## Inopeki 3 years ago ab=c Solve for a Do i need to divide both sides by b?

1. Zed

yes

2. Inopeki

And that is all?

3. Zed

yes, so when you do that what will you get?

4. Inopeki

a=a?

5. Zed

$\frac{ab}{b}=\frac{c}{b}$ $a=\frac{c}{b}$

6. Inopeki

Oh! now i get it, can you set up another example so i can try?

7. Zed

Sure ax+y=c, solve for x

8. Inopeki

Ok, divide ax with a, then divide c with a getting ax/a+y=c/a. You then simplify that getting x+y=c/a. Then you take both sides minus y getting x=c/a-y

9. Inopeki

Is this correct?

10. Zed

Really close but we need to either get rid of the y first before divide or also divide the y by a

11. Inopeki

YES! Thanks for teaching me :)

12. Inopeki

aw :(

13. abdul_shabeer

No, first start by subtracting y from both sides

14. Inopeki

Alright, so ax+y=c ax=c-y x/a=c/a-y x=c/a-y

15. Zed

remeber that when you divide the a you need to do it to all the terms

16. Inopeki

Where did you get z?

17. Inopeki

Oh right.. so x=c-y/a

18. zbay

$x=\frac{c-y}{a}$

19. Zed

zbay is correct

20. zbay

Inopeki got the answer i just put it in the equation maker for ease of the viewers

21. Inopeki

Yay! :D

22. Zed

inopeki you need to write x=(c-y)/a just to make it clearer. i see it now good work

23. Inopeki

Thanks for the help :) Is there any way you can make it a little harder to give me a challenge?

24. zbay

yea i will come up with one for you

25. Inopeki

Thanks.

26. zbay

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ Solve for T_2

27. Inopeki

umm.. ill try

28. FoolForMath

Nobody mentioned division by zero ?!!

29. abdul_shabeer

Good observation @ Foolformath

30. Inopeki

Well to begin with i need to multiply both sides with P2V2, getting P1V1/T1*P2V2, right?$P _{1}V _{1}divT _{1}+P _{2}V _{2}=T2$

31. zbay

I'll lay it out by steps for you

32. Inopeki

NOO. wait. Its

33. abdul_shabeer

Cross multiplication

34. Inopeki

P1V1/T1*P2V2=T2

35. zbay

lcd = T_1T_2 $T_2P_1V_1=V_2P_2T_1$ $T_2=\frac{V_2P_2T_1}{P_1V_1}$

36. Inopeki

Oh right, i see it now. You wanted me to put the things i was multiplying by(V2P2) in by the other things that were multiplied. That would mean my answer is right, but yours is simplified.

37. Inopeki

No?

38. zbay

No because you need to multiply both sides by the lcd of $T_1T_2$ first then we cancel one of the varibles out because they are the same on top and bottom. then we can break T_2 apart from the rest of the left side by division, your answer would look close but in the end it would be wrong

39. Inopeki

What is an lcd? What makes T1 jump up?

40. zbay

lcd= lowest common denominator

41. Inopeki

What is that?

42. Inopeki

Is a denominator the 1 in T1?

43. zbay

The denominator is what is on the bottom of the fraction, so when we do the left side we get $\frac{T_1T_2P_11V_1}{T_1}$ and because we have the same thing on the top and bottom we can cancel out the denominator leving us with $T_2P_1V_1$ and we do the same operation on the right side

44. Inopeki

Why do we get T1 on both sides of the line?

45. zbay

because you are multiplying by t1t2 but on each side one of them will cancell

46. Inopeki

Ah

47. zbay

Good job that was a hard one

48. Inopeki

Thanks, when am i going to get to learn this in school?

49. Zed

50. Inopeki

8th

51. zbay

Not sure when you will cover that but it's a chemistry equation compariing similar gasses to determine preasure, temp, or volume. You might touch it in highschool sometime.

52. Inopeki

Oh, cool. What should i do now?

