## pooja195 one year ago @mathmate

1. pooja195

@mathmate

2. pooja195

@Nnesha its not spam ;)

3. pooja195

@Nnesha its not spam ;)

4. mathmate

Yes!

5. mathmate

6. pooja195

O_o question? XD 2+2=?

7. mathmate

Not allowed to give direct answers. Check the drawing: |dw:1433043776037:dw|

8. pooja195

4 LOL

9. pooja195

@mathmate ;-;

10. mathmate

That means: 1. i dont like this and 2. i dont understand this OR 3. i dont want to do this Which one/ones?

11. pooja195

3 +_+

12. mathmate

gimme a minute!

13. mathmate

Calculate $$\large \frac{1}{2}+\frac{2}{3}$$. Show work.

14. pooja195

7/6

15. mathmate

Calculate 12+23. $$\huge \color{red}{Show work.}$$

16. pooja195

Why are we doing this? .-.

17. mathmate

Because it is the same way you do this as with rational fractions. This is a pretest!

18. pooja195

$12+23=35$

19. pooja195

idk how to show work for that .-.

20. pooja195

2+1=3 3+2=5

21. mathmate

ok, that's how it works: Calculate $$\large \frac{1}{2}+\frac{2}{3}$$, show work.

22. mathmate

$$\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}$$

23. mathmate

Do you know why we multiply the first fraction by three, and the second by two?

24. pooja195

to get a common denominator

25. mathmate

Exactly!

26. mathmate

Just a review of terms: 1/2 = 2/4 because they are __________________

27. pooja195

equivalent

28. mathmate

exactly "equivalent fractions"! That's a good start! You know your stuff.

29. pooja195

:)

30. mathmate

The idea of adding and subtracting rational fraction is exactly the same!

31. mathmate

We multiply top and bottom of each term by a factor so that the denominator becomes the common denominator.

32. mathmate

Since each term remains an equivalent fraction, the answer will not change, BUT...

33. mathmate

we now only have to add the numerators of the equivalent fractions, just like the numerical example.

34. mathmate

$$\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}$$

35. mathmate

Sorry it took a while!

36. pooja195

this looks easy :)

37. mathmate

The next one will be equally easy! gimme a minute.

38. mathmate

Calculate $$\large \frac{1}{x}+\frac{2}{x^2}$$ show work!

39. pooja195

$\huge~\frac{ 1*x }{ x*x }+\frac{ 2 }{ x^2}=\frac{ 1x+2 }{ x^2 }$

40. mathmate

Excellent! Just remember that 1x is usually written as x, except when showing work.

41. mathmate

Now: Calculate $$\large \frac{2}{(x-1)}-\frac{1}{(x-1)^2}$$ show work!

42. pooja195

$\frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }=\frac{ 1(x-1) }{ (x-1) }=1$

43. mathmate

The denominator is still meant to be (x-1)^2 ? You need to do the math on the numerators. Give it another shot!

44. mathmate

T_T not allowed!

45. pooja195

;-;

46. pooja195

i dont get it

47. mathmate

Good! I'll explain!

48. mathmate

$$\large \frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }$$ is perfect!

49. mathmate

At this point, both denominators are identical, so we can do the math just on the numerators $$\large \frac{ 2*(x-1) -1 }{ (x-1)^2 }$$

50. mathmate

following so far?

51. pooja195

yes

52. mathmate

Now can you do the math (on the numerator) and finish?

53. pooja195

$\frac{ x(x-1) }{ (x-1)^2 }$

54. mathmate

You need to distribute the first term, and add like terms, 2*(x-1)-1 = 2x-2 -1 =2x-3

55. mathmate

following?

56. pooja195

Yes

57. mathmate

58. pooja195

$\huge\frac{ x(x-1) }{ 2x-3 }$

59. pooja195

>_<

60. pooja195

wrong spot...

61. pooja195

$\frac{ 2x-3 }{ (x-1)^2 }$

62. mathmate

Whew! Yes, excellent!

63. mathmate

One thing we did without saying it is...nothing. We did "nothing" to factorize the numerator, because it was "obvious" there are no factors.

64. mathmate

In general, after the arithmetic on the numerator, we need to factor the numerator, can you tell me why?

65. pooja195

:/ so its easier to multiply or is it to break down the numbers? :/

66. mathmate

So that we can cancel (with condition) IF there are common factors between the numerator and denominator, just like what we did before in the simplification. Does that make sense?

