 View Question
Q: SYSTEM OF EQUATIONS, ALGEBRA, DIFFERENTIAL EQUATIONS ( No Answer,   2 Comments ) Question
 Subject: SYSTEM OF EQUATIONS, ALGEBRA, DIFFERENTIAL EQUATIONS Category: Science > Math Asked by: baroqqque-ga List Price: \$60.00 Posted: 19 Oct 2006 05:54 PDT Expires: 18 Nov 2006 04:54 PST Question ID: 774994
 ```Thanks for taking the time to check this. I have been struggling to solve the following system of equations: ma+ (ma^(ra/rb))*(1-(1/rb)-y)*(y^(1/rb))*((1-y)^(-1-(1/rb))) = (mb^(rb/ra))* (x^((1/ra)-1))*((1-x)^(-1-(1/ra)))*(1/(ra da)) and the symmetric equation mb+ (mb^(rb/ra))*(1-(1/ra)-x)*(x^(1/ra))*((1-x)^(-1-(1/ra))) = (ma^(ra/rb))* (y^((1/rb)-1))*((1-y)^(-1-(1/rb)))*(1/(rb db)) where 0 There is no answer at this time. ```I don't understand the notations: 1/(ra da) in the RHS of the first equation, and 1/(rb db) in the RHS of the second equation. Although the Subject of the Question refers to differential equations, I don't see how a differential expression, if that's what these are, can be combined with the rest of the equations. regards, mathtalk-ga```
 ```Perhaps 1/(ra*da) and 1/(rb*db) are equivalents, as da,db are described as "parameters". I think this means da,db are given scalar quantities, as are ra,rb. -- mathtalk-ga``` 