geerky42 one year ago .

1. geerky42

@DarkBlueChocobo

2. geerky42

$$\dfrac{2x}{4}-\dfrac{3}{4}\quad\Longleftrightarrow\dfrac{2x-3}{4}$$

3. DarkBlueChocobo

Hay there

4. DarkBlueChocobo

so then I went

5. DarkBlueChocobo

$\frac{ 2x-3 }{ 4 }=\frac{ 10-15x }{ 6 }$

6. geerky42

Yes, then you cross-multiply

7. DarkBlueChocobo

$12x-18=40-60x$

8. DarkBlueChocobo

$72x-18=40$

9. geerky42

yeah

10. DarkBlueChocobo

$72x=58$

11. geerky42

Is that enough to tell what kind of equation it is?

12. geerky42

you can tell that x is going to equal something, so it wont be indentity.

13. geerky42

two other options are conditional equation and inconsistent equation

14. DarkBlueChocobo

Inconsistant i believe?

15. geerky42

conditional equation true for certain value of x, where if you simplify equation, you end up with x=4 or something inconsistent equation false equation, where you cannot solve for x.

16. DarkBlueChocobo

Oh so conditional is just like a usual equation that works out ?

17. geerky42

If you have something like x+1 = x+2, it's inconsistent equation because you end up with 1=2

18. geerky42

yeah

19. DarkBlueChocobo

lel wow the facepalms are coming out today boys

20. DarkBlueChocobo

1c. 0=8 so this is false or inconsistent?

21. geerky42

pretty much both

22. DarkBlueChocobo

I am having many issues on 2a

23. geerky42

inconsistent equation is kind of fancy way to say false equation lol

24. geerky42

you would need to combine fraction on right side. To do so, you need to get denominator to be same

25. geerky42

you have $$y-2$$ and $$y+1$$ So you just multiply numerator and denominator by $$y+1$$ for first fraction and $$y-2$$ for second

26. geerky42

$\dfrac{1}{y-2}-\dfrac{2}{y+1}\quad\longrightarrow\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}$

27. geerky42

Does that make sense? Basically you just need to get denominator to be same, so you just multiply numerator and denominator by something that would get you to desired denominator

28. geerky42

kinda hard for me to explain lol

29. DarkBlueChocobo

so 1/(y+1)(y-2) - 4/(y+1)(y-2) ?

30. geerky42

$\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\\~\\= \dfrac{y+1}{(y+1)(y-2)} - \dfrac{2(y-2)}{(y+1)(y-2)}\\~\\=\dfrac{y+1-2(y-2)}{(y+1)(y-2)}$

31. geerky42

ok let's start with something simply. $$\dfrac{1}{6}+\dfrac{1}{2}$$ How can we combine fraction here?

32. DarkBlueChocobo

multiply by 3/3

33. DarkBlueChocobo

so 1/6+1/2*3/3

34. DarkBlueChocobo

so you would get 1/6+3/6

35. geerky42

right, so for problem 2a. it's same idea, just with variable.

36. geerky42

get what I am saying here?

37. DarkBlueChocobo

Yes I do its just weird for me to visualize and see it right away t-t idk why

38. geerky42

because it's something new for you, I guess?

39. DarkBlueChocobo

Well like the bigness of the equation kinda like put me in my place kinda deal

40. geerky42

ok, so now we have $\dfrac{y+1-2(y-2)}{(y+1)(y-2)}$ simplify numerator

41. DarkBlueChocobo

$y+1-2y+4$

42. geerky42

CLEP test seem little harsh lol...

43. geerky42

don't forgot combine like term

44. DarkBlueChocobo

yeh was just showing steps

45. DarkBlueChocobo

-y+5

46. DarkBlueChocobo

It does seem harsh :/ I'm gonna like work myself to the point of no return so I do well on it . I want this

47. geerky42

Right. Now cross multiply $\dfrac{7}{y^2-y-2} = \dfrac{-y+5}{(y-2)(y+1)}\longrightarrow~?$

48. DarkBlueChocobo

so we have to do the same with the other side then get (y-2)(y+1) as denominator?

49. DarkBlueChocobo

oh i guess we dont

50. geerky42

Not really. denominator doesn't matter if they are on different side of equation.

51. DarkBlueChocobo

$y^2(-y+5) ,-y(-y+5),-2(-y+5)$ would this work?

52. DarkBlueChocobo

so I got -y3 + 5y^2 y^2 + 5y 2y+10

53. DarkBlueChocobo

before combining

54. geerky42

-y^3 + 5y^2 y^2 - 5y 2y-10

55. DarkBlueChocobo

damn i messed up multiplying

56. geerky42

you just messed up with +/- signs

57. DarkBlueChocobo

$-y^2+6y^2-3y-10$

58. geerky42

yeah

59. DarkBlueChocobo

now the other side

60. DarkBlueChocobo

let me delete that i need to slow down

61. DarkBlueChocobo

7(y+1) 7(y-2) 14y+7, 14y-14 14y+14y, 7-14

62. geerky42

none. left side should start with $$7(y+1)(y-2)$$

63. DarkBlueChocobo

so just keep it as is o.o?