53. zbay

PV=nrt solve for n

54. Zed

And then try this $V=\frac{4}{3}\pi r^3$ Solve for r

55. Inopeki

Divide both sides by rt, getting PV/rt=n, am i right? By the way, why is PV upper case?

56. Inopeki

Zed, isnt that for volúmes of spheres or something? exept r should be ^3

57. zbay

good question and i don't have an answer for you but in that equation they are all upercase but the little n but i made a mistake and didn't capitilize the RT

58. Inopeki

But was i right?

59. Inopeki

Zed, im going to do yours now.

60. zbay

yes you were right and that was the ideal gas law by the way

61. Inopeki

Oh, that was a ^3! Divide both sides with (pi)r^2 getting V/(pi)r=4/3r, am i right sofar?

62. Inopeki

What does the law state?

63. Zed

okay so first step is to get r^3 by itself. what do you get?

64. zbay
65. Inopeki

Wait. I take back my first answer. First you divide both sides by 4/3(pi) to isolate r^3

66. Zed

good. have you done powers before and their inverses?

67. Inopeki

I think so, what do you mean?

68. Zed

okay so we have $r^3=\frac{3V}{4}$ how do I get rid of the cubed part?

69. Inopeki

Oh, i have done powers but i dont recall doing inverses, is that like squareroot and cubicroot? I guess we need to get the "cubicalroot" out of r^3.

70. Inopeki

We should be able to just divide r by r 2 times.

71. Inopeki

3* and by itself i mean the root.

72. Zed

not quite see let's try this example to help explain it $x^2=4$$x=\sqrt{4}$$x=4^{\frac{1}{2}}$$x=2$

73. Inopeki

What? Can you explain that?

74. Zed

okay so when we bring powers (eg 3) over the equation sign they turn to 1/powers (eg 1/3). if we look at squared when we bring it over it turns to 1/2 which is also known as the square root.

75. Inopeki

Why does it become 1/2?

76. Zed

the may be abit hard to follow but this is why. Let f(x)=y $y=x^7$ $\ln y=7 \ln x$ $lnx=(lny)/7$ $x=e^{(lny)/7}$ $Inverse = f^{-1}(x)=e^{(lnx)/7}$ $f^{-1}(x)=e^{lnx^\frac{1}{7}}$$=x^\frac{1}{7}$

77. Zed

Now usually when learning this we just accept this as fact that $x^n=x^{\frac{1}{n}}$ where n is a positive integer

78. Zed

That was probably way too much information at once just use the rule in the above post and try to solve $r^3=\frac{3V}{4}$

79. Inopeki

So if r^2 becomes 1/2, then r^3 should become 1/3, right?

80. Zed

Correct

81. Inopeki

That means that i have to maake r^3 to 1/3 and convert 1/3 to 4/12(to merge it with the right side). i then convert the right sides fractions too, getting 9V/12. Now i should be able to merge them, right? so i get r=13/12?

82. Inopeki

I screwed that up right...

83. Inopeki

What?

84. Zed

Sorry we want to raise both sides by the power of (1/3)

85. Inopeki

r^1/3=3V^1/3/4?

86. Zed

so the left hand side will be $(r^3)^{\frac{1}{3}}=r^{\frac{3}{3}}=r^1=r$

87. Inopeki

I dont understand.

88. Zed

okay the right side is good. we have $(\frac{3V}{4})^\frac{1}{3}$

89. Zed

remember on the left hand side we had r^3 and when raised it to the power of 1/3 the powers need to be multiplied.

90. Inopeki

Oh! So its r^3*1/3?

91. Zed

yes exactly it's one of the rules for powers. here are others if you wanted to learn and practice http://www.math.com/school/subject2/lessons/S2U2L2DP.html

92. Inopeki

93. Zed

yes and that simplifies to r=(3V/4)^1/3

94. Inopeki

Oh!

95. Zed

Well I'm off to sleep. I'll be on tomorrow if you want more examples or help :)

96. Inopeki

Thanks for everything!