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# Cartesian equations watch

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

is this a correct equation

y=x-2 ?
2. (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.
3. 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.
4. what I did was

x=1 +(1/t) I multiplied by t so that it gives x= t2 + 1 ?
5. (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.
6. (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 >?
7. (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
8. (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.
9. (Original post by PhysicsGal)
On a side note, if you did multiply out by t, you did the RHS fine
Are you sure?
10. (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
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 ?
11. (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?
12. (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.

Multiply the right hand side by 2.

13. In any event and this is what you typed in post 4.
14. (Original post by Mr M)
Are you sure?
Of course not
xt = t + 1/t^2

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

Au revoir.
15. (Original post by PhysicsGal)
Of course not
xt = t + 1/t^2
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.
16. (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
17. (Original post by PhysicsGal)
trying to learn how to write French letters...
3D printer?

18. (Original post by Mr M)
3D printer?

this one
Attached Images
19. Scan0216.pdf (280.6 KB, 58 views)
20. (Original post by Mr M)
3D printer?

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

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