## Inopeki Group Title In linear equations, how do i know if i should subtract from the right side or the left side? 2 years ago 2 years ago

1. karatechopper Group Title

example..

2. meverett04 Group Title

I like my coefficient for x to remain positive; therefore, if I have 2x + 10 = 12x I subtract 2x from the left side and if I have 12x + 10 = 2x I subtract 2x from the right side

3. Inopeki Group Title

3x + 7 = 10x +5 7 = 7x +5 - ^here? ^ or here?

4. TuringTest Group Title

it really doesn't matter so long as you keep the goal in mind: get x by itself pick one and we can continue from there

5. Inopeki Group Title

And how much do i subtract? the whole number or so it remains +/-0?

6. Inopeki Group Title

Ok

7. Inopeki Group Title

so i got 0=7x-2

8. Inopeki Group Title

now i need to divide -2 by 7?

9. TuringTest Group Title

no, we need the 2 on the other side of the equation that's what I mean about getting x by itself

10. Inopeki Group Title

Ok so i dont do 0=7x-2?

11. TuringTest Group Title

if we divide by 7 now we get$0=x-\frac{2}{7}$which is okay, we can still solve it from here; but it's a good policy to leave fractions for the end if possible. so instead add 2 to both sides to make x a little bit more alone.

12. Inopeki Group Title

I dont get it, i get like 0.29..

13. TuringTest Group Title

leave the answer in fraction form

14. Inopeki Group Title

Oh i see, so i get 2=7x?

15. TuringTest Group Title

right, now we divide by 7 to isolate x what is x=? (leave as a fraction)

16. Inopeki Group Title

I dont know.. 2/7?

17. TuringTest Group Title

you do know! lol shall we do another?

18. phi Group Title

19. TuringTest Group Title

that's what he's working from phi

20. Inopeki Group Title

What?

21. TuringTest Group Title

Another?

22. Inopeki Group Title

Sure

23. Inopeki Group Title

Well ill be damned.. it was right!

24. TuringTest Group Title

got one you want to post? or should I make one up?

25. Inopeki Group Title

wait, ill try the next one. 6x+3=2x+7

26. TuringTest Group Title

just go ahead and see how far you get

27. Inopeki Group Title

So i subtract 2x from both sides, getting 4x+3=7

28. TuringTest Group Title

good, then?

29. Inopeki Group Title

Then i subtract 3 from both sides, getting 4x=4

30. TuringTest Group Title

good, then?

31. Inopeki Group Title

And then i divide 4 by 4 and get 1 as the final answer :D

32. karatechopper Group Title

:D

33. karatechopper Group Title

well taught turing:)

34. Inopeki Group Title

TT, you are an awesome teacher!

35. TuringTest Group Title

tadah! nice see how you didn't need to ask me which was better, subtract 7 or 3? you knew by looking that 3 was the smart way to go :) practice practice... thanks for the compliment btw

36. Inopeki Group Title

So where do i go on from here?

37. TuringTest Group Title

have you done ALL the exercises from this section already?

38. Inopeki Group Title

Soon. you mean until i can choose another section right?

39. TuringTest Group Title

Yes. It is imperative that you do every problem. This needs to become second nature for you when you can look at 5x-3=2x+12 and tell me the answer very quickly, with almost no work shown, then you can continue

40. Inopeki Group Title

3/4?

41. Inopeki Group Title

actually 1/3

42. TuringTest Group Title

nope, better keep working on it. practice practice... don't try to get ahead of yourself, if you need to write it and go slowly, do it. Don't try to shortcut past that.

43. Inopeki Group Title

alright. but i couldnt figure the last one out, it was 6x+6=0

44. Inopeki Group Title

I didnt know what to do there. was it 1 or 0? maybe -6?

45. TuringTest Group Title

just try the problem and show me your steps you don't need to tell me what each step is, btw, I can see by looking

46. Inopeki Group Title

it was 9x+9=3x+3 6x+9=3 6x+6=0 6x= -6?

47. TuringTest Group Title

so now what is the last step to get x all alone?

48. karatechopper Group Title

tht was correct now show us your next and final step

49. Inopeki Group Title

6/-6?

