## Inopeki Group Title Is this right? Q: Solve 13+s/p=2 for p A: p(13)=2(13)+2=28? What is s? Ill say 2. 28/13≈2.15 2 years ago 2 years ago

1. UnkleRhaukus Group Title

nope

2. Inopeki Group Title

Which part isnt right?

3. Inopeki Group Title

Goddammit this is just hopeless..

4. TuringTest Group Title

hi Inopeki do not despair! practice practice...

5. TuringTest Group Title

what made you feel like putting a 13 there? we were supposed to get a function like before

6. Inopeki Group Title

And didnt i?

7. TuringTest Group Title

you can't start plotting points like that until we've solved it for one of the variables. The answer to the question I was looking for should have 2 variables.

8. TuringTest Group Title

13+s/p=2 what do you get multiplying both sides by p? do nothing else but that.

9. Inopeki Group Title

I can go check

10. TuringTest Group Title

no, I was wrong, nevermind

11. UnkleRhaukus Group Title

13+s/p=2 for p 11+s/p=0 11=-s/p p=-s/11

12. Inopeki Group Title

See? This guy knows it. Why cant i?

13. Inopeki Group Title

Whatever. give me a new one

14. UnkleRhaukus Group Title

You can with Time Inopeki

15. TuringTest Group Title

well, there's one way. btw Rhauckus I made up this problem for Inopeki to teach him how to solve for a variable @Inopeki, he has many years doing this, you need to practice basics. It will be like that for a while.

16. Inopeki Group Title

Well that sucks.. Still, if i really need to learn this i guess i will. Give me a new example so i can practise please

17. TuringTest Group Title

a new one? solve$A=2\pi r h$ for A, r, and h I need to get you to stop thinking in numbers.

18. TuringTest Group Title

that's three problems, by the way.

19. Inopeki Group Title

is $\pi$ 3.14 here or is it a variable?

20. TuringTest Group Title

no, it's about 3.14159..., that's why I didn't make it something to solve for. but don't think of it as a number, leave it as pi. The number does not matter.

21. TuringTest Group Title

A=2(pi)rh

22. Inopeki Group Title

Ok, so the answer should not be a number?

23. TuringTest Group Title

not at all

24. Inopeki Group Title

Interesting..

25. TuringTest Group Title

it should be a function like r=5t+4 it has some numbers, but not all.

26. Inopeki Group Title

One question. If i, in the answer have to give you As "definition", how can that not be a number?

27. TuringTest Group Title

The formula I gave you is for the surface area of the side of a cylinder. The surface area of any particular cylinder depends on its radius and length. So the formula covers all possible cylinders, but only for some particular cylinders will you get a certain value for the surface area. It is a function, this time of both r and h.

28. TuringTest Group Title

if I tell you the radius and height of our particular cylinder, only then can you get a number.

29. Inopeki Group Title

I just studied that in school! I thought we were talking algebra here..

30. Akshay_Budhkar Group Title

@Turning you train guys well I will need your help for multivariable calculus, Laplace and stuff I will let you know then :D Great job is all I can say :D

31. TuringTest Group Title

Solving for a variable is algebra. They are all connected as you will learn. Thanks Akshay!

32. Inopeki Group Title

But that still doesnt answer my question, how should i define A?

33. TuringTest Group Title

in this case, A is already defined by$A=2\pi rh$that is basically the definition of the surface area of the side of a circular cylinder. No more definition is needed. I wanted you to catch that the first problem, solve for A, is already done for you: we already have A=(something) so it IS solved for A. So now solve it for r, if you can. what can we divide by to get r by itself?

34. Inopeki Group Title

We can divide h with A, right?

35. Inopeki Group Title

No.. that cant be it.

36. TuringTest Group Title

when you say 'divide h with A' I'm not sure what you mean. Do you mean divide both sides by A or divide both sides by h ??? One is okay, one is not. You know that division is the inverse of multiplication. So ask yourself 'what is r being multiplied by?' and divide by that. Well, what is r being multiplied by?

37. Inopeki Group Title

The height.

38. TuringTest Group Title

is that all?