67. pooja195

yes

68. mathmate

Are there any points to clarify, or are we good?

69. pooja195

We're good.

70. mathmate

ok, now, try this: Calculate $$\large \frac{3}{x}-\frac{2}{x-1}$$ show work!

71. pooja195

ummm O_O

72. mathmate

Look at the denominators.

73. mathmate

there are no common factors! So you just "cross multiply", which is a short cut when there are no common factors. like $$\large \frac{1}{2}+\frac{3}{5}= \frac{1*5}{2*5}+\frac{2*3}{x*5}=\frac{1*5+2*3}{2*5}=\frac{11}{10}$$

74. mathmate

* 2*5 in denominator

75. pooja195

O_o

76. pooja195

i thought u can only do that if there is an equal sighn :/

77. mathmate

It's not a real cross multiplication, but it works in a similar way.

78. mathmate

The pattern helps when there are no common factors. You can use that in the new problem.

79. mathmate

Calculate $$\large \frac{3}{x}-\frac{2}{x-1}$$, show work! Remember that when there is no common factors in the denominators, the common denominator is just the product, namely x(x-1)

80. pooja195

:?

81. mathmate

$$\large \frac{3}{x}-\frac{2}{x-1}=\frac{3(x-1)}{x(x-1)}-\frac{2x}{(x-1)x}=\frac{3(x-1)-2x}{x(x-1)}$$

82. pooja195

hmm ok makes a bit more sense

83. mathmate

Good! you can finish it?

84. pooja195

$\frac{ x-1 }{ x(x-1) }$

85. mathmate

brb

86. mathmate

IF it was x(x-1)/(x-1), you would have simplified it to x, x$$\ne$$1. But 3(x−1)−2x=3x-3 -2x (distribute 3(x-1) to 3x-3 first, then add -2x) =x-3

87. mathmate

With x-3 as numerator, what would the answer be then?

88. pooja195

>_< idk :(

89. mathmate

the common denominator does not change, so still x(x-1) Answer would then be: $$\huge \frac{ x-3 }{ x(x-1) }$$ nothing to further simplify!

90. mathmate

Here's one for you to think about when I'm off for a few minutes. Calculate $$\large \frac{2}{(x+1)}+\frac{1}{(x-1)}$$ , show work! Do remember to work out the math in the numerator, and the common denominator (product of x+1 and x-1) does not change.

91. mathmate

@pooja195

92. pooja195

$\frac{ (x-1)*2 }{ (x-1)*(x+1) }+\frac{ (x+1)*1 }{ (x+1)*x-1) }=\frac{ 3(x+1)(x-1) }{ (x-1)(x+1) }$

93. mathmate

The first part is very good!

94. mathmate

Recall that since the denominator is the same, and does not change, we need to ADD the numerators (and not multiply).

95. mathmate

So it would read: $$\LARGE \frac{ (x-1)*2 + (x+1)*1}{ (x-1)(x+1) }$$ $$=\LARGE \frac{ 2X-2 + X+1}{ (x-1)(x+1) }$$ $$=\LARGE \frac{ 3X-1 }{ (x-1)(x+1) }$$, OR $$=\LARGE \frac{ 3X-1 }{ x^2-1 }$$

96. pooja195

O_O

97. mathmate

Your steps are correct. It's just the last step of expanding and adding that there was a correction to make. Are you ok with the correction, or confused?

98. pooja195

x_x

99. mathmate

Well, if it's the part of expansion perhaps you could use some help. (x-1)*2 + (x+1)*1 is the same as 2(x-1) + (x+1) .... a coefficient of 1 is understood. Then we distribute, basically a half FOIL 2*x + 2*(-1) + x+1 ..... parentheses after a plus sign can be removed without change that makes 2x -2 +x +1 Now add like terms 2x+x -2+1 =3x-1 this is the numerator, since the denominator does not change, we have as the answer: =$$\Large \frac{ 3X-1 }{ x^2-1 }$$

100. pooja195

seems easier... :)

101. mathmate

Good! Would you like to do one while I'm offline, probably in less than 5 minutes?

102. pooja195

Sure i gotta go too xD

103. mathmate

$$\large \frac{3}{x-1}-\frac{2x}{x^2-1}$$

104. pooja195

huh? O_o

105. pooja195
106. mathmate

Simplify $$\large \frac{3}{x-1}-\frac{2x}{x^2-1}$$. Show work!

107. pooja195

im not sure how to do this .-.