50. karatechopper Group Title

perfect

51. karatechopper Group Title

now simplify that

52. Inopeki Group Title

-1?

53. karatechopper Group Title

correct

54. Inopeki Group Title

Yes!

55. TuringTest Group Title

yes, very good what was the last step? dividing by...?

56. karatechopper Group Title

that wasn't too bad now was it!

57. Inopeki Group Title

The number on the right?

58. Inopeki Group Title

dividing the quantity of x by the number on the right side of =?

59. Inopeki Group Title

Other*

60. TuringTest Group Title

no, that is what I was afraid of... though you got the right answer, the -6 should have stayed on top. 6x=-6 what do we divide by to isolate x?

61. Inopeki Group Title

-6 with 6?

62. TuringTest Group Title

it's just one answer: divide by 6

63. TuringTest Group Title

let's try something a little different... solve xy+7=10 for x

64. Inopeki Group Title

Ok

65. TuringTest Group Title

Note that above for 6x=-6 we divided both sides by the number next to the x (i.e. 6), that is what I wanted you to see. We do not divide by the number on the right. You will see why in this next problem

66. Inopeki Group Title

1.5 or 2 for x

67. Inopeki Group Title

Ok..

68. TuringTest Group Title

so now xy+7=10 you got a number answer? what happened to y? please show steps.

69. karatechopper Group Title

@turing does he know what he is solving for?

70. TuringTest Group Title

solve for x thanks karate

71. karatechopper Group Title

anytime turing:)

72. Inopeki Group Title

I thought that x times y would be the answer. since 1.5x2 is 3 (which was needed to make 10) i thought that x would be 1.5 or 2.(the other one being y)

73. TuringTest Group Title

you are over-thinking Inopeki just take each step one-by-one what is the first step?

74. Inopeki Group Title

Determening what i need to make to complete 10

75. TuringTest Group Title

no the first step is the same as the others xy+7=10 subtract 7 from both sides

76. TuringTest Group Title

what do we have after?

77. Inopeki Group Title

Ah.. xy=3

78. TuringTest Group Title

good! now again, we want x by itself. it looks like x is being multiplied by y, so how can we get x by itself?

79. Inopeki Group Title

Removing y from the equation?

80. TuringTest Group Title

y cannot leave the equation entirely, we can only shift it to one side. I mentioned that x was being multiplied by y. Do you know the inverse operation of multiplication?

81. Inopeki Group Title

division?

82. TuringTest Group Title

right! and in this case what do we need to divide by to get x by itself?

83. Inopeki Group Title

y with 3?

84. TuringTest Group Title

close, but what do you mean "divide by y with 3"? that makes no sense again it can only be one answer divide by...?

85. TuringTest Group Title

just to make it visible again: xy=3 what do we divide by on both sides (one answer to this question)

86. Inopeki Group Title

Ah, divide x and 3 by y?

87. TuringTest Group Title

YES!!! and what do we have after? write the expression.

88. Inopeki Group Title

How can i do that when i dont even know what y is?

89. TuringTest Group Title

Good question. write the answer first and I will tell you what it means after.

90. Inopeki Group Title

You mean x/y=3/y?

91. TuringTest Group Title

close, but on the left x should be by itself, that's why we divided by y right? so it should be...

92. Inopeki Group Title

I GET IT NOW! because the left side was x times y we needed to isolate x so we divided x with y to reverse the effect. now it is x=3/y

93. TuringTest Group Title

YES!!! very good! but what does it mean...? as you said y can be any number (except zero, but we'll discuss that later) so we have x=3/y that means we have infinite solutions to this problem. To find what they are plug in various numbers for y. So what is x if y=1 y=3 y=10 ???

94. Inopeki Group Title

1.5?

95. TuringTest Group Title

no, its different for each one. Remember that I said there were infinite solutions. just plug in the numbers into the equation there is no one answer for x=3/y so plug in a number for y like y=1 what is x?

96. Inopeki Group Title

3

97. TuringTest Group Title

98. Inopeki Group Title

1

99. TuringTest Group Title

Excellent :) and y=10?

100. Inopeki Group Title

0.3333333?

