## pooja195 one year ago @mathmate

1. pooja195

@mathmate

2. mathmate

Finally, got here!

3. mathmate

4. pooja195

ok :)

5. mathmate

Solve $$x^2 -6x +9=0$$

6. pooja195

a=1 b=-6 c=9

7. mathmate

ok.

8. pooja195

$\huge~x=\frac{ -(-6)\pm \sqrt{(-6)^2-4(1)(9)} }{ 2(1) }$

9. mathmate

Great, now can you simplify that, because it will be greatly simplified when you put the numbers in there.

10. pooja195

$\huge~x=\frac{ 6\pm \sqrt{0} }{ 2}$

11. pooja195

$\huge~\frac{ 6 }{ 2 }=3~~~~~x=3$

12. mathmate

Yes, that is the same as {3,3}, or 3 (twice)

13. mathmate

It's called a double root, andthe graph looks like this:|dw:1433211976391:dw| the vertex just touches the x-axis.

14. mathmate

ok so far?

15. pooja195

ye

16. mathmate

If you factored, you would have got (x-3)^2=0, which is the same as saying (x-3)=0, or x=3 (twice), or x={3,3}

17. mathmate

Was wondering you still have the online textbook?

18. mathmate

Can you dig some questions out of there, which will be more like what you have learned.

19. pooja195

T_T but there are answers ,_,

20. mathmate

You don't have to look at them, in any case, we are looking at the work, can check the answers at the end.

21. mathmate

There may be word problems, which take more time to make. Using ready-made problems can help us do more.

22. pooja195

23. mathmate

Have you done anything like this? "A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?"

24. pooja195

:/

25. pooja195

no O_O

26. mathmate

Always the numerical equations, and nothing like word problems?

27. pooja195

no ._.

28. mathmate

ok, so we can skip that.

29. pooja195

:)

30. mathmate

Whenever you solve quadratics, there are three ways to do it. If the teacher does not specify, you can use any method you want. If he/she specifies completing the square, you have no choice. the three are:

31. mathmate

quad formula (just did) factoring (did over the weekend) completing the square Do youknow how to do the last one?

32. pooja195

no :/ we havent learned it

33. mathmate

ok, so we skip that too. Anything else you'd like to do on quadratic equations/

34. pooja195

Nope.

35. mathmate

So we will move onto system of linear equations. Have you done word problems?

36. pooja195

yes but dont like those :/

37. mathmate

How many ways do you know to solve system of linear equations?

38. pooja195

Elmination Substituion :/

39. mathmate

comparison?

40. pooja195

eh?

41. mathmate

When you have y=2x+3 y=4x-2 then you equate the right-hand sides (since they both equal y) to give 4x-2=2x+3 and solve for x. It is a form of elimination, but makes life easier. (called method of comparison)

42. pooja195

i know that :P

43. mathmate

Ok, so let go: Solve y = 36 – 9x 3x + y/3 = 12

44. mathmate

*let's

45. pooja195

3x+(36-9x)/3=12 6x+36/3=12 6x+12=12 x=0

46. mathmate

...not finished! need y, and the final answer!

47. pooja195

y=36-9(0) y=36-0 y=36? (0,36)

48. mathmate

Excellent!

49. mathmate

Solve the following system by substitution. 2x – 3y = –2 4x + y = 24 When you see a singleton y, or singleton x, then you can try substitution, as in the above case.

50. pooja195

y=-4x+24 2x-3y=-2 2x-3(-4x+24)=-2 2x+12x-72=-2 14x-72=-2 14x=70 x=5 y=-4(5)+24 y=20+24 y=44 Final answer (5,44)

51. mathmate

Something wrong here, otherwise all correct. y=-4(5)+24 y=20+24 y=44

52. pooja195

y=4

53. mathmate

yep! Solve the following system using addition (a kind of elimination) 2x + y = 9 3x – y = 16

54. pooja195

x=5 2(5)+y=9 10+y=9 y=-1 (5,-1)

55. mathmate

That was fast! You solved for x by inspection! But show work in exams!

