To solve this problem, set up and solve a system of equations. The variables b and m will represent a bread loaf and milk jug, respectively:
[tex] \left \{ {{3b + m = 9} \atop {b + 2m =5.5}} \right. [/tex]
I would solve using substitution. Take one of the equations and set it equal to one of the variables, for example:
[tex]3b + m =9 \\ m = 9 -3b[/tex]
Now, plug this into the other equation for m and solve for b:
[tex]b + 2m =5.5 \\ b + 2(9-3b)= 5.5 \\ b + 18 - 6b = 5.5 \\ -5b + 18 = 5.5 \\ -5b = -12.5 \\ b = 2.5[/tex]
We now know that a loaf of bread costs $2.50. Plug this value in for b in the first equation and solve for m:
[tex]3b + m = 9 \\ 3(2.5) + m = 9 \\ 7.5+m = 9 \\ m = 1.5[/tex]
One jug of milk costs $1.50 and one loaf of bread costs $2.50.