39. Inopeki Group Title

Actually no

40. TuringTest Group Title

what else?

41. Inopeki Group Title

pi^2

42. Inopeki Group Title

right?

43. TuringTest Group Title

where did you see a squared term?

44. Inopeki Group Title

I have no idea.. i realize thats wrong now cause that is for getting the volume of a cylinder..

45. Inopeki Group Title

And that concludes that i am an utter and complete retard

46. TuringTest Group Title

it's okay. so again, what, in all, is r being multiplied by?

47. TuringTest Group Title

c'mon, being hard on yourself never helped anything.

48. Inopeki Group Title

2 and then pi

49. TuringTest Group Title

and...?

50. Inopeki Group Title

height?

51. Inopeki Group Title

wait!

52. TuringTest Group Title

right, it's being multiplied by 2(pi)h so that is what we need to divide by no don;t take it back!

53. Inopeki Group Title

if you divide both sides by h, and both sides by pi you should isolate it so you have 2r, right?

54. TuringTest Group Title

true!

55. Inopeki Group Title

Yeah.. pretty much what you said

56. Inopeki Group Title

But that doesnt mathematically give us anything..

57. Inopeki Group Title

Then you just divide both sides by 2 and get r

58. TuringTest Group Title

sure it does, it gives us the answer not all answers in math are numbers, in fact I hate using numbers if I can help it. about your last post, yes, so what do we get for r=?

59. Inopeki Group Title

What do you mean?

60. TuringTest Group Title

what is on the other side of the equation when we solve for r?

61. TuringTest Group Title

what did we divide both sides by?

62. Inopeki Group Title

2(åi)h?

63. Inopeki Group Title

(pi)*

64. TuringTest Group Title

right. so what will A be divided by?

65. Inopeki Group Title

well we get r on both sides right?

66. Inopeki Group Title

2(pi)h

67. TuringTest Group Title

no, just one and all alone, that's what it means to solve for r again, yes to your last post so write out r=...?

68. Inopeki Group Title

1

69. TuringTest Group Title

where did you get 1 ?

70. Inopeki Group Title

Nevermind, my mind is messing up my mind, if you know what i mean

71. TuringTest Group Title

If you are tired then take a break here is the answer, watch is closely: $A=2\pi rh$divide both sides by$2\pi h$giving$r=\frac{A}{2\pi h}$If you want to keep going then try just something simpler with NO numbers:$ab=c$solve for a

72. Inopeki Group Title

Maybe i should catch some sleep, its 4:50 am here..

73. TuringTest Group Title

that is a very good idea then. sleep on it and try it in the morning. I can't tell you how many problems I've solved just by sleeping on them. Don't worry, step-by-step. You will get there I know. Goodnight!

74. Inopeki Group Title

Good night, thank you for all the help!

75. TuringTest Group Title

anytime :)

76. Inopeki Group Title

A=2πrh divide both sides by 2πh giving r=A2πh If you want to keep going then try just something simpler with NO numbers: ab=c solve for a When you say "solve for a", do you mean i should isolate it? I guess we would divide both sides by b and then c.

77. Inopeki Group Title

Actually, just b.

78. Inopeki Group Title

I would then get a/b=c/b Simplified, a=c/b

79. TuringTest Group Title

^^^correct! now get some sleep dude! the mind needs to be fresh.

80. Inopeki Group Title

I did sleep, that answer was today.

81. TuringTest Group Title

lol I thought you stayed up all night. Told you sleeping on it would help. now solve ab=c for b

82. Inopeki Group Title

Thats easy, both sides divided by a, you get b/a=c/a Simplify: b=c/a

83. Inopeki Group Title

no!

84. TuringTest Group Title

yes^^

85. Inopeki Group Title

actually, yes

86. TuringTest Group Title

that's what I wanted to hear since it's so easy A=2πrh for r

87. Inopeki Group Title

Divide both sides with 2(pi)h, you get A/2(pi)h=2(pi)rh/2(pi)h Simplified: A/2(pi)h=r

88. TuringTest Group Title

awesome! I know you got it but you should write it as A/(2π)h=r parentheses make it clear what is on the bottom.

