Simultaneous equatios

[Oaky]

Hey guys, having a lot of trouble with a system of equations. I know that the pronumerals (\(\displaystyle a, b, x\) and \(\displaystyle y\)) are all positive.

1. \(\displaystyle y(a^2b^2 - 2b^2x^2 - a^2y^2) = 0\)
2. \(\displaystyle x(a^2b^2 - b^2x^2 - 2a^2y^2) = 0\)

So far what I've done...

Looking at (1), I know \(\displaystyle y\neq0\).
Therefore \(\displaystyle a^2b^2 - 2b^2x^2 - a^2y^2 = 0\)
Therefore \(\displaystyle 2b^2x^2 = a^2b^2 - a^2y^2\)
Therefore \(\displaystyle x^2 = \frac{a^2(b^2 - y^2)}{2b^2}\)
Therefore \(\displaystyle x = \frac{a}{\sqrt{2}b}\sqrt{b^2-y^2}\)

And you can put the root 2 on top and then put it inside the main square root, but it's still messy. I can't find an easy way to substitute it into the second equation, but I know the answer is very neat (\(\displaystyle x=\frac{a}{\sqrt{3}}\) and \(\displaystyle y=\frac{b}{\sqrt{3}}\)). Have I gone wrong/is there an easier way to do it?

Thanks!

 

[tkhunny]

Hey guys, having a lot of trouble with a system of equations. I know that the pronumerals (\(\displaystyle a, b, x\) and \(\displaystyle y\)) are all positive.

1. \(\displaystyle y(a^2b^2 - 2b^2x^2 - a^2y^2) = 0\)
2. \(\displaystyle x(a^2b^2 - b^2x^2 - 2a^2y^2) = 0\)

So far what I've done...

Looking at (1), I know \(\displaystyle y\neq0\).
Therefore \(\displaystyle a^2b^2 - 2b^2x^2 - a^2y^2 = 0\)
Therefore \(\displaystyle 2b^2x^2 = a^2b^2 - a^2y^2\)
Therefore \(\displaystyle x^2 = \frac{a^2(b^2 - y^2)}{2b^2}\)
Therefore \(\displaystyle x = \frac{a}{\sqrt{2}b}\sqrt{b^2-y^2}\)

And you can put the root 2 on top and then put it inside the main square root, but it's still messy. I can't find an easy way to substitute it into the second equation, but I know the answer is very neat (\(\displaystyle x=\frac{a}{\sqrt{3}}\) and \(\displaystyle y=\frac{b}{\sqrt{3}}\)). Have I gone wrong/is there an easier way to do it?

Thanks!

1) What's wrong with "messy"?
2) Why does it have to be "easy"?
3) Have you tried substituting x^2, rather than x?

 

[HallsofIvy]

Hey guys, having a lot of trouble with a system of equations. I know that the pronumerals (\(\displaystyle a, b, x\) and \(\displaystyle y\)) are all positive.

1. \(\displaystyle y(a^2b^2 - 2b^2x^2 - a^2y^2) = 0\)
2. \(\displaystyle x(a^2b^2 - b^2x^2 - 2a^2y^2) = 0\)

So far what I've done..

Looking at (1), I know \(\displaystyle y\neq0\).
Therefore \(\displaystyle a^2b^2 - 2b^2x^2 - a^2y^2 = 0\)

Yes, and \(\displaystyle a^2b^2- b^2x^2- 2a^2y^2= 0\)

I would now subtract the first equation from the second: \(\displaystyle b^2x^2- a^2y^2= 0\) so that \(\displaystyle b^2x^2= a^2y^2\) so that \(\displaystyle x= \pm\frac{a}{b}y\).

Now put that into either of the two equations: \(\displaystyle a^2b^2- b^2\left(\frac{a^2}{b^2}x^2\right)- a^2y^2= 0\).
\(\displaystyle 2a^2y^2= a^2b^2\) is easy to solve.

