A Difficult simultaneous Equation.

Separate the variables completely
You have to completely separate the variables in one equation to substitute into the other one. You are on the right track by converting the equations into

1) 2ab = 2b - 3a
2) b^2 - ab - a^2 = 5a^2b^2

Now, you need to solve equation one for a

1.1) 2ab + 3a =2b
1.2) a(2b+3) = 2b
1.3) a = 2b / (2b + 3)

Now you can substitute 1.3 into 2 (and you get a mess, but it simplifies down). After a lot of simplifying, you end up with a quadratic that you can solve. Post back if you need further help.

It may be easier to come up with the quadratic if you substitute for ab

ab=(2b=3a)/2 from 2=2/a=5/b

into

(b-a)(b+a)=ab(5ab+1) from 1/a^2-1/(ab)=1/b^2=5

But it looks like you're going in the right direction.

Rap

I think I managed to solve for a and b. Check me please.


Part 1


Part 2



So finally a = 1/4
b = 2/14


Is that what you got?

I've not read the solutions above, but if you substitute p=1/a and q=1/b you get a quadratic in either p or q
whence p=7+or- 2sqrt(10)

(NB I dont think the your solution satisfies the first equation).

LATER EDIT.

My solution seems to work.
(a,b) = (.075, 0.12) or (1.47,-4.63) appx

Congrats


You posted them backwards (b=1/4, a=1/7), but yes, those work. Fresco's idea of substituting p=1/a and q=1/b is also a very good idea. Take the recipricols out of the problem and I bet you would have blazed through this problem.

Yes that rational solution does work. Are there more than 2 solutions ? I didn't seem to be getting rational numbers....let me check it out.

Ah yes....mistake in my cross multiplication.

P^2 -14p+49=0
(p-7)(p-7)=0
p=7

q=(14-2)/3
q=4

So there is one solution...interesting !

Yeah b = 2/14 and a = 1/4 My mistake. I'm going to try your way out too fresco. Thanks for your help.