101. TuringTest Group Title

better to leave it as a fraction, but yes

102. Inopeki Group Title

YES! I get it now!

103. TuringTest Group Title

Great! so what is going on here? it looks like we can put any value for y (except zero) and we get a number for x. THIS IS YOUR FIRST FUNCTION I told you earlier that a function takes a number as an input (in this case what we put for y is our input) and out pops another number. So x=3/y is a formula for x as a function of y. denoted x=f(y)=3/y here is a graph of your function http://www.wolframalpha.com/input/?i=f%28y%29%3D3%2Fy this line denotes all the points that are solutions. notice that some of these points (x,y) are (1,3) (3,1) (10,3/10) etc. just like you found :)

104. Inopeki Group Title

Whoa, whoa, whoa. You just made this 10x more complicated.

105. TuringTest Group Title

look all the deal is is that you put in a number for y, you get a number back for x that's what a function is

106. Inopeki Group Title

Oh! I just made a function?

107. TuringTest Group Title

Yes. Well actually it was always a function, buit you made it workable. in this case we would write that x is a function of y. f(y)=3/y

108. Inopeki Group Title

So f(x) means that you want to pop something for y?

109. TuringTest Group Title

exactly! if you want to find f(x) then solve$xy+7=10$for x instead of y

110. TuringTest Group Title

sorry, I meant y instead of x above

111. Inopeki Group Title

So thats what ill be using in quadratic equalities?

112. TuringTest Group Title

it's one thing you will need in order to understand them. in general though, you should not be thrown by seeing f(x)=5x+7 that is just the same situation we had. they are saying that f(x) depends on what number you put in for x. you may see 'find when f(x)=10 for f(x)=5x+5' in which case you just use the equation, and put the value of f(x) on the other side of the = sign: 10=5x+5 5=5x x=1 we can do that for any value of f(x)

113. Inopeki Group Title

Shouldnt x be 2 there?

114. TuringTest Group Title

where?

115. TuringTest Group Title

the problem? no follow it closely

116. Inopeki Group Title

Oh nevermind

117. Inopeki Group Title

Can you give me one so i can try?

118. TuringTest Group Title

sure, solve 5t+2r=3r-4 for r

119. Inopeki Group Title

I see what you are doing, you are trying to flutter up my train of thoughts by replacing x and y with t and r.. hah, itll never work.

120. Inopeki Group Title

flutter?

121. TuringTest Group Title

good! better get used to it, all kinds of funny symbols in physics :P

122. TuringTest Group Title

flutter is how they translate the f-word lol

123. Inopeki Group Title

5t+2r=3r-4 5t=r-4 Am i right so far?

124. TuringTest Group Title

yes, now what?

125. Inopeki Group Title

Well i need you to tell me what t is. Ill just assume it is 1. 5t=9(r)-4?

126. TuringTest Group Title

no you don't. we are going to get a function out of this because we have two variable and only one eqn. so how do you isolate r?

127. TuringTest Group Title

5t=r-4

128. Inopeki Group Title

5t-4=r?

129. TuringTest Group Title

perfect! so do you think you can tell me 'what is a function of what' in this case?

130. Inopeki Group Title

f(t)?

131. TuringTest Group Title

yes, that's right! sorry btw, you made a mistake earlier: 5t=r-4 ADD 4 to both sides: 5t+4=r but you got the fact that this is a function of t, very good. which variable is a function of t?

132. Inopeki Group Title

r

133. Inopeki Group Title

It must be because 4 is an insteger, right?

134. TuringTest Group Title

perfect :) (the integer thing makes no difference. If it was$r=5t+\sqrt2$that would be a function as well. we can write our function as r=f(t)=5t+4 or we can also write r(t)=5t+4 both mean that r is a function of t. Same exact thing. Now I'm going to use the notation r(t)=5t+4 and I am asking you, what is f(0)=? f(1)=? f(-1)=?

135. TuringTest Group Title

sorry, I meant r(t)=5t+4 what is r(0)=? r(1)=? r(-1)=?

136. Inopeki Group Title

I thought we determined that r(0) cant be a variable?

