Question 158222
you don't have enough information to solve this for one unique solution.
you have multiple possible solutions based on the fact that a common multiple of $1.25 and $3.00 is $15.00
at $15.00, you can sell 12 beverages at $1.25 or you can sell 5 sandwiches at $3.00
some examples of possible solutions are:
solution 1:
44 sandwiches and 2 beverages = 44*$3.00 + 2*1.25 = $132.00 + $2.50 = $134.5
to get other solutions, all you have to do is subtract 5 sandwiches and add 12 beverages each time.
solution 2:
39 sandwiches and 14 beverages = $117 + $17.5 = $134.5
solution 3:
34 sandwiches and 26 beverages = $102 + $32.5 = $134.5
etc................................
normally this type of problem says:
the number of beverages and sandwiches sold in total is 60 (for example)
each beverage costs $1.25 and each sandwich costs $3.00
now you have enough information.
you would then set up your formula to say
s + b = 60 (first equation)
s*3 + b*1.25 = 134.5 (second equation)
to solve you would find s in terms of b, or b in terms of s.
solving in the first equation for s, we get s = 60-b
substituting 60-b for s in the second equation, we get
3*(60-b) + 1.25*b = 134.5
solving for b, this equation becomes
180 - 3*b + 1.25 * b = 134.5 which becomes
180 - 134.5 = 3*b - 1.25*b which becomes
45.5 = 1.75*b  which becomes
b = 45.5/1.75 = 26
if this looks familiar it's because i took the sum of sandwiches sold and beverages sold from solution 3 above.
if s + b = 60, then s = 60 - 26 = 34, so
s = 34, and
b = 26
solving in the second equation, we get
34*3 + 26*1.25 = $102 + $32.5 = $134.5 
this confirms the answer as correct.