56. mathmate

The admission fee at a small fair is $1.50 for children and$4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? 57. pooja195 O_O 58. pooja195 Nooooooooooooooooo 59. mathmate With word problems, always define variables first, so you don't get confused at the end. 60. mathmate I'll do this one, ok? 61. mathmate Define variables: Number of children = c Number of adults = a (we'll could get confused using x,y, but never this way) 62. mathmate Admission for children =$1.50 Admission for adults = $4 63. mathmate "2200 people enter the fair and$5050 is collected" means c+a = 2200 1.50c + 4a=5050 = 15c+40a = 50500 (keep integers as long as possible Substitute (because c has a coefficient of 1) c=2200-a so 15(2200-a) + 40a = 50500 expand 33000 -15a + 40a = 50500

64. mathmate

40a-15a = 50500-33000 25a = 17500 a=17500/25 = 700 c=2200-700=1500 Check 1.5c+4a=2250+2800=5050 good!

65. mathmate

You will find that once the variables are defined, everything else look easy. What do you think?

66. pooja195

It seems so complicated >_<

67. pooja195

*-*

68. mathmate

Do you want to tell me which part is harder to understand?

69. mathmate

It looks complicated probably because I show all the work. If I skipped or jumped some steps, it will look easy, but a little harder to follow.

70. pooja195

hmm.... :/

71. mathmate

Now your turn! The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number

72. mathmate

First define variables, use meaningful names, like t = tens digit u = units digit. so far so good?

73. mathmate

When you have a number like 25, then t=2, u=5, and 10t+u = 25.

74. mathmate

"The sum of the digits of a two-digit number is 7" can you form an equation with that?

75. pooja195

7u+t=27?

76. mathmate

Here, we are saying the equivalent of 2+5=7, so t+u=7 will do. now "When the digits are reversed, the number is increased by 27. "

77. pooja195

t+u=27?

78. mathmate

10t+u is the original number 10u+t is the number reversed (units digit becomes 10's digit) so 10u+t = 10t+u + 27 (increased by 27)

79. mathmate

Simplifying this equation, we get 9u-9t=27 or u-t=3 combined with u+t=7 we solve for t=2, u=5, or the original number is 25.

80. mathmate

Check: 52-25=27, ok.

81. pooja195

:/

82. mathmate

@pooja195

83. pooja195

.

84. mathmate

I'm typing up an example.

85. mathmate

$$\large \frac{5}{x^2-1}+\frac{4}{x^2+2x+1}$$

86. mathmate

So first factorize the denominators:

87. mathmate

$$\frac{5}{(x+1)(x-1)}+\frac{4}{(x+1)^2}$$

88. mathmate

Can you find the common denominator? between (x+1)(x-1) and (x+1)^2

89. pooja195

it would be (x-1)

90. pooja195

and then x+1 on the other side?

91. mathmate

"common" denominator of two expressions must contain ALL the factors, and is the same for all the terms. So start with one of the two , say (x+1)(x-1) and add factors to it so that it contains every expression.

92. mathmate

Say we start with (x+1)(x-1), surely it contains (x+1)(x-1) , which is the first expression.

93. mathmate

Question is, does it contain the second expression (x+1)^2?

94. mathmate

the answer is no. What do we need to do? Multiply by another (x+1), so we end up with a common denominator of (x+1)(x-1)(x+1).

95. mathmate

So going back to the question: $$\Large \frac{5\color{red}{(x+1)}}{(x+1)(x-1)\color{red}{(x+1)}}+\frac{4\color{blue}{(x-1)}}{(x+1)^2\color{blue}{(x-1)}}$$ notice that now both denominators are identical.

96. mathmate

$$\Large \frac{5\color{red}{(x+1)}}{(x+1)^2(x-1)}+\frac{4\color{blue}{(x-1)}}{(x+1)^2(x-1)}$$

97. mathmate

and that's the same as $$\Large \frac{5\color{red}{(x+1)}+4\color{blue}{(x-1)}}{(x+1)^2(x-1)}$$

98. mathmate

Next step is to expand and simplify the numerator: $$\Large \frac{5x+5+4x-4}{(x+1)^2(x-1)}=\frac{9x+1}{(x+1)^2(x-1)}$$ and that's the end of the calculations.

99. pooja195

I read these but only half .-. @mathmate

100. mathmate

Did you understand as far as you read?

101. pooja195

yes but now we can start studying for the finals :P >_<

102. mathmate

Here's the diagnostic test, as far as I have done. I will add other things on.

103. mathmate

@pooja195 you there?

104. pooja195

i iz here

105. mathmate

ok, let's get started. Any questions on chapter 1?

106. mathmate

Say, 4^d=64, what is "d".

107. mathmate

need help?

108. pooja195

yeah :/

109. mathmate

4^0=?

110. pooja195

1

111. mathmate

Very good! 4^1=

112. pooja195

4

113. mathmate

Just two rules: Anything raised to the power of 0 is 1, anything raised to the power of 1 is the base itself, like 4^1=4

114. mathmate

4^2=

115. pooja195

16

116. mathmate

Remember, $$4^2=4.4$$, $$4^3=4.4.4$$, $$4^4=4.4.4.4$$, etc.

117. mathmate

so 4^3=?

118. pooja195

64

119. pooja195

so 3?

120. mathmate

compare 4^3=64, 4^d=64 so what is d? yes, d=3

121. mathmate

Great! Now try 8=2^z. what is z?

122. pooja195

2*2*2 2^3

123. mathmate

and z=?

124. pooja195

3

125. mathmate

Good! Now a difficult one: $$-3^2=$$?

126. pooja195

-9? or if its (-3)^2 it would be 9

127. mathmate

Exactly, that's where many students make mistakes. when in doubt, use PEMDAS which says that exponentiation goes before subtraction (negative), so what you did was correct. (-3)^2 would raise the power of -3, so (-3)*(-3)=+9. Well done!

128. mathmate

and -3^2=-9.

129. mathmate

Try the following: 1^3= 3=3^j, j= 1=4^h, h= 3^r=81, r= (-2)^3=

130. mathmate

Laws of exponent: 2^n = 2.2.2.....2 (n times) $$a^m \times a^n = a^{m+n}$$ $$\large a^{-m}=\frac{1}{a^m}$$ $$\large a^{\frac{1}{2}}=\sqrt a$$ $$\Large a^{-\frac{1}{2}}=\frac{1}{\sqrt a}$$

131. mathmate

Now practice: $$\Large (\frac{1}{2})^2=$$ $$\Large (0.2)^3=(\frac{2}{10})^3=(\frac{1}{5})^3=\frac{1}{125}$$ Express $$\large 3^{-8}$$ with a positive exponent. $$\large 3^{-8}$$ =

132. pooja195

O_O

133. mathmate

|dw:1433336557931:dw|

134. pooja195

What is that??

135. mathmate

Helps you see where numbers belong. R includes all numbers you're studying (i.e. does not include complex numbers) Inside of R is divided into Q (rational) and Q' (irrationals) Q' (irrationals) include numbers like sqrt(2), $$\pi$$, sin(10$$^\circ$$), etc. Q (rationals) include $$all$$ integers (Z), and all naturals (N). Decimals (2.5) repeating decimals (2.33333...), fractions (7/3) all belong to rationals, but not Z and N. All N belong to Z, but negative integers are in Z but not in N. See if you can read these relations from the diagram.

136. mathmate

@pooja195

137. pooja195

:)

138. mathmate

Solve x+4<7 (page 324)

139. pooja195

x<3

140. mathmate

ok... too easy for you?

141. pooja195

Yes xD

142. mathmate

Write an inequality to show the max. no. of apples I can buy if I have $4.8 and apples cost$0.60 each.

143. mathmate

* possible number of apples

144. pooja195

O_O i dunno this...

145. pooja195

x+0.60<4.8?

146. mathmate

I have $4.80, and apples cost 0.60 each. So the maximum number of apples I can buy is$4.80/0.60 =?

147. pooja195

8

148. mathmate

Good, and the minimum?

149. pooja195

would i subtract?

150. mathmate

Well, the minimum would be zero. I don't have to buy any. So the answer would be:

151. pooja195

x=8?

152. pooja195

x<8

153. mathmate

That's the maximum number of apples. If we ask for the possible number, then it would be:

154. pooja195

4? o.L

155. mathmate

I can buy anything from 0 to 8, so $$\large 0\le x \le 8$$, where x is the number of apples.

156. pooja195

ooo >_<

157. mathmate

Does that make sense?

158. pooja195

Yea

159. mathmate

Word problems!

160. mathmate

Try 4x-3=21

161. mathmate

solve.

162. pooja195

x=6

163. mathmate

Excellent! now solve -5x+10=30

164. pooja195

x=-4

165. mathmate

Good!

166. mathmate

now solve $$-5x+10\le30$$

167. pooja195

$\huge~x \ge-4$

168. mathmate

Very good!!! you actually know the rules!

169. mathmate

Why is it $$\ge$$ and not $$\le$$ ?

170. pooja195

because when you divide by a negative you always need to flip the signs

171. mathmate

Excellent! What if you multiply by a negative?

172. pooja195

Keep the sign as it is DO NOT change it :)

173. mathmate

Is x/(-1) equal to x*(-1)? why would the rule be different?

174. pooja195

idk its just like that .-.

175. mathmate

Actually, whether you multiply or divide by a negative, you flip the direction of the > or < sign. You can remember that precisely because x*(-1) is the same as x/(-1). So can you now solve

176. mathmate

$$\Large \frac{5}{3}(x-5)=6$$

177. pooja195

$\frac{ 5 }{ 3 }x-\frac{ 25 }{ 3 }=6$

178. mathmate

There is a trick to solving equations with fractional coefficients!

179. mathmate

I would multiply by the denominator, or LCM of the denominator if there are many.

180. mathmate

This way, we end up with an equation with integer coefficients.

181. mathmate

$$\Large \frac{5}{3}(x-5)=6$$ $$\Large 3*\frac{5}{3}(x-5)=3*6$$ $$\Large 5(x-5)=18$$ and then go from here

182. pooja195

$\huge~5x-25=18$ $\huge~5x=43$ $\huge~x=\frac{ 43}{ 5 }$

183. mathmate

Exactly!

184. mathmate

Any questions before we move on to linear equations?

185. mathmate

...and graphing?

186. pooja195

no, but lets skip graphing :/

187. mathmate

You're willing to forfeit the points?

188. pooja195

nvm -_-

189. mathmate

What is a linear equation in slope-intercept form?

190. pooja195

y=mx+b

191. mathmate

Good! Can you find a line with a slope of 2 that passes through (2,4)?

192. pooja195

Point slope formula $\huge~y-y1=m(x-x1)$ $\huge~y-4=2(x-2)$ $\huge~y-4=2x-4$ $\huge~y=2x$

193. mathmate

Excellent. What form of equation did you use?

194. mathmate

I call it point-slope form. Not many school teach this form.

195. mathmate

*schools

196. pooja195

i wrote it on top xD

197. mathmate

Sorry, I didn't see it!

198. pooja195

itz okz :-)

199. mathmate

Now how about a line that passes through 2 points. Can you find it?

200. mathmate

The points are (2,5), (7,15)

201. pooja195

Find slope first $\huge~\frac{ y2-y1 }{ x2-x1 }$ $\huge~\frac{ 15-5 }{ 7-2}=\frac{ 10 }{ 5 }=5$ Point slope formula $\huge~y-y1=m(x-x1)$ Pick a point right? $\huge~y-5=5(x-2)$ $\huge~y-5=5x-10$ $\huge~y=5x-5$

202. mathmate

you got 90% for this one!

203. pooja195

-_-

204. mathmate

Spot the error!

205. pooja195

i dont see it .-.

206. pooja195

Wait i simplified

207. mathmate

All the steps are perfect, but 10/5=2 !!!!

208. pooja195

>_<

209. pooja195

Are you going to make me redo all the work????? ;-;

210. mathmate

That's no problem. That's how I would correct, more for steps.

211. mathmate

No!

212. pooja195

:)

213. mathmate

If your steps are good, that's what's important.

214. mathmate

But do be careful in tests and exams.

215. pooja195

ok :)

216. mathmate

Translate into a mathematical inequality: Ken's age (k) is at least four years older than Pete's (p).

217. pooja195

$\huge~K \ge 4+p$

218. pooja195

:/

219. mathmate

Great! It wasn't easy for you, but you did it!

220. mathmate

Now try: Liz (L) has at least 4 times less flowers than Priya (P).

221. mathmate

At least 4 times could mean 5 times!

222. mathmate

* could be

223. pooja195

Idk know this one .-.

224. mathmate

When in doubt, put numbers in. Say Liz has $10, how much should Priya have? 225. mathmate$40, $50,$60, anything $40 or more, right? 226. mathmate Sorry, change flowers to$...:(

227. mathmate

So if L=10, P=40, so we write $$\Large 4L \le P$$

228. pooja195

i knew it!

229. mathmate

That's what you would have done?

230. pooja195

i was thinking something similiar

231. pooja195

4p

232. mathmate

Good. Be careful, many students would write L$$\le$$ 4P.

233. pooja195

XD

234. mathmate

so using numbers, you can check your work.

235. pooja195

Right

236. mathmate

How are we doing so far?

237. pooja195

Great :)

238. mathmate

Pace? 1=too slow, 10=too fast.

239. pooja195

5

240. mathmate

ok, we're on the right track.

241. mathmate

Gimme a minute, I have to flip the pages. You can go get a glass of water, you talked a lot! lol

242. pooja195

XD

243. mathmate

244. pooja195

yes

245. mathmate

Have you done interval notation?

246. pooja195

no

247. mathmate

Like (5,8(, or ]5, 8]

248. pooja195

nope

249. mathmate

They are usually used to solve compound inequalities.

250. mathmate

Since you haven't done it, we'll describe by words.

251. pooja195

You mean and and or ?

252. pooja195

$x+3<5~~or~~~5<x+3$

253. pooja195

like that?

254. mathmate

Yes.

255. mathmate

Can you try to solve x<-5 or x>-4

256. pooja195

257. mathmate

Excellent!

258. mathmate

Change to and: Can you try to solve x<-5 and x>-4

259. pooja195

still the same :/ ?

260. mathmate

No, I'll draw the number line.

261. mathmate

|dw:1433387503479:dw|

262. mathmate

Solution for or is shown in the graph. But for and, there is no number that satisfies the and condition. No number can be less than -5 and greater than -4 at the same time, so answer is "no solution".

263. mathmate

How about solving: x>4 and x<10 Use the number line if necessary

264. mathmate

Feel free to show intermediate steps progressively. You don't have to wait till you typed up everything.

265. pooja195

|dw:1433387789387:dw|

266. mathmate

The answer is correct graphically. How would you write it mathematically?

267. pooja195

x>4 and x<10

268. mathmate

You can write it as $$4\lt x\lt10$$ whe it is a range

269. mathmate

But be careful about the difference between $$\le$$ and $$\lt$$ etc

270. mathmate

Moving onto 6.6 unless you have questions.

271. mathmate

Solve |x|=2

272. pooja195

+2 -2

273. mathmate

Very good. I see that your book does not write the answer in set notation, for example: x={-2,2}

274. mathmate

275. pooja195

Nope

276. mathmate

ok, no problem. Solve |x|<2

277. pooja195

x<2 and x>−2

278. mathmate

Again, these two "and" inequalities make a range between -2 and +2 so we write -2<x<2. ok?

279. pooja195

ok

280. mathmate

solve |x-3| =5

281. pooja195

x−3=5 x=8 x−3=−5 x=-2

282. mathmate

So x=8 or -2 Good, you know your stuff! This is encouraging! what do you think?

283. pooja195

Its easy :)

284. mathmate

Even better!

285. mathmate

now solve |2x-7| -3=6

286. pooja195

add 3 to both sides 2x-7=9 x=8 2x-7=-9 x=-1

287. mathmate

Very good! It's still easy?

288. pooja195

yes

289. mathmate

We move onto abs. inequalities.

290. mathmate

solve |4-3x| -2 $$\le$$ 7

291. pooja195

Do you want me to show the work?

292. mathmate

Not in this case, will save some time.

293. pooja195

$\huge~x \ge \frac{ -5 }{ 3} ~~~and ~~~x \le \frac{ 13 }{ 3 }$

294. mathmate

Can you check if it is x>=13/3

295. pooja195

−3x+4−4≥−9−4 I subtracted 4 from both sides −3x≥−13 i divided and flipped the signs it should be $\huge~\frac{ 13 }{ -3 }$

296. mathmate

I have 4-3x <= -9 (second case) -3x <= -13 3x >= 13 so x>=13/3 do you agree?

297. pooja195

oh right the negatives cancel

298. mathmate

yes, they do (after flipping sign)

299. pooja195

and then u flipped .-.

300. mathmate

xD

301. pooja195

LOL

302. mathmate

Looks like you're good till 6.7 (perhaps as you expected).

303. mathmate

Sorry, didn't achieve 2 chapters, not even one, but we're efficient.

304. mathmate

No dull moment. :)

305. pooja195

:)

306. mathmate

If we have time later on, we may go back to word problems.

307. pooja195

NO

308. mathmate

Sure, we'll continue with 6.8 then.

309. mathmate

The NO is a tell-tale sign! :)

310. pooja195

-_-

311. pooja195

$$\color{blue}{\text{Originally Posted by}}$$ @mathmate If we have time later on, we may go back to word problems. $$\color{blue}{\text{End of Quote}}$$ NONONONONONON!!!

312. mathmate

Not today. Want to do word problems when you're fresh. We'll move on to graphics in 2 variables. p.367

313. mathmate

Can you tell me in a few words about graphing linear inequalities in two variables?

314. pooja195

:/

315. mathmate

Use words like region, feasible, vertical, horizontal lines, etc

316. mathmate

ok, so I'll do it.

317. mathmate

... another time! :)

318. pooja195

yes :)

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