137. TuringTest Group Title

that was only when the 'independent variable' (that's what we call the t in this case) was in the denominator, because any number divided by zero is undefined (infinity basically) here we don't have that problem, so just plug in the numbers and see what you get.

138. TuringTest Group Title

r is called the 'dependent variable' btw, because it 'depends' on what we put for t.

139. Inopeki Group Title

I see

140. TuringTest Group Title

so find r(0) r(1) and r(-1) for r=5t+4

141. Inopeki Group Title

r(t)=5t+4 what is r(0)=-1/4 r(1)=1/9 r(-1)=-5

142. Inopeki Group Title

Thats probably really wrong..

143. TuringTest Group Title

no, show your work... r(0)=5(0)+4=4 do the same for the others

144. Inopeki Group Title

OH! so (this) is what equals t?

145. TuringTest Group Title

right!

146. Inopeki Group Title

r(t)=5t+4 what is r(0)=5(0)+4=4 r(1)=5(1)+4 So it equals 9! r(-1)=5(-1)+4=-1 ((-5)+4=-1)

147. Inopeki Group Title

That should do it.

148. TuringTest Group Title

excellent!

149. Inopeki Group Title

YEEEEEEEEES!!!!!!!!

150. TuringTest Group Title

great, so now you have dealt with two functions today. do you know how to plot points on a graph?

151. Inopeki Group Title

|dw:1325547966414:dw|

152. TuringTest Group Title

yes! now do you know what r(0)=4 means as a point?

153. Inopeki Group Title

As a point?

154. TuringTest Group Title

yes, it is the point (0,4) we can call the up-down direction on our graph r, and the horizontal direction t. this way we get a point out of every value of t we put in. the points (t,r) that we found are then r(0)=4-->(0,4) r(1)=9--->(1,9) r(-1)=-1-->(-1,-1) ^^^don't continue until you understand all this, or at least have tried.

155. Inopeki Group Title

Ok, give me an example and ill try it. like r(2)=4---->2,4?

156. TuringTest Group Title

here is our graph marked from 0 to 4 in all directions:|dw:1325548510639:dw|can you plot two of the points? If you want to try some more first that's even better: -find the point (actually it's called an 'ordered pair') that corresponds to r(-2) for r(t)=5t+4

157. TuringTest Group Title

Actually, just do the question for now: -find the ordered pair that corresponds to r(-2) for r(t)=5t+4

158. Inopeki Group Title

r(-2)=5(-2)+4=-6

159. TuringTest Group Title

good, so what is the ordered pair (t,r) ?

160. Inopeki Group Title

-2,5?

161. TuringTest Group Title

-2 is right, but where did 5 come from? that's not the value of r(-2)... and please put it in parentheses, that is how ordered pairs are written.

162. Inopeki Group Title

I see, (-2,-6)?

163. TuringTest Group Title

perfect :D wanna try to graph it yet, or find more ordered pairs?

164. Inopeki Group Title

One more ordered pair would be good.

165. TuringTest Group Title

ok, find the ordered pairs that correspond to r(-3) r(3) and r(5) for r=5x+4

166. TuringTest Group Title

sorry, you said one more you can only do one if you want.

167. Inopeki Group Title

Thats alright, ill do all of them. r(-3)=5(-3)+4=-11 The corresponding pair is (-3,-11) r(3)=5(3)+4=19 The corresponding pair is (3,19) r(5)=5(5)+4=29 The corresponding pair is (5,29)

168. TuringTest Group Title

wow!!! you are a very fast learner Inopeki, you'll be doing calculus before you know it ;) so shall we try to graph it now?

169. Inopeki Group Title

YES! Thanks, the teacher is just as important though! Sure.

170. TuringTest Group Title

thanks for that :D so let's lay out some points that might fit on our graph: r(-1)=-1->(-1,-1) r(0)=4-->(0,4) can you plot those points onto this graph?|dw:1325549886229:dw|(use 'reply with drawing')

171. Inopeki Group Title

|dw:1325549850006:dw|

172. TuringTest Group Title

just put dots at their coordinates:|dw:1325550068136:dw|now because this graph is linear, what will we connect those two points by?

