# System of equations

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Prasiortle

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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?

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

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?

**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?

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

(Original post by

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.

**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.

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

**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?

First clear the denominators in both equations to obtain:

(i) b

^{2}+3a

^{2}= 2018a

^{3}b

^{2 }and (ii) 3b

^{2}+a

^{2}=290b

^{3}a

^{2 }

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) b

^{3}+3a

^{2}b = 2018a

^{3}b

^{3 }and (iv) 3b

^{2}a +a

^{3}=290b

^{3}a

^{3 }

and subtracting (iv) from (iii) then gives

(v) (b-a)

^{3}= 2018a

^{3}b

^{3}- 290b

^{3}a

^{3 }= 1728 a

^{3}b

^{3 }

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

(Original post by

Here is another way in case you are interested:

First clear the denominators in both equations to obtain:

(i) b

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) b

and subtracting (iv) from (iii) then gives

(v) (b-a)

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

I hope this wasn't the cubic you were referring to!

**bluenotewitt**)Here is another way in case you are interested:

First clear the denominators in both equations to obtain:

(i) b

^{2}+3a^{2}= 2018a^{3}b^{2 }and (ii) 3b^{2}+a^{2}=290b^{3}a^{2 }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) b

^{3}+3a^{2}b = 2018a^{3}b^{3 }and (iv) 3b^{2}a +a^{3}=290b^{3}a^{3 }and subtracting (iv) from (iii) then gives

(v) (b-a)

^{3}= 2018a^{3}b^{3}- 290b^{3}a^{3 }= 1728 a^{3}b^{3 }From here on it should be clear how to finish off.

I hope this wasn't the cubic you were referring to!

Anyway, thanks for your input.

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DFranklin

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#6

(Original post by

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?

**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?

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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