A Weirdly Defined Function

The function is defined for all real $x$ , so we can substitute $x$ as anything and the equation will still hold.

I will plug in $x=10$ , because I'm looking for $f(10)$ . According to the equation,

$f(10)+2f(a-10)=60$

I will also plug in $x=a-10$ , in the hopes that the $2f(a-x)$ will simplify to $f(10)$ . The equation then becomes,

$f(a-10)+2f(10)=6a-60$

or

$2f(a-10)+4f(10)=12a-120$

These simultaneous equations

$f(10)+2f(a-10)=60$ $4f(10)+2f(a-10)=12a-120$

And

$3f(10)=12a-180$

or

$f(10)=4a-60$

So $m=4$ and $n=60$ , so $m+n=\boxed{64}$

let, f(x)= kx+ C, C= Constant number so, f(x)+ 2 f(a- x)= 6 x= k x+ C+ 2 (k (a- x)+ C)= 6 x 2 a k- k x+ 3 C= 6 x k= -6, C= 4 a, f(x)= -6 x+ 4 a so, f(10)= -6* 10+ 4 a= 4 a- 60= a*m- n so, m= 4, n= 60 m+ n= 4+ 60= 64