Here's the question you clicked on:
Inopeki
In linear equations, how do i know if i should subtract from the right side or the left side?
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
3x + 7 = 10x +5 7 = 7x +5 - ^here? ^ or here?
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
And how much do i subtract? the whole number or so it remains +/-0?
now i need to divide -2 by 7?
no, we need the 2 on the other side of the equation that's what I mean about getting x by itself
Ok so i dont do 0=7x-2?
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.
I dont get it, i get like 0.29..
leave the answer in fraction form
Oh i see, so i get 2=7x?
right, now we divide by 7 to isolate x what is x=? (leave as a fraction)
you do know! lol shall we do another?
If you want more info, watch this and the next 2 videos http://www.khanacademy.org/video/simple-equations?playlist=Algebra
that's what he's working from phi
Well ill be damned.. it was right!
got one you want to post? or should I make one up?
wait, ill try the next one. 6x+3=2x+7
just go ahead and see how far you get
So i subtract 2x from both sides, getting 4x+3=7
Then i subtract 3 from both sides, getting 4x=4
And then i divide 4 by 4 and get 1 as the final answer :D
well taught turing:)
TT, you are an awesome teacher!
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
So where do i go on from here?
have you done ALL the exercises from this section already?
Soon. you mean until i can choose another section right?
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
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.
alright. but i couldnt figure the last one out, it was 6x+6=0
I didnt know what to do there. was it 1 or 0? maybe -6?
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
it was 9x+9=3x+3 6x+9=3 6x+6=0 6x= -6?
so now what is the last step to get x all alone?
tht was correct now show us your next and final step
now simplify that
yes, very good what was the last step? dividing by...?
that wasn't too bad now was it!
The number on the right?
dividing the quantity of x by the number on the right side of =?
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?
it's just one answer: divide by 6
let's try something a little different... solve xy+7=10 for x
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
so now xy+7=10 you got a number answer? what happened to y? please show steps.
@turing does he know what he is solving for?
solve for x thanks karate
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)
you are over-thinking Inopeki just take each step one-by-one what is the first step?
Determening what i need to make to complete 10
no the first step is the same as the others xy+7=10 subtract 7 from both sides
what do we have after?
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?
Removing y from the equation?
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?
right! and in this case what do we need to divide by to get x by itself?
close, but what do you mean "divide by y with 3"? that makes no sense again it can only be one answer divide by...?
just to make it visible again: xy=3 what do we divide by on both sides (one answer to this question)
Ah, divide x and 3 by y?
YES!!! and what do we have after? write the expression.
How can i do that when i dont even know what y is?
Good question. write the answer first and I will tell you what it means after.
close, but on the left x should be by itself, that's why we divided by y right? so it should be...
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
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 ???
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?
yes! what about when y=3?
Excellent :) and y=10?
better to leave it as a fraction, but yes
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 :)
Whoa, whoa, whoa. You just made this 10x more complicated.
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
Oh! I just made a function?
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
So f(x) means that you want to pop something for y?
exactly! if you want to find f(x) then solve\[xy+7=10\]for x instead of y
sorry, I meant y instead of x above
So thats what ill be using in quadratic equalities?
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)
the problem? no follow it closely
Can you give me one so i can try?
sure, solve 5t+2r=3r-4 for r
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.
good! better get used to it, all kinds of funny symbols in physics :P
flutter is how they translate the f-word lol
5t+2r=3r-4 5t=r-4 Am i right so far?
Well i need you to tell me what t is. Ill just assume it is 1. 5t=9(r)-4?
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?
perfect! so do you think you can tell me 'what is a function of what' in this case?
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?
It must be because 4 is an insteger, right?
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)=?
sorry, I meant r(t)=5t+4 what is r(0)=? r(1)=? r(-1)=?
I thought we determined that r(0) cant be a variable?
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.
r is called the 'dependent variable' btw, because it 'depends' on what we put for t.
so find r(0) r(1) and r(-1) for r=5t+4
r(t)=5t+4 what is r(0)=-1/4 r(1)=1/9 r(-1)=-5
Thats probably really wrong..
no, show your work... r(0)=5(0)+4=4 do the same for the others
OH! so (this) is what equals t?
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)
great, so now you have dealt with two functions today. do you know how to plot points on a graph?
yes! now do you know what r(0)=4 means as a point?
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.
Ok, give me an example and ill try it. like r(2)=4---->2,4?
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
Actually, just do the question for now: -find the ordered pair that corresponds to r(-2) for r(t)=5t+4
good, so what is the ordered pair (t,r) ?
-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.
perfect :D wanna try to graph it yet, or find more ordered pairs?
One more ordered pair would be good.
ok, find the ordered pairs that correspond to r(-3) r(3) and r(5) for r=5x+4
sorry, you said one more you can only do one if you want.
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)
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?
YES! Thanks, the teacher is just as important though! Sure.
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')
just put dots at their coordinates:|dw:1325550068136:dw|now because this graph is linear, what will we connect those two points by?
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.
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?
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?
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)=?
r(0.8)=t(0.8)? r(0.8)=t(0)+0.8?
oh wait, ill take that bacl
that's a point in the middle, not a comma, and it's negative: t=-0.8 go again!
Should i follow r=5t+4?
Ohhh.. r(-0.8)=(-0.8)+4=3.2?
what happened to the 5? r=5t+4
Oh right, i knew i forgot something. r(-0.8)=5(-0.8)+4=0!
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!
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
Alright :( Well thanks for all your help! It was really fun! Where do i start at khan now?
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!
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
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
Wait, on the first one i need to do this: 13= s/p-2 right?
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!
p(13)=2(13)+2=28? What is s? Ill say 2. \[28/13\approx2.15\]
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?