89. Inopeki Group Title

Yes! Oh ok.

90. TuringTest Group Title

sorry, meant A/(2πh)=r

91. TuringTest Group Title

anyway, slightly different one: solve 5t=t/2+s for s, then for t

92. Inopeki Group Title

So the whole bottom part in parenthesis?

93. TuringTest Group Title

that's two problems^^^

94. Inopeki Group Title

Multiply both sides by 2+s, right?

95. TuringTest Group Title

right, because everything in parentheses is on the bottom if you wrote 1/2x I don't know whether you mean$\frac{1}{2x}$or$\frac{1}{2}x$so parentheses are nice to clear things up think you can do the other? solve $5t=\frac{t}{2}+s$ for s a good first step is to subtract t/2 from both sides. Then s will be isolated, right?

96. Inopeki Group Title

Oh, i thought you meant t/(2+s)

97. TuringTest Group Title

I would have used parentheses if I meant that. if I had meant that then you would have been right, so you'll get one like that soon. but for now solve it as it is for s

98. Inopeki Group Title

Right, so now i have 5t-t/2=s, am i right?

99. TuringTest Group Title

totally, but can you simplify the left a little?

100. Inopeki Group Title

5t-t/(2)?

101. Inopeki Group Title

It cant be 4t/2

102. TuringTest Group Title

so I mean what is 5t-t/2 as a fraction?$5t-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{9t}{2}$do you know how to simplify fractions in this way?

103. Inopeki Group Title

Nope

104. TuringTest Group Title

well did you know that you can only add or subtract the numerators when the denominators are the same?

105. Inopeki Group Title

Yeah

106. Inopeki Group Title

Where did you get the 10?

107. TuringTest Group Title

so we had to get the 5 to have a 2 in the denominator. How can we do this without changing our expression? the trick is to multiply the top and bottom by the same thing, in this case 2/2. Notice that 2/2=1 so it won't change our expression, because multiplying by 1 changes nothing.$5t-\frac{t}{2}=5t*\frac{2}{2}-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{9t}{2}$make sure you understand the above. here's a couple pages on fractions http://www.mathsisfun.com/fractions_addition.html http://www.mathsisfun.com/numbers/fractions-mixed-addition.html

108. Inopeki Group Title

Oh, i knew that, i still dont see where the 10 and 9 come from...

109. Inopeki Group Title

Wait, i see now! 5t *2/2 should become 10/10 though...

110. Inopeki Group Title

No, wait, since the top is the only "t" part that is the only one being multiplied. THEN we take that minus t/2 and that simplifies it to 9t/2

111. TuringTest Group Title

no it doesn't because the original denominator is 1$5t-\frac{t}{2}=\frac{5t}{1}*\frac{2}{2}-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{10t-t}{2}=\frac{9t}{2}$everything is secretly over 1, times 1, and to the power of 1. we just don't write$a=\frac{1a^1}{1}$because it is redundant, but they are the same. You figured it out as I was typing.

112. TuringTest Group Title

now solve it for t

113. Inopeki Group Title

5t=t/2+s?

114. Inopeki Group Title

I have to isolate t by getting away the 5, right?

115. TuringTest Group Title

yeah, solve it for t you can take it from$\frac{9t}{2}=s$there is nothing wrong with starting there if you want. It's up to you

116. Inopeki Group Title

Divide both sides by 5?

117. TuringTest Group Title

my first step if I start from 5t=t/2+s would be to multiply both sides by 2 to get rid of the fraction can you do that?

118. Inopeki Group Title

10t=t+s, i guess?

119. TuringTest Group Title

what about s? didn't it get multiplied by 2 as well?

120. Inopeki Group Title

oh right, 10t=t+2s

121. TuringTest Group Title

so now you should be able to solve for t...

122. Inopeki Group Title

Now divide both sides by 10?

123. Inopeki Group Title

Then we get t=t+2s/(10)

124. TuringTest Group Title

no, because the t on the right would be divided by 10 as well let's see what happens if we do that: t=t/10+s/5 I don't think that helps... any other ideas?

125. Inopeki Group Title

I really dont know...