173. Inopeki Group Title

|dw:1325550114671:dw|

174. TuringTest Group Title

good, but the line goes on to infinity in both directions:|dw:1325550191308:dw|the arrows mean it keeps going. Good job though you have the idea. Here is a plot of our graph on two different scales: http://www.wolframalpha.com/input/?i=r%28t%29%3D5t%2B4 it doesn't look exactly like ours, but that is only because they made the increments of r larger because the graph is so steep. They wanted to fit more on there.

175. Inopeki Group Title

So now what?

176. TuringTest Group Title

you can see on the graph I linked you to that the point (1,9) is on there, just as it should be. It also looks like the graph shows that the point (-0.8,0) should be as well. Why don't you see if you can show that the point (-0.8,0) is on our graph. Any idea how to do that?

177. Inopeki Group Title

|dw:1325550664006:dw|

178. Inopeki Group Title

NO! WAIT!

179. TuringTest Group Title

no, we want to do it mathematically, not graphically. Questions should only be answered graphically if they specifically request it. got a better way to show that (-0.8,0) is a point?

180. Inopeki Group Title

Oh...

181. TuringTest Group Title

well that is the graphic solution, and it looks like you are right. But we want to PROVE that you are right. so how can we do that? think about this: if (-0.8,0) is a point on the graph of r=5t+4 then what should r(-0.8)=?

182. Inopeki Group Title

r(0.8)=t(0.8)? r(0.8)=t(0)+0.8?

183. Inopeki Group Title

oh wait, ill take that bacl

184. TuringTest Group Title

that's a point in the middle, not a comma, and it's negative: t=-0.8 go again!

185. Inopeki Group Title

186. TuringTest Group Title

yep :)

187. Inopeki Group Title

Ohhh.. r(-0.8)=(-0.8)+4=3.2?

188. TuringTest Group Title

what happened to the 5? r=5t+4

189. Inopeki Group Title

Oh right, i knew i forgot something. r(-0.8)=5(-0.8)+4=0!

190. TuringTest Group Title

which makes sense right? the ordered pair corresponding to that is (-0.8,0) which is what we wanted to prove, that this point is on the graph. Awesome job!

191. Inopeki Group Title

Thanks!

192. TuringTest Group Title

I'm sure you've learned a lot today, please continue on your own working off khan academy. I do have to eat and stuff, so I'll catch you later. Again, super job! Keep practicing :D

193. Inopeki Group Title

Alright :( Well thanks for all your help! It was really fun! Where do i start at khan now?

194. TuringTest Group Title

did you finish ALL of that other lecture and problems? if you really finished it all, then you should do the next-to-last one that James gave you, then the first one he gave you. Most importantly, keep working on moving the symbols around. If you run out of exercises here are some to keep you busy. solve: 13+s/p=2 for p 5qr+6=qr/2 for q z/t+z=4t for z You shouldn't be able to do those so easily, as they involve factoring and such. If they give you too much trouble continue with the links James sent you. See you later tonight maybe, otherwise perhaps tomorrow. Good luck!

195. TuringTest Group Title

Here it is a little more clear. Again, if you can't do it it's okay. solve$13+\frac{s}{p}=2$for p$5qr+6=\frac{qr}{2}$for q$\frac{z}{t}+z=4t$for z

196. Inopeki Group Title

The 3 he gave me were linear equations 3, intro to the quadratic equation and simple equations, im soon finished with linear and then im gonna start with quadratic

197. Inopeki Group Title

Wait, on the first one i need to do this: 13= s/p-2 right?

198. TuringTest Group Title

actually the first step on the first one is 1)multiply both sides by p the last one above will have a 'squared' term in it, so you may need the lectures first, I don't know. you're on the way to being a physicist, step-by-step again, good luck!

199. Inopeki Group Title

Thanks :)

200. Inopeki Group Title

p(13)=2(13)+2=28? What is s? Ill say 2. $28/13\approx2.15$

201. Inopeki Group Title

5qr+6=qr2 4qr+6=qr I can now determine that q*r is negative because otherwise 4qr+6=qr The possibilities of what q and r are endless, what now?

202. Inopeki Group Title

TT?