System of equations

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Prasiortle
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#1
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#1
Let a and b be real numbers such that 1/(a^2) + 3/(b^2) = 2018a
and 3/(a^2) + 1/(b^2) = 290b. What is (ab)/(b-a)?

By playing around with it, I was eventually able to get a cubic in (a/b), but the roots are extremely hideous, and so this method of solution isn't really practical, especially as this problem is taken from a non-calculator competition, and the final answer is supposed to be a fairly nice fraction, so there should be a nice solution. Any suggestions?
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DFranklin
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#2
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#2
(Original post by Prasiortle)
Let a and b be real numbers such that 1/(a^2) + 3/(b^2) = 2018a
and 3/(a^2) + 1/(b^2) = 290b. What is (ab)/(b-a)?

By playing around with it, I was eventually able to get a cubic in (a/b), but the roots are extremely hideous, and so this method of solution isn't really practical, especially as this problem is taken from a non-calculator competition, and the final answer is supposed to be a fairly nice fraction, so there should be a nice solution. Any suggestions?
Not guaranteeing this is correct, but I started by writing A = 1/a, B = 1/b, giving A^2 + 3B^2 = 2018/A, 3A^2+B^2 = 290/B.

Get rid of the fractions and subtract one from the other; it should be plain sailing from there.
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Prasiortle
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#3
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#3
(Original post by DFranklin)
Not guaranteeing this is correct, but I started by writing A = 1/a, B = 1/b, giving A^2 + 3B^2 = 2018/A, 3A^2+B^2 = 290/B.

Get rid of the fractions and subtract one from the other; it should be plain sailing from there.
Thanks so much! I subtracted them to get (A-B)^3 = 1728 so the answer is 1/(A-B) = 1/12.

I spent three days working on this problem and didn't think of substituting for 1/a and 1/b as you did. What was the motivation behind that?
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bluenotewitt
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#4
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#4
(Original post by Prasiortle)
Let a and b be real numbers such that 1/(a^2) + 3/(b^2) = 2018a
and 3/(a^2) + 1/(b^2) = 290b. What is (ab)/(b-a)?

By playing around with it, I was eventually able to get a cubic in (a/b), but the roots are extremely hideous, and so this method of solution isn't really practical, especially as this problem is taken from a non-calculator competition, and the final answer is supposed to be a fairly nice fraction, so there should be a nice solution. Any suggestions?
Here is another way in case you are interested:
First clear the denominators in both equations to obtain:

(i) b2 +3a2 = 2018a3 b2 and (ii) 3b2 +a2 =290b3 a2

These terms strongly remind me of a cubic, but we are not quite there yet. Multiplying (i) by b and (ii) by a should do the trick. We then have

(iii) b3 +3a2 b = 2018a3 b3 and (iv) 3b2 a +a3 =290b3 a3

and subtracting (iv) from (iii) then gives

(v) (b-a)3 = 2018a3 b3 - 290b3 a3 = 1728 a3 b3

From here on it should be clear how to finish off. I don't see a problem in taking the cube root of 1728 ( it's 12).

P.S.
I hope this wasn't the solution you were already aware of. If so... nevermind
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Prasiortle
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#5
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#5
(Original post by bluenotewitt)
Here is another way in case you are interested:
First clear the denominators in both equations to obtain:

(i) b2 +3a2 = 2018a3 b2 and (ii) 3b2 +a2 =290b3 a2

These terms strongly remind me of a cubic, but we are not quite there yet. Multiplying (i) by b and (ii) by a should do the trick. We then have

(iii) b3 +3a2 b = 2018a3 b3 and (iv) 3b2 a +a3 =290b3 a3

and subtracting (iv) from (iii) then gives

(v) (b-a)3 = 2018a3 b3 - 290b3 a3 = 1728 a3 b3

From here on it should be clear how to finish off.

I hope this wasn't the cubic you were referring to!
No, the cubic I was referring to was a cubic in (b/a), from which I thought I could find b/a = some constant, so b = some constant times a, then substitute to solve for a and b, and finally work out the required answer.

Anyway, thanks for your input.
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DFranklin
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#6
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#6
(Original post by Prasiortle)
Thanks so much! I subtracted them to get (A-B)^3 = 1728 so the answer is 1/(A-B) = 1/12.

I spent three days working on this problem and didn't think of substituting for 1/a and 1/b as you did. What was the motivation behind that?
To be honest just "clearing the fractions feels less intimidating, somehow". The fact that the desired result was also a "nice" expression in terms of A,B also made me feel it was worth pursuing.

I'm not sure a substitution of this sort makes a "real" difference (bluenotewitt found essentially the same solution without substituting), but sometimes it worth it just to get a form you're more comfortable with.
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