126. TuringTest Group Title

hint: combine like terms, we want the t's together 10t=t+2s

127. Inopeki Group Title

subtract t from both of them?

128. AravindG Group Title

turing can u hlp me?

129. TuringTest Group Title

@Inopeki yes! @arvind no I have to do 2 weeks of spanish homework in one day, I'm multitasking sorry.

130. Inopeki Group Title

TT, shouldnt you do your homework first?

131. TuringTest Group Title

I am doing it, this doesn't really distract me much, but arvind's problems can be kinda tricky, so that would distract me.

132. Inopeki Group Title

Oh, ok. Now he have 9t=2s

133. TuringTest Group Title

good, now what?

134. Inopeki Group Title

We have no other choice but to divide them by 9. Do we get t=-4.5s then?

135. TuringTest Group Title

how did it change to negative? and you need to leave it as a fraction for now. What did you do to both sides?

136. TuringTest Group Title

you say you divided by 9 but $2/9\neq -4.5$

137. Inopeki Group Title

Yeah, i have no idea what happened there..

138. TuringTest Group Title

so go again... 9t=2s

139. TuringTest Group Title

leave as a fraction

140. Inopeki Group Title

How do i divide 2 by 9?

141. TuringTest Group Title

just like I wrote it: 2/9

142. Inopeki Group Title

wait!

143. Inopeki Group Title

Oh nevermind.. so t=2/9s?

144. TuringTest Group Title

yes, good, but it is better to write t=2s/9 that way I don't think you put the s in the denominator now do the same starting from the equation when it was solved for s: s=9t/2 solve for t

145. Inopeki Group Title

2s=9t?

146. TuringTest Group Title

good, what next?

147. Inopeki Group Title

Divide by 2?

148. TuringTest Group Title

that would solve it for s, right? we wanted to solve for t...

149. Inopeki Group Title

Oh right

150. Inopeki Group Title

Divide by 9

151. TuringTest Group Title

yes, which gives what as the answer??

152. Inopeki Group Title

2s/9=t

153. TuringTest Group Title

which is what we got before, right? so mathematics is consistent, hooray! slightly different one: yz/(z+4)=7 solve for y

154. Inopeki Group Title

I see!

155. Inopeki Group Title

I have to eat, brb

156. TuringTest Group Title

me too, good idea, breakfast!

157. Inopeki Group Title

Hah, i hade dinner.. You there?

158. TuringTest Group Title

yep, where were we? yz/(z+4)=7 solve for y

159. Inopeki Group Title

yz/(z+4)=7 I need to isolate y so maybe i should multiply both sides with z+4. then im left with yz=7z+4

160. TuringTest Group Title

good but you forgot to distribute the 7 yz=7(z+4) do you know what I mean by distribute?

161. Inopeki Group Title

the parenthesis?

162. TuringTest Group Title

yes, what do you get by distributing the 7? remember$a(b+c+d)=ab+ac+ad$you need to multiply each term in the parentheses by what is on the outside

163. Inopeki Group Title

I knew that :D

164. TuringTest Group Title

so what should you have gotten after you multiplication? yz/(z+4)=7

165. Inopeki Group Title

Oh well, yz/(z+4)=7----->yz=7(z+4) ----->y=7(4)? Can i do that?

166. TuringTest Group Title

no you dropped the z take it from yz=7(z+4) and remember to distribute the 7 to all terms in the parentheses

167. Inopeki Group Title

i have 7z and 7*4=yz

168. Inopeki Group Title

7z+7*4=yz

169. TuringTest Group Title

good, now feel free to multiply the 7 and 4...

170. Inopeki Group Title

28

171. TuringTest Group Title

write thew whole expression every time

172. TuringTest Group Title

the*

173. Inopeki Group Title

yz=7z+28?

174. Inopeki Group Title

This must be like teaching physics to a mentally challenged monkey..

175. Inopeki Group Title

Without all the throwing of feces

176. TuringTest Group Title

stupid server is very slow today :( yz=7z+28 what can we do next to solve for y ?

