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

1. NickR

where do you get 10q from?

2. cbm

12-2, then I distributed by parenthesis

3. cbm

Is that wrong NickR?

4. etemplin

you need to distribute first

5. theEric

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

6. theEric

That's the start, right?

7. theEric

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

8. jayz657

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

sorry q<56

10. theEric

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

11. cbm

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

I wish you could give more than one best response

13. theEric

Yep! Because you multiplied or divided by a negative!

14. cbm

@theEric thanks!

15. theEric

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

16. cbm

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

17. theEric

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

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

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

21. cbm

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

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

23. cbm

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

24. cbm

idn

25. theEric

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

Yep, thats the question.

27. theEric

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

I got 246,9 so P≥246.9

29. theEric

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

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

31. cbm

Yes, it's supposed to be an inequality

32. theEric

That's what I would do to!

33. cbm

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

34. theEric

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

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

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

37. theEric

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

38. theEric

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

Oh, ok thanks! Maybe (10x20)/.81≥p

40. theEric

I would say $p\ge 246.0$

41. theEric

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

42. cbm

LOL, that's the truth.

43. theEric

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

44. theEric

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

45. cbm

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

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

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

So maybe that's the case here.

49. cbm

true, I hope so.

50. theEric

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

51. cbm

alright, Thanks again!!

52. theEric

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

53. cbm

:)

54. theEric

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

55. cbm

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

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

57. theEric

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

58. theEric

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

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!