## Inopeki Group Title ab=c Solve for a Do i need to divide both sides by b? 2 years ago 2 years ago

1. Zed Group Title

yes

2. Inopeki Group Title

And that is all?

3. Zed Group Title

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

4. Inopeki Group Title

a=a?

5. Zed Group Title

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

6. Inopeki Group Title

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

7. Zed Group Title

Sure ax+y=c, solve for x

8. Inopeki Group Title

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 Group Title

Is this correct?

10. Zed Group Title

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

11. Inopeki Group Title

YES! Thanks for teaching me :)

12. Inopeki Group Title

aw :(

13. abdul_shabeer Group Title

No, first start by subtracting y from both sides

14. Inopeki Group Title

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

15. Zed Group Title

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

16. Inopeki Group Title

Where did you get z?

17. Inopeki Group Title

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

18. zbay Group Title

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

19. Zed Group Title

zbay is correct

20. zbay Group Title

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

21. Inopeki Group Title

Yay! :D

22. Zed Group Title

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

23. Inopeki Group Title

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

24. zbay Group Title

yea i will come up with one for you

25. Inopeki Group Title

Thanks.

26. zbay Group Title

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

27. Inopeki Group Title

umm.. ill try

28. FoolForMath Group Title

Nobody mentioned division by zero ?!!

29. abdul_shabeer Group Title

Good observation @ Foolformath

30. Inopeki Group Title

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 Group Title

I'll lay it out by steps for you

32. Inopeki Group Title

NOO. wait. Its

33. abdul_shabeer Group Title

Cross multiplication

34. Inopeki Group Title

P1V1/T1*P2V2=T2

35. zbay Group Title

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

36. Inopeki Group Title

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 Group Title

No?

38. zbay Group Title

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 Group Title

What is an lcd? What makes T1 jump up?

40. zbay Group Title

lcd= lowest common denominator

41. Inopeki Group Title

What is that?

42. Inopeki Group Title

Is a denominator the 1 in T1?

43. zbay Group Title

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 Group Title

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

45. zbay Group Title

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

46. Inopeki Group Title

Ah

47. zbay Group Title

Good job that was a hard one

48. Inopeki Group Title

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

49. Zed Group Title

50. Inopeki Group Title

8th

51. zbay Group Title

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 Group Title

Oh, cool. What should i do now?

53. zbay Group Title

PV=nrt solve for n

54. Zed Group Title

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

55. Inopeki Group Title

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

56. Inopeki Group Title

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

57. zbay Group Title

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 Group Title

But was i right?

59. Inopeki Group Title

Zed, im going to do yours now.

60. zbay Group Title

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

61. Inopeki Group Title

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

62. Inopeki Group Title

What does the law state?

63. Zed Group Title

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

64. zbay Group Title
65. Inopeki Group Title

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

66. Zed Group Title

good. have you done powers before and their inverses?

67. Inopeki Group Title

I think so, what do you mean?

68. Zed Group Title

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

69. Inopeki Group Title

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 Group Title

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

71. Inopeki Group Title

3* and by itself i mean the root.

72. Zed Group Title

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 Group Title

What? Can you explain that?

74. Zed Group Title

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 Group Title

Why does it become 1/2?

76. Zed Group Title

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 Group Title

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 Group Title

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 Group Title

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

80. Zed Group Title

Correct

81. Inopeki Group Title

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 Group Title

I screwed that up right...

83. Inopeki Group Title

What?

84. Zed Group Title

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

85. Inopeki Group Title

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

86. Zed Group Title

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

87. Inopeki Group Title

I dont understand.

88. Zed Group Title

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

89. Zed Group Title

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 Group Title

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

91. Zed Group Title

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 Group Title

93. Zed Group Title

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

94. Inopeki Group Title

Oh!

95. Zed Group Title

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

96. Inopeki Group Title

Thanks for everything!