64. geerky42

So firstly, you expand $$(y+1)(y-2)$$

65. DarkBlueChocobo

why not work out?

66. geerky42

You only distribute something with plus sign. $$a(b+c) = ab+ac$$ But $$a(bc) = abc$$

67. DarkBlueChocobo

Alrights

68. geerky42

make sense?

69. DarkBlueChocobo

Yes it does

70. DarkBlueChocobo

y^2+-2y+y-2?

71. geerky42

Yeah

72. DarkBlueChocobo

so we have $y^2-y-2=-y^3+6y^2-3y-10$

73. geerky42

you forgot 7

74. geerky42

$7(y^2-y-2)=-y^3+6y^2-3y-10$

75. DarkBlueChocobo

damn

76. DarkBlueChocobo

so do we expand it out with 7 now?

77. geerky42

yup

78. geerky42

$7y^2-7y-14=-y^3+6y^2-3y-10$

79. DarkBlueChocobo

I got that

80. geerky42

now take all terms to right side; $$0 = -y^3-y^2+4y-4$$

81. DarkBlueChocobo

can confirm got that after working out

82. DarkBlueChocobo

lel sorry I doing this in a note book :p

83. geerky42

know what to do from here?

84. geerky42

Notice something you can factor out? Hint: $0 = \boxed{-y^3-y^2}+\boxed{4y-4}$

85. DarkBlueChocobo

we can simplify ?

86. DarkBlueChocobo

87. geerky42

Oops should be $$0 = -y^3-y^2+4y+4$$

88. DarkBlueChocobo

would we simplify the y's?

89. geerky42

well, actually we try to factor, so we can easily figure out solutions.

90. DarkBlueChocobo

-1(y^2+4)(y+1)

91. DarkBlueChocobo

Would that be how we factor it?

92. geerky42

close. should be $$(y+1)(4-y^2)$$

93. geerky42

Then you can factor $$4-y^2$$

94. geerky42

or you can "factor out" -1 from $$4-y^2$$ So you would have something like $$(-1)(y^2-4)$$

95. DarkBlueChocobo

-y^2+4. Doesnt this just put us back where we were?

96. geerky42

Yeah, just to make it slightly easier to factor out, but never mind.

97. geerky42

factoring -1 out is just optional.

98. geerky42

You got $$0 = (y+1)(2-y)(y+2)$$?

99. DarkBlueChocobo

Yes that would be it fully factored

100. geerky42

yeah, now can you find solutions?

101. geerky42

I was thinking, maybe we need to do something more efficient with time. we spend a lot of time on few problems lol.

102. DarkBlueChocobo

indeed. Im just confusedwhere to go from here like.... you are dealing with 3 exponents of y and you need to get it to 1 y to solve for the question

103. geerky42

you are confused on how to get solution?

104. DarkBlueChocobo

Yes... like right now I feel like you could take -y^2 add to both sides and square both sides so you could get y=sqrt-y^3+4y-4

105. DarkBlueChocobo

But that wouldn't work so I just am straight confused where to go from here

106. geerky42

while solving for something, we want equation to be fully factored, so we can apply something (I forgot name) that states that If ab = 0, then either a = 0 or b = 0

107. geerky42

So we can "split" equation into $$0 = y+1$$ or $$0=2-y$$ or $$0=y+2$$

108. geerky42

So from here, just solve for y from each equation Does that make sense?

109. geerky42

Who gave you medal? lol

110. DarkBlueChocobo

Lol I have no clue, and Yeah that makes sense in order to get it fully factored when you have many parts

111. DarkBlueChocobo

y=-1 or y=2 or y=-2

112. geerky42

hmm Now plug y = -1 or y = 2 back into original equation

113. geerky42

what happens?

114. DarkBlueChocobo

i got 7/0=1/0-2/3

115. geerky42

yeah. zero in denominator. So we are left with y = -2 That it lol

116. DarkBlueChocobo

wait how we left with -2?

117. DarkBlueChocobo

s: we still had -2/3 left ove

118. geerky42

y = -1 and y = 2 get us zero in denominator, so they are invalid solutions.

119. DarkBlueChocobo

oh that makes sense nvm so we would try y=-2

120. DarkBlueChocobo

and baddabing there it comes out

121. geerky42

You can. and you will get true equation.

122. DarkBlueChocobo

Yeh Imma try that second one alone and check it laters just to make sure I understand this

123. geerky42

alright

124. geerky42

|dw:1434844521381:dw|

125. geerky42

ehh not right reaction, but whatever

126. DarkBlueChocobo

loool wowsers

127. DarkBlueChocobo

Alright so with the first word problem., Ralph = 6-2n? Ralph + Tommy= 30?

