Question 1192262
<br>
First an informal solution using logical reasoning....<br>
(1) take away the one "extra" meat, leaving 14 items for a total of $46, with equal numbers of meat and cheese.
(2) The numbers of meat and cheese are the same, and the number of items remaining is 14; that means the number of breads is even.
(3) Since the number of breads is even, make groups of 2 breads each, with a cost of 2($3) = $6 per group.  And since the numbers of meats and cheeses are the same, make groups of one meat and one cheese, with a cost of $5+$2 = $7 per group.
(4) So we need to make a total of $46 using groups of $6 and $7.  There is only one way to do this: 3 groups at $6 each and 4 groups at $7 each for a total of 3($6)+4($7) = $18+28 = $46.<br>
So we have 4 groups of one meat and one cheese, plus the "extra" meat we accounted for at the beginning; so the number of meat you are to purchase is 5.<br>
ANSWER: 5 meats<br>
CHECK: 6($3)+5($5)+4($2) = $18 + $25 + $8 = $51<br>
A formal algebraic solution is easier than the informal one....  But solving a problem like this using logical reasoning is good brain exercise.<br>
A typical formal algebraic solution....<br>
x = # of cheese
x+1 = # of meat (one more than the number of cheese)
15-(x+x+1) = 14-2x = # of bread (the total number of items, 15, minus the total number of meat and cheese)<br>
