Isolating variable confounded by cursed reciprocals

[Atomised]

Homework Statement

Given

1) xy = 1

2) x(t^2)-y = (t^3)-1/t

Express x in t, y in t
2. The attempt at a solutionx/t = [(tx+1)(td-1)] / [(t^2)-1] but I still can't separate t and x, driving me mad it is.

Also subtracted 1) from 2) to obtain

(t^2)x - (t^3) +1/t -1/x = 0

but no progress from there.

 

[AlephZero]

(t^2)x - (t^3) +1/t -1/x = 0

That's a good start. You can get rid of the fractions by multiplying everything by tx.

Then, if you know t and you want to find x, what sort of equation do you have?

 

[Atomised]

(x - t) (x t^3 + 1) = 0, therefore x = t, and (1) implies y = 1/t.

Solved thank you very much. I am self studying and this forum is incredibly helpful.

 

[Ray Vickson]

Homework Statement

Given

1) xy = 1

2) x(t^2)-y = (t^3)-1/t

Express x in t, y in t



2. The attempt at a solution


x/t = [(tx+1)(td-1)] / [(t^2)-1] but I still can't separate t and x, driving me mad it is.

Also subtracted 1) from 2) to obtain

(t^2)x - (t^3) +1/t -1/x = 0

but no progress from there.

If your second equation is
t^2 x - y = t^3 - \frac{1}{t}
then, substituting y = 1/x from the first equation gives you
t^2 x - \frac{1}{x} = t^3 - \frac{1}{t}
so if you divide through by t you find that the 'variable' $z = xt$ obeys
z - \frac{1}{z} = t^2 - \frac{1}{t^2}
There are two solutions for $z$.

 

[AlephZero]

(x - t) (x t^3 + 1) = 0, therefore x = t.

There are two solutions to that quadratic equation. not one. if AB = 0, either A = 0, or B = 0.

 

[Atomised]

Full Solution

There are two solutions to that quadratic equation. not one. if AB = 0, either A = 0, or B = 0.

x_1=t\\ y_1=1/x\\ x_2=-1/t^3\\ y_2=-t^3

 

[Atomised]

If your second equation is
t^2 x - y = t^3 - \frac{1}{t}
then, substituting y = 1/x from the first equation gives you
t^2 x - \frac{1}{x} = t^3 - \frac{1}{t}
so if you divide through by t you find that the 'variable' $z = xt$ obeys
z - \frac{1}{z} = t^2 - \frac{1}{t^2}
There are two solutions for $z$.

It is not immediately apparent to me what the advantage of making this substitution is. Is it that it makes apparent that $$z=t^2$$ and therefore $$x=t$$

Or could it be that it is suggesting the difference of two squares identity, which I cannot see how to use?




.

 

[Ray Vickson]

It is not immediately apparent to me what the advantage of making this substitution is. Is it that it makes apparent that $$z=t^2$$ and therefore $$x=t$$

Or could it be that it is suggesting the difference of two squares identity, which I cannot see how to use?




.

You can do it however you want; it all amounts in the end to finding the two solutions of a quadratic equation. However, the form $z - 1/z = t^2 - 1/t^2$ has nice "symmetry" that allows one to spot the two solutions right away, with almost no work. The solutions are $z = t^2$ and $z = -1/t^2$. The point is that both sides of this equation involve differences of terms that are reciprocals of each other, and that makes the solution easy to write down.

I cannot figure out why you refuse to acknowledge that there are two solutions.

 

[Atomised]

You can do it however you want; it all amounts in the end to finding the two solutions of a quadratic equation. However, the form $z - 1/z = t^2 - 1/t^2$ has nice "symmetry" that allows one to spot the two solutions right away, with almost no work. The solutions are $z = t^2$ and $z = -1/t^2$. The point is that both sides of this equation involve differences of terms that are reciprocals of each other, and that makes the solution easy to write down.

I cannot figure out why you refuse to acknowledge that there are two solutions.

Fully acknowledge two solutions - it was hare brained of me to omit in earlier post. I will experiment with substitutions more from now on.