128. geerky42

R = 2T - 6 R+T = 30

129. DarkBlueChocobo

Could we set the equation to 2t-6=30? Cause we know the total it could be

130. geerky42

|dw:1434844843534:dw|

131. geerky42

yeah you could do that

132. DarkBlueChocobo

|dw:1434844913540:dw|

133. DarkBlueChocobo

you have beautiful mouse control lol

134. geerky42

thanks lol

135. DarkBlueChocobo

3t=36 36/3 t=12

136. DarkBlueChocobo

so we know Tommy scored 12 12*2 R=24-6=18

137. geerky42

|dw:1434845125612:dw|

138. DarkBlueChocobo

lol Loki pls

139. DarkBlueChocobo

So this one is worded weirdly for me it sounds like we are using multiple variables?

140. DarkBlueChocobo

OH wait no

141. DarkBlueChocobo

twice the 3rd is 86

142. DarkBlueChocobo

86/2

143. geerky42

no

144. geerky42

let's go ahead and set up equation

145. DarkBlueChocobo

Alright

146. geerky42

we can let first age be $$a$$ So second age would be $$a+2$$ third age is $$a+4$$ Because we are given that three ages arethree even CONSECUTIVE integers

147. geerky42

"If the sum of the 1st, four times the 2nd, and twice the 3rd is 86, what are the three ages?" So we have $$a + 4(a+2)+2(a+4) = 86$$ Make sense?

148. DarkBlueChocobo

Yep. As I fail reading it. It sounded really weird to me egh.

149. geerky42

150. DarkBlueChocobo

a+4a+8+2a+8=86 7a+16=86 7a=70 a=10 First Kid=10 Second Kid= 12 Third Kid = 14

151. geerky42

Yep!

152. DarkBlueChocobo

The next one is 20%*n=960?

153. DarkBlueChocobo

or .2n=960?

154. DarkBlueChocobo

n being the original price

155. geerky42

$$20\%$$ off means you only pay 80% of original price. So that would be .8n = 960

156. DarkBlueChocobo

wow

157. geerky42

wow such deal amazing price wow

158. DarkBlueChocobo

well screw me lol

159. DarkBlueChocobo

I was like yeh man that makes sense. Then like boom .8 is the original price payed

160. DarkBlueChocobo

n=1200 That makes sense cause 960/.2 is like 4000 so yeh thats a lot more sense

161. geerky42

yeah lol

162. DarkBlueChocobo

and 1200-240=960

163. DarkBlueChocobo

3y=2x-5 y=2/3x-5/3

164. geerky42

correct

165. DarkBlueChocobo

or y=.67x-1.67

166. geerky42

often we prefer to just leave it as fraction

167. DarkBlueChocobo

Alrights will do :p

168. DarkBlueChocobo

Next one would it look like x(x-5-7)

169. geerky42

get all term to one side So you have $$x^2-5x-14$$

170. DarkBlueChocobo

Well damn I just realized I can't cut it in half cause 5 is a prime number so I would have to multiply it by 2 and that would be 2*5 and thats 10 egh

171. DarkBlueChocobo

the most you can do its take x out?

172. geerky42

well, we just factor into something like $$(x-a)(x+b)$$

173. DarkBlueChocobo

Oh gawd more facepalms rolling out

174. DarkBlueChocobo

Cause we legit just did that like 2 problems ago ._.

175. DarkBlueChocobo

well 3

176. DarkBlueChocobo

So how would we deal with 5 ?

177. DarkBlueChocobo

cause other than the 5 i know how to solve this

178. geerky42

well, you can add itself by 2x, then you also subtract by 2x, to keep" balance" So you have $$x^2-5x-14+ 2x-2x~~~\Rightarrow~~~x^2+2x-7x-14$$ Now can you factor?

179. DarkBlueChocobo

(x+2)(x+7)

180. geerky42

Or to factor, you can ask yourself ab = 14 b+a = -5 So what are a and b?

181. geerky42

close. it's $$(x+2)(x-7)$$

182. DarkBlueChocobo

oh damn didnt realize the subtraction signs

183. DarkBlueChocobo

So for the next one can we do 6x^2 +3x + 4x +6?

184. DarkBlueChocobo

i mean +2

185. geerky42

yeah

186. DarkBlueChocobo

Got it !(3x+2)(2x+1)

187. geerky42

yay lol

188. DarkBlueChocobo

so to complete the square of the next problem you would do b=2 (2/2)^2 = 1 x^2 + 2x +1 = 5+1 (x+1)^2 =6

189. DarkBlueChocobo

i think

190. DarkBlueChocobo

i had to like triple check this lel

191. geerky42

Actually it's $$(x+1)^2=-4$$

192. DarkBlueChocobo

Wow hows?

193. DarkBlueChocobo

Where did I go wrong

194. geerky42

You already have 5 there, and you want 1, so you subtract both sides by 4

195. DarkBlueChocobo

Oh welp that would make sense

196. geerky42

so as you can see. no real root.

197. geerky42

because when you square any real number, it is always positive. But here, you have -4/

198. DarkBlueChocobo

So for the next one then so you divide both sides by 2 to x^2 alone so x^2-7/2x+3=0

199. DarkBlueChocobo

((7/2)/2)^2 so 3.1 is the decimal

200. geerky42

you only need b/2 So you would have 7/4 $$(x-\frac74)^2$$

201. DarkBlueChocobo

The like rules they giving me is to get x^2 secluded s: