Question 1135529
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The other response you have so far shows one of many good algebraic solutions.<br>
If an algebraic solution is not required, you can get some good brain exercise solving the problem with logical reasoning.<br>
Suppose the problem were posed slightly differently so that the two girls originally planned to buy 2 hamburgers and 1 carton of milk, using the whole $5, but then they decided instead to get 2 cartons of milk and one hamburger, saving 40 cents.<br>
So taking away that 2nd burger and replacing it with a 2nd carton of milk changed the total cost downwards by 40 cents.  That means the hamburger cost 40 cents more than the carton of milk.<br>
Now use that knowledge, along with either version of the order, to determine the cost of each hamburger and each carton of milk.<br>
The second version of their order was 1 burger and 2 cartons of milk, for a total of $4.60.  The burger cost 40 cents more than each carton of milk; that means 3 cartons of milk would cost $4.60-$0.40 = $4.20.  And that makes the cost of each carton of milk $4.20/3 = $1.40.<br>
Then you can find the cost of each burger using that in either of the original orders.<br>
And here is another solution by logical reasoning that works nicely for this problem.<br>
We know...
(1) 2 burgers and 1 milk cost $5.00
(2) 1 burger and 2 milks cost $4.60<br>
From this we can see that exchanging a burger for a milk reduces the total cost by $0.40; or exchanging a milk for a burger increases the cost by $0.40.<br>
But that means<br>
(3) 3 burgers and no milk would cost $5.00+$0.40 = $5.40 -- making the cost of each burger $5.40/3 = $1.80; and
(4) 0 burgers and 3 milk would cost $4.60-$0.40 = $4.20 -- making the cost of each milk $4.20/3 = $1.40