177. Inopeki Group Title

Yeah :/ subtract 1z from both sides, getting y=6z+28

178. TuringTest Group Title

you can't subtract z because it is being multiplied by y what is the inverse of multiplication again?

179. Inopeki Group Title

Division, right. Divide both zs by 2?

180. TuringTest Group Title

will that isolate y ? why divide by 2 ?

181. Inopeki Group Title

I dont really know.. Maybe divide by z?

182. TuringTest Group Title

yes, see you do know!

183. Inopeki Group Title

yz=7z+28----->y1=7z+28?

184. Inopeki Group Title

y*

185. TuringTest Group Title

almost, but you didn't divide the right by z... always gotta do BOTH sides

186. Inopeki Group Title

Ah,, y=7+28=35?

187. TuringTest Group Title

again, almost, but you need to divide each term by z you didn't divide 28 and look now there is no z in our expression! we can't be dropping variables like that. try again and make sure to divide every term yz=7z+28

188. Inopeki Group Title

y=7z+28/z?

189. TuringTest Group Title

what about the 7z now you forgot to divide that term by z... gotta do both

190. Inopeki Group Title

But you said i shouldnt drop that z? y=7+28/z?

191. TuringTest Group Title

you didn't you still have z in the expression, so that is right. good job, but I'm gonna make you do it again for practice please show every step, and try to write it all out in one post. so from the top: yz/(z+4)=7 solve for y please show each step clearly.

192. Inopeki Group Title

yz/(z+4)=7 yz=7(z+4) Here i took multiply both sides by (z+4). yz=7z+28 Here i simplify 7(z+4). y=7+28/z Here i divide both sides by z. y=35/z Here i simplify the right side.

193. TuringTest Group Title

that's good but you can't simplify the right side like that at the end. you also don't need to tell me what each step is, just show. I can see what you are doing. yz/(z+4)=7 yz=7(z+4) yz=7z+28 y=7+28/z ^^^ this is sufficient as far as the simplifications you tried, I will show you a way to change the expression on the right that are valid:$y=7+\frac{28}{z}=\frac{7z}{z}+\frac{28}{z}=\frac{7z+28}{z}$this doesn't really make it more simple though, so there's not much point in it. remember we can only add when the denominators are the same, so here we had to multiply 7 by z/z first. still, good job! another: 4s(1-1/s)=5s+t solve for s (hint: first step is to distribute)

194. Inopeki Group Title

Awesome! Ok i got this..

195. Inopeki Group Title

4s(1-1/s)=5s+t 4(1-1/)=5s+t if i take the s in 4s and put it into the same parenthesis that has 1-1/s the two s'es should "negate" eachother since one is divided by and one is multiplied by, it would be like saying this: 2*4/2. Am i right so far?

196. Inopeki Group Title

Im sorry for the delay, openstudy crashed for me

197. TuringTest Group Title

no, you lost an s we should practice distribution I think 4s(1-1/s)=4s-4s(1/s)=4s-4 one of the s's cancelled because it was multiplied by 1/s the other did not new exercise distribute: 5r(2+1/r+r)

198. Inopeki Group Title

5(2+1/r+r)

199. TuringTest Group Title

where did the r on the outside go? you didn't multiply anything in the parentheses at all :( watch very closely every step: 5r(2+1/r+r)=5r(2)+5r(1/r)+5r(r)=10r+5+5r^2 another distribute 2x(x-1/x)

200. Inopeki Group Title

2x(x)-1/x?

201. TuringTest Group Title

not quite, you keep forgetting to multiply EVERY term. write out the middle step: 2x(x-1/x)=2x(x)-2x(1/x)=??? continue from there

202. Inopeki Group Title

2x(x)(-2x)(1/x)?

203. TuringTest Group Title

no, sorry, I think you need to practice division.. 2x(x-1/x)=2x(x)-2x(1/x)=2x^2-2 what is a(1/a)=?

204. Inopeki Group Title

a? I dont know.. This is just hopeless! Do i really need all this for physics?!

205. Inopeki Group Title

1

206. TuringTest Group Title

there you go anything multiplied by its reciprocal is 1 a(1/a)=1 x(1/x)=1 3(1/3)=1 what about something like a(2/a)=?

