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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
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.
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
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.
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
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)
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?
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?
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 :)
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)
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?
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)=?
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.
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.
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
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)
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')
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)=?
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
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
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!