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Cartesian equations

x = t + (1/t) , y = t - (1/t)

is this a correct equation

y=x-2 ?

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Original post by otrivine
x = t + (1/t) , y = t - (1/t)

is this a correct equation

y=x-2 ?


Absolutely not.

Try finding x+y and x-y.
Reply 2
that does not look right... it is equivalent to x - y = 2 or 2/t =2 which says t = 1... but t can have many values.
Reply 3
what I did was

x=1 +(1/t) I multiplied by t so that it gives x= t2 + 1 ?
Original post by otrivine
what I did was

x=1 +(1/t) I multiplied by t so that it gives x= t2 + 1 ?


Why? And why do you think it gives that?

I have told you how to proceed.
Reply 5
Original post by Mr M
Why are you randomly multiplying one side of an equation by t?

I have told you how to proceed.


Yep I got it
so x+y

so x+y=2t and t= (x+y)/2 and then sub to either x or y >?
Original post by otrivine
what I did was

x=1 +(1/t) I multiplied by t so that it gives x= t2 + 1 ?


On a side note, if you did multiply out by t, you did the RHS fine but not the LHS. x multiplied by t is xt :smile:
Original post by otrivine
Yep I got it
so x+y

so x+y=2t and t= (x+y)/2 and then sub to either x or y >?


Just find the product of (x+y)(x-y) and no substitution is required.
Original post by PhysicsGal
On a side note, if you did multiply out by t, you did the RHS fine :smile:


Are you sure?
Reply 9
Original post by PhysicsGal
On a side note, if you did multiply out by t, you did the RHS fine but not the LHS. x multiplied by t is xt :smile:


But in the example in the book

y= 3+2(2-x/x+3)
____________ they mutlpied by x+3 to get rid of it and they did not
1+(2-x/x+3) multiply the y with x+3 ?
Reply 10
Original post by Mr M
Just find the product of (x+y)(x-y) and no substitution is required.


oh ok, thanks!


one other question please Mr M

x=t2-1 , y= t-t3
for -2<equal t <equal to 2

In the book to get the limits in terms of t , they used x=-1 and x= 0 why?
Original post by otrivine
But in the example in the book

y= 3+2(2-x/x+3)
____________ they mutlpied by x+3 to get rid of it and they did not
1+(2-x/x+3) multiply the y with x+3 ?


Can you use brackets or, better still, LaTex?

Is it possible to scan this example as you are seem to have a fairly horrific misconception?

If it was possible just to arbitrarily multiply one side of an equation by something you could do stuff like this.

x=1x=1

Multiply the right hand side by 2.

x=2x=2

1=21=2
(edited 11 years ago)
In any event t(1+1t)t2+1\displaystyle t (1+\frac{1}{t}) \neq t^2 +1 and this is what you typed in post 4.
Original post by Mr M
Are you sure?

Of course not:colondollar:
xt = t + 1/t^2:blushing:

Edit: Happy April Fool's!! Bwahahahaa....

As we all know, x*t = xt
Thus (t + 1/t)*t = t^2 + t/t = t^2 + 1

:smile:
Au revoir.
(edited 11 years ago)
Original post by PhysicsGal
Of course not:colondollar:
xt = t + 1/t^2:blushing:


That isn't right either but the matter is confused by the fact that otrivine copied down the original question incorrectly in post 4. Never mind.
Original post by Mr M
That isn't right either but the matter is confused by the fact that otrivine copied down the original question incorrectly in post 4. Never mind.


Sorry, I'm trying to be helpful whilst trying to learn how to write French letters...clearly neither is happening well :/

And fair do's, looks like you're giving sound advice anyhow :smile:
Original post by PhysicsGal
trying to learn how to write French letters...


3D printer?

:tongue:
Reply 17
Original post by Mr M
3D printer?

:tongue:


this one :smile:
Original post by Mr M
3D printer?

:tongue:


LOL!!:angelblush: I should've seen that one arriving my way :P
Obviously, I meant actual letters written to people living in France, clearly :ahee:
Original post by otrivine
this one :smile:


The book multiplied the right hand side by (x+3)/(x+3).

Now 5/5 = 1, and x/x = 1, anything divisible by itself = 1, thus (x+3)/(x+3) = 1.

So if we multiply the LHS by this 'fraction', we get y multiplied by (x+3)/(x+3) = y multiplied by 1 = y
:smile:

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