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the calculus of functions. 
them equal by pairs, the result shall be given : a particular 
case would be the equation 
F { x,y, z, v, w, r, + (x,y, z, v,w, r, &c. } = F { x, v, r } i=t> J 
Instead of making y equal to x, and % to v and so on ,y might 
become a given function of x, and % a given function of v> 
&c. thus : ___ 
Y{x,y i z,v,w,r,^{x,y,%,v y zv,rf^c.^=¥[x ) v i r, 1 \ j 
a few examples will sufficiently explain the method to be 
pursued in treating these equations. 
Problem XXVIII. 
Given the equation 
4' 3 ' 1 Cj = t3 
This Problem, of which without the condition ofy being made 
equal to x we could not find even a particular case, readily 
admits of solution in its present state. Since i] is 
only equal to ^ r » 3 f,y), wheny = x we may put for the given 
equation 
4* 3j 1 (x,y) = ? { y)>$ x > 2 (*>y) } (0 
provided that when y is equal to x, the latter side of the equation 
shall become 4 1 ’ 2 (a?,y ), and this fully satisfies the condition of 
the Problem. If, therefore, we can find such a value of <p, 
the equation ( 1 ) may be treated as a common functional equa- 
tion of two variables, and may be solved by the rules already 
given. 
Nor is it at all difficult to find such a value of <p ; if we make 
4* 1 ’ 2 (<xqy)==%, <p must be such a function that 
<p(ayy, %)===* D==<1 
