27. theEric Group Title

Alright! Then (10*20) is definately your area of the patio. And you wanna see how many .81 area portions will fit into that. How many times does .81 go into (10*20) is the question to get you to your answer. That's a division thing.

28. cbm Group Title

I got 246,9 so P≥246.9

29. theEric Group Title

That's just one way to look at it. There's another way in which you build the equation and solve.. I got the same number after rounding. Is this supposed to be inequality?

30. cbm Group Title

I'm not sure if I'm writing the inequality right or not (10ftx20ft)/.81=p

31. cbm Group Title

Yes, it's supposed to be an inequality

32. theEric Group Title

That's what I would do to!

33. cbm Group Title

To me it works and gives the answer when you work it.

34. theEric Group Title

I don't see why it would be inequality! There must be something I'm not thinking of.. I thought it would want an exact number...You want so many pavers.... OR! Do you want at least so many pavers? Meaning that many or more?

35. cbm Group Title

Do you want me to create a problem so I can give you a medal? I'm not sure what the medal do.

36. cbm Group Title

I was trying to figure out why it want come out to a whole number also unless it's ≥

37. theEric Group Title

They're just sort of recognition you can count. I'm not worried about it. Thanks though!

38. theEric Group Title

Yeah, I think maybe that's the idea of it.. why they want inequality.. Although you'd think rounding it up to 247 would be good enough! Maybe the question was just not appropriate for this subject!

39. cbm Group Title

Oh, ok thanks! Maybe (10x20)/.81≥p

40. theEric Group Title

I would say $p\ge 246.0$

41. theEric Group Title

You would want more pavers rather than less. At least you get your patio done that way :P

42. cbm Group Title

LOL, that's the truth.

43. theEric Group Title

Pavers area things should be greater than or equal to the patio area... to get the patio done :P Yeah!

44. theEric Group Title

Using just that sentence, you can set up the inequality. $p_{NumberOfPavers}\ge A_{PatioArea}$

45. cbm Group Title

I've been trying everything to get an even number. I don't understand why it would be a decimal. I tried breaking it up into inches, etc. Thanks again for your help.

46. theEric Group Title

It might just not come out to be an even number. It happens. We could solve it to be a fraction instead of a decimal. With fractions, your always exact (no rounding).

47. theEric Group Title

I have five cookies to share between us. How many can we each have? 2.5. Books don't always make problems come out to nice numbers either..

48. theEric Group Title

So maybe that's the case here.

49. cbm Group Title

true, I hope so.

50. theEric Group Title

Same. Good luck with that! If you solve it as a fraction, note that .81 = 81/100.

51. cbm Group Title

alright, Thanks again!!

52. theEric Group Title

$\frac{10*20}{.81}=\frac{10*20}{\frac{81}{100}}$ Np!

53. cbm Group Title

:)

54. theEric Group Title

Just in case... From there, it's $\frac{10*20*100}{81}=\frac{20000}{81}=246\frac{74}{81}$

55. cbm Group Title

Yep, that works good or at least it looks better. I figured I could write the equation as (p)x0.81≥200. I'm still working it to see if it will work.

56. theEric Group Title

That's great, and it was the other way I would think of the equation to solve!

57. theEric Group Title

And you don't have to change the sign because 0.81 is not negative.|dw:1351404148861:dw|

58. theEric Group Title

All other operations keep the inequality the same direction. Adding (positive number) to both sides: Each side gets the same bump up Subtracting (positive number) from bothe sides: Each side gets the same cut down Multiplying (number greater than 1) to both sides: Each side gets larger proportional to its size Dividing both sides by (number greater than one): Each side gets smaller proportional to its size

59. theEric Group Title

I threw in number specifications, just because adding a negative is like subtracting a positve. Then each side gets cut down the same. Same for multiplication/division. Multiplying by a fraction less than one is like dividing by that fraction's reciprocal$x*\frac{1}{2}=x\div 2$and so each side gets proportionally smaller. Just some extra, if you wanna think about it. Take care! It was nice working with you!