Therefore \(\displaystyle 2b^2x^2 = a^2b^2 - a^2y^2\)
Therefore \(\displaystyle x^2 = \frac{a^2(b^2 - y^2)}{2b^2}\)
Therefore \(\displaystyle x = \frac{a}{\sqrt{2}b}\sqrt{b^2-y^2}\)

And you can put the root 2 on top and then put it inside the main square root, but it's still messy. I can't find an easy way to substitute it into the second equation, but I know the answer is very neat (\(\displaystyle x=\frac{a}{\sqrt{3}}\) and \(\displaystyle y=\frac{b}{\sqrt{3}}\)). Have I gone wrong/is there an easier way to do it?

Thanks!

 

[Deleted member 4993]

Hey guys, having a lot of trouble with a system of equations. I know that the pronumerals (\(\displaystyle a, b, x\) and \(\displaystyle y\)) are all positive.

1. \(\displaystyle y(a^2b^2 - 2b^2x^2 - a^2y^2) = 0\)
2. \(\displaystyle x(a^2b^2 - b^2x^2 - 2a^2y^2) = 0\)

Noting that x is attached to b and y is attached a, we define two new variables

r = x*b and s = a*y

then the equations become:

s/a * (a²b² - 2r² - s²) = 0

and

r/b * (a²b² - r² - 2s²) = 0

since it is given that \(\displaystyle a, b, x\) and \(\displaystyle y\) are all positive - I assume it to mean a,b,x,y >0.

a²b² - 2r² - s² = 0 ..................................................................(1)

a²b² - r² - 2s² = 0 ..................................................................(2)

which gives r = s = \(\displaystyle \pm\dfrac{ab}{\sqrt{3}}\) ..... and continue

So far what I've done...

Looking at (1), I know \(\displaystyle y\neq0\).
Therefore \(\displaystyle a^2b^2 - 2b^2x^2 - a^2y^2 = 0\)
Therefore \(\displaystyle 2b^2x^2 = a^2b^2 - a^2y^2\)
Therefore \(\displaystyle x^2 = \frac{a^2(b^2 - y^2)}{2b^2}\)
Therefore \(\displaystyle x = \frac{a}{\sqrt{2}b}\sqrt{b^2-y^2}\)

And you can put the root 2 on top and then put it inside the main square root, but it's still messy. I can't find an easy way to substitute it into the second equation, but I know the answer is very neat (\(\displaystyle x=\frac{a}{\sqrt{3}}\) and \(\displaystyle y=\frac{b}{\sqrt{3}}\)). Have I gone wrong/is there an easier way to do it?

Thanks!

.

 

[Deleted member 4993]

My proposed solution was presented as a different way to skin the same cat.

Since the student lost way in complex "arithmatic", my alternate way was presented.

By the way, when my teachers showed a more elegant method - that method got "burned" in my head.....

There is a story behind it....

Long time ago (well long long time ago), in my vector algebra class - I faced a homework that took me about 3 pages of manipulations to prove the conjecture.

When I complained to my teacher - expressing how tedious that problem was - he said he could do it 10 lines. I was shocked to see that he was right - maximum 10 lines it was.

Then long time ago (notice I dropped one "long") - I noticed that my son was struggling with the same problem and whining as I was that long-long time ago. I casually informed him that the problem was very simple - as a matter of fact can be done in 10 lines. He challenged - and I did.

Since then my title became "super-dooper math looper" dad.

Well, next year - after gloating for one full year - I told him the whole story .... and lost my title....

By the way, now he is a Research Director of a computer company. I think my challenge brought that "I'll show you" spirit in him.

 

[Deleted member 4993]

Not all students have minds like yours. .

However, this student is quite advanced. S/he had posted problems dealing with partial differentiation, in prior threads.

I suspect, the original equations were product of some problem regarding extrema of two variable function.

 

[Oaky]

Thanks for all of the help!

I didn't "give up" or "quit" because it got messy, it was just that the teacher implied to us that the solution wasn't as drawn out as I was making it and that there was probably a simpler way that I was overlooking.

And yes, the equations were only a part of a bigger question. We had to find the volume of largest rectangular box that could fit inside the ellipsoid \(\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\).

Appreciate the response from everybody.