We can solve this by setting up a system of equations. We will use x for the $3 boxes and y for the $5 boxes:
x+y=34
3x+5y=130
We can use the elimination method by multiplying the first equation by 3:
3x+3y=102
3x+5y=130
0x-2y=-28
y=14
We can then plug this number back into one of the original equations to solve for x:
x+14=34
x=20
We can check these answers by plugging them into the equations:
20+14=34
34=34
(3)(20)+(5)(14)=130
60+70=130
130=130
Therefore, there are 20 of the $3 boxes, since we originally said that x would be the $3 boxes.