207. Inopeki Group Title

2a?

208. TuringTest Group Title

yes, good! here are a few for practice x(5/x)=? 3y(1/y)=? z(2/3z)=?

209. Inopeki Group Title

5x 3y 2/3z

210. TuringTest Group Title

inopeki a(2/a)=2 the a's cancel sorry I couldn't correct myself, but the lag... so now that you know that try x(5/x)=? 3y(1/y)=? z(2/3z)=? again, sorry for the confusion.

211. Inopeki Group Title

5 1 2/3 I dont know.. Do i need to know much more before i cant start with trigo or calc?

212. TuringTest Group Title

The first and last are right, but 3y(1/y)=3 and I hope you don't get discouraged, but you're gonna have to master basically all of algebra, which will take at least a year. the year after you will have to master trigonometry and pre-calculus. You will not be in calculus for some time, so get prepared to settle into algebra for a while...

213. Inopeki Group Title

That sucks... Are you sure that it will take a year?

214. TuringTest Group Title

maybe not if you really delve into it, but if you want to do that you better get an algebra book and read it cover-to-cover. That's what I did over the summer when I was 15. Like I said I knew very little before that and always considered myself bad a math.

215. Inopeki Group Title

What book do you reccommend?

216. Inopeki Group Title

Is all the algebra i need included in khanacademy?

217. TuringTest Group Title

I actually just used the algebra book from my high school, I don't know what it was called. But any decent textbook on algebra should be fine. ask around. I think much algebra is in khan, but I doubt that by itself is enough. You need to practice a LOT of problems to get good. I'm not sure khan has enough.

218. Inopeki Group Title

Khans practice place has unlimited problems, i think. Or did you mean that the subjects there arent enough?

219. TuringTest Group Title

If it has unlimited practice problems than do at least 75 from each section. I don't know khan I don't use it. I still recommend getting a book though. You need good reference material.

220. Inopeki Group Title

I will, we have one universal book for all the math subjects we are supposed to cover that year. Should i just google algebra?

221. Inopeki Group Title

Algebra book*

222. TuringTest Group Title

Maybe, but I think you should post the question on OS 'what is a good algebra book from basics to advanced algebra' I'm sure somebody has an opinion on that here.

223. Inopeki Group Title

Yeah, ill do that.

224. Inopeki Group Title

Where were we, algebra wise?

225. TuringTest Group Title

pretty basic honestly. solving linear equations, inverse operations, distribution, etc. we haven't even done factoring yet, let alone quadratics.

226. Inopeki Group Title

So im pretty much still at square 1?

227. TuringTest Group Title

for the most part, yes. don't get discouraged though, I remind you I refused to look at math until I was 15 there is plenty of time.

228. Inopeki Group Title

But even this is so hard.. How do you expect me to get advanced level algebra when i barely understand linear equations..

229. TuringTest Group Title

You need to learn the basic rules a bit better, then you will see it makes more sense. You are trying to bite off more than you can chew at once. Learn all these rules like your very name: http://www.capitan.k12.nm.us/teachers/shearerk/basic_rules_of_algebra.htm that is the first step. it will take more than 1 day to commit all this to memory, so be patient and practice along the way with khan as you try.

230. Inopeki Group Title

Just the first part? (Algebra properties)

231. TuringTest Group Title

No, everything on the page. Dividing fractions, adding unlike denominators, everything. All the basic rules are there, so one you know them the rest will be much much easier.

232. TuringTest Group Title

once you know them*

233. Inopeki Group Title

Alright.

234. Inopeki Group Title

I think i know all the first ones

235. TuringTest Group Title

I bet you have seen them before, but this time you really need to commit all the rules to memory, not just get the basic idea. Like I said, keep working practice problems in the meantime. When you get stuck, look at the list and try to apply a rule, that way they will have more meaning. Just memorizing the list out of context is pretty difficult and seemingly pointless.

236. Inopeki Group Title

Thanks again for all your help!

237. TuringTest Group Title

you're welcome I'll see you when this site starts working properly again. Keep it up, and good luck!

238. Inopeki Group Title

Thanks :)