108. mathmate

I would factor the second denominator, x^2-1 into (x+1)(x-1) using difference of 2 squares (note 1=1^2). This gives $$\large \frac{3}{x-1}-\frac{2x}{(x+1)(x-1)}$$ the proceed as before.

109. pooja195

$\frac{3*x+1 }{ x-1*x+1 }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 6x(x+1) }{ (x^2-1 }$

110. mathmate

A minor correction by adding parentheses: (using PEMBAS) $$\Large\frac{3*(x+1) }{ (x-1)*(x+1) }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 3x+3-2x }{ (x^2-1) }=?$$

111. mathmate

The first step was perfect. Then you would distribute(expand) and add/subtract as needed. The final step I leave to you, if possible.

112. pooja195

Ew factoring

113. pooja195

no wait

114. mathmate

in this case, there is nothing to factor! 3x-3-2x = 3x-2x-3=x-3 so answer is $$\Large \frac{x-3}{x^2-1}$$

115. pooja195

i was gonna say that XD

116. mathmate

of course! I knew that! xD

117. pooja195

lol u are learning all mai tricks >.>

118. mathmate

Well, this is either called learning, or contamination, depending on the point of view! xD

119. pooja195

contamination >.> because now i cant use it :P

120. mathmate

Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam! Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!

121. pooja195

xD

122. mathmate

Just another contamination!

123. pooja195

LOL

124. mathmate

Ok, would you feel sad if I say...

125. pooja195

Are we done for the day? :o

126. pooja195

Not at all

127. mathmate

nope, but we're done with 11.2 to 11.6 ?

128. pooja195

We are done? :o YAY NO MORE MATH

129. mathmate

130. mathmate

So now we're onto Rational equations! Interesting stuff!

131. pooja195

T_T

132. mathmate

Solve for x: $$\large x+1=\frac{72}{x}$$

133. pooja195

i dont think i have learned this .-.

134. mathmate

If not, now you will!

135. pooja195

O_O

136. pooja195

wait im starting to remeber things >.>

137. mathmate

Any idea where to start?

138. pooja195

$\huge~\frac{ x+1 }{ 1}=\frac{ 72 }{ x }$

139. mathmate

Excellent! Memories are coming back!

140. mathmate

next?

141. pooja195

$\huge~72*1=x(x+1)$ $\huge~72=x^2+1x$

142. mathmate

Excellent again! next?

143. pooja195

$\huge~x^2+1x-72=0$

144. mathmate

Beautiful! Pray continue!

145. pooja195

|dw:1433116854133:dw| $\huge~(x+9)(x-8)$

146. mathmate

You're almost done. Equate each factor to zero to find the answers.

147. pooja195

x=-9 x=8

148. mathmate

100% Excellent job!

149. pooja195

I think i have this section lets not do anymore :D ?

150. mathmate

Just two more! Solve for x: $$\large \frac{10}{x+4}=\frac{15}{4(x+1)}$$ This is easier than FOIL.

151. pooja195

NO T_T

152. mathmate

There is no factoring involved, you'd do it in 10 seconds, + 30 minutes of typing.

153. pooja195

$\huge~40(x+1)=15(x+4)$

154. mathmate

Too fast.... but everything is right!

155. pooja195

40x+40=15x+60 40x=15x+20 25x=20 x=20/25

156. mathmate

157. pooja195

x=20/25

158. mathmate

... or 4/5

159. pooja195

OR 20/25.

160. mathmate

I think most teachers would tolerate this (not simplifying), although I don't. I would give 95% instead of 100%. So it depends on your teacher!

161. pooja195

smh

162. mathmate

You're in trouble when I learn more of these! Well, you need to give extra credit to those who do a perfect job, that's my reasoning!

163. pooja195

xD

164. mathmate

Now the last one, you'd be sad when it's done!

165. pooja195

WAIT

166. pooja195

After this no more math??? :O

167. pooja195

T_T <---tears of joy

168. mathmate

Yeah, after this last problem, no more math (until your quiz) unless you have questions.

169. pooja195

i have 2 questions on my hw thing i dont get at all : /

170. mathmate

We'll look at that after this question. Perhaps by that time you will know how to do. Solve $$\large 3x=\frac{5x+6}{2x+3}$$. Show work.

171. pooja195

$\huge~\frac{ 3x }{ 1 }=\frac{ 5x+6 }{ 2x+3 }$ $\huge3x(2x+3)=1(5x+6)$

172. mathmate

Excellent, so far so good. You're living up to your reputations!

173. pooja195

$6x^2+9x=5x+6$ $6x^2+4x-6=0$ |dw:1433118120725:dw|

174. mathmate

So far so good! Pray continue!

175. mathmate

Actually, to make you life simpler, divide the equation by 2 to get 3x^2+2x-3=0 This will give smaller numbers, so easier to work with.

176. mathmate

|dw:1433118297396:dw|

177. pooja195

None.

178. mathmate

Excellent. What's the next step?

179. pooja195

idk

180. mathmate

181. pooja195

OMG NO T_T

182. pooja195

So much work

183. pooja195

lets just skip this one ;-; lets do one that doesnt have this

184. mathmate

The reason I choose this one is because in the real world (including exams), answers don't always come out in integers.

185. mathmate

Just to make sure you're prepared for the quadratic formula. You could type the discriminant faster than lightning, why not the quad. formula?

186. pooja195

What would the equation be?

187. pooja195

:/

188. mathmate

$$6x^2+4x-6=0$$, or half of that: 3x^2+2x-3 Either one will do, doesn't make a differce with the quad. formula. Makes a big diff. with factoring.

189. pooja195

a=3 b=2 c=-3 $\huge~x=\frac{ -2 \pm \sqrt{(-2)^2-4(3)(-3)} }{ 2(3) }$

190. mathmate

Correct! Now pull out your calculator, or do it mentally, to find the final solution.

191. pooja195

$\huge~x=\frac{ -2\pm~\sqrt{40} }{ 6 }$

192. pooja195

$\huge~x=\frac{ -2 \pm \sqrt{4\times10} }{ 6 }$

193. pooja195

$\huge~x=\frac{ -2 \pm 2\sqrt{10} }{ 6 }$

194. mathmate

Yes, yours is correct (again 95%), unless ... $$\Large~x=\frac{ -2 \pm \sqrt{40} }{ 2(3) }$$ $$\Large~x=\frac{ -1 \pm \sqrt{10} }{ 3 }$$

195. pooja195

T_T

196. mathmate

lol, don't worry about the 5%, but I give you credit for coming that close, especially the part going from sqrt(40) to sqrt(10), excellent job!

197. pooja195

:)

198. mathmate

Well, in case you have nightmares with + and - signs poking at you, you can wake up and try this. It's not as hard as it seems. $$\Large \frac{3c}{c^2-4}+\frac{1}{c-2}=\frac{2}{c+2}$$ You will need to simplify the left-hand side first, then do the equation solving.

199. pooja195

O_O where did u get that?

200. mathmate

Does it look too difficult? BTW, the answer is -3, so you don't have to cheat! lol

201. pooja195

noo it looked fimiliar

202. mathmate

Did you get it from your teacher? Well, I'll show you a gold mine. If you can't sleep, go to http://www.regentsprep.org/regents/math/algtrig/ate11/RationalEqPract.htm and practise all those problems!

203. pooja195

omg no way T_T

204. mathmate

With this one, do the simplification on the left-hand side and then solve. Do not follow the solution provided.

205. pooja195

hmmmmk

206. mathmate

You said you had two hw questions!

207. pooja195

nvm :)

208. pooja195

Can we practice more on everything once yr back? :)

209. mathmate

sure!

210. pooja195

211. pooja195

@mathmate

212. mathmate

213. mathmate

or you have specific questions?

214. pooja195

No lets just go with 11.2 etc

215. mathmate

|dw:1433125010350:dw| name the variation.

216. pooja195

inverse variation

217. mathmate

Good! x 1 2 3 4 5 y 2 4 6 8 10 what is x when y=14

218. pooja195

7? :/

219. mathmate

Yep, good!

220. mathmate

x 1 5 10 y 20 4 2 What is y when x=40

221. pooja195

80

222. mathmate

Hmmm....

223. pooja195

10?

224. mathmate

We see that y is decreasing as x increases. See if you can find the rule!

225. pooja195

0

226. mathmate

no, that's too small. The rule for direct variation is y=kx, where k is a constant. In the first example, x 1 2 3 4 5 y 2 4 6 8 10 we see that the rule is y=2x, so it is a direct variation.

227. mathmate

The rule for inverse variation is y=k/x. For the second example, we have x 1 5 10 y 20 4 2 so the rule is y=20/x (20=20/1, 4=20/5, 2=20/10, ...) so what is y when x=40?

228. pooja195

2

229. mathmate

Another way to write the rule is xy=k For example 2, k=20, so 1*20=20 4*5=20...

230. mathmate

According to the rule, y=20/40 = 1/2, so when x=40, y=1/2 or use the alternate rule, xy=20 40y=20, or y=1/2

231. mathmate

x 2 7 12 y 6 21 36 Name the variation, and find x when y=120

232. pooja195

i dont like there >.>

233. pooja195

these

234. mathmate

You've done those before?

235. mathmate

I know we did not go over it, but never too late! :(

236. mathmate

I'll show a few examples first. Direct variation: rule y=kx (when x is greater, y is greater) Inverse variation: rule xy=k (when x increases, y decreases, that's why inverse) so far so good?

237. mathmate

|dw:1433125957505:dw|

238. mathmate

Notice that inverse variation is not a straight line. It is a curve.

239. mathmate

ok so far?

240. pooja195

Yea

241. mathmate

Ok, now, to find the rule, this is what you do.

242. mathmate

First find if the function is increasing, or decreasing. ok

243. mathmate

do you know how to tell if a function is increasing?

244. pooja195

If the line is going up?

245. pooja195

|dw:1433126223857:dw|

246. mathmate

Yes. An excellent example. If it is a straight line, we know it is not inverse.

247. pooja195

Right :)

248. mathmate

Is it automatically direct variation?

249. pooja195

yes? :/

250. mathmate

Unfortunately, there are other variations. The one you drew is called a partial variation. It is a straight line that does not pass through the origin. A straight line that passes through the origin AND increasing, then it's direct!

251. pooja195

Lets skip this.

252. mathmate

|dw:1433126603203:dw|

253. mathmate

It has to do with the rule. A partial is y=mx+b where b$$\ne$$0 A direct is when y=kx (i.e. b=0) For example, x 2 5 8 y 6 12 18 is partial because y=2x+2 while x 2 5 8 y 4 10 16 is direct because y=2x

254. pooja195

hmmmm makes a bit more sense :3

255. mathmate

Good! We'll get back to the last problem. x 2 7 12 y 6 21 36 Name the variation, state the rule, and find x when y=120

256. pooja195

T_T 120=k/x

257. mathmate

Can you start with the variation? (is it increasing? If it is, it is not inverse)

258. pooja195

increasing .-.

259. mathmate

so it cannot be inverse. The rule you gave is for inverse.

260. pooja195

so y=mx+b?

261. mathmate

When you give a rule, you have to check that it works for at least two points.

262. pooja195

mm ;-; <---

263. mathmate

and you need to find k. Let's say we decide that it is direct variation, then y=kx substitute (2,6) we have 6=2k, or k=3 Let's try it for another point: (12,36), so that is 36=12k, k=3 again. So we have found the rule y=3x. If y=120, then 120=3x, what is x?

264. pooja195

40

265. mathmate

Exactly. So the rule is: 1. examine and assume a distribution. 2. substitute one of the points in the rule (of the assumed distribution) 3. check with another point, to see if the rule is still valid. 4. Give the answer.

266. pooja195

mmm k

267. mathmate

Now, try x 2 4 6 y 6 3 2 State 1. the variation 2. the rule 3. value of x when y=24

268. pooja195

1) PArtial 2) y=mx+b 3) Lets skip it ;-;

269. mathmate

Well, it's decreasing

270. mathmate

so it is not direct. Is it a straight line, let's check the slope.

271. mathmate

x 2 4 6 y 6 3 2 slope = (y2-y1)/(x2-x1) between 2 and 4, slope = (3-6)/(4-2) = -3/2 = -1.5 between 4 and 6, slope = (2-3)/(6-4) = 1/2 Since the slope is not constant, it is not a straight line.

272. mathmate

So it's not partial. Try inverse: x 2 4 6 y 6 3 2 2*6=12 4*3=12 6*2=12 so k=12 for inverse!

273. mathmate

274. pooja195

y=12x :/

275. mathmate

Variation: inverse Rule: either xy=12, or y=12/x Now when y=24, what would x be?

276. pooja195

2

277. mathmate

Hin: Rule: either xy=12, or y=12/x

278. pooja195

weve spent about an hour on this .-. y=2? ??

279. mathmate

xy=12, y=24 so x=12/y=12/24=1/2 lol Yes, almost an hour.

280. mathmate

and it was a good thing, because now you won't miss the easy questions.

281. pooja195

@mathmate

282. mathmate

k, where do we go from here? (it's laggy)

283. mathmate

Want to start a new post?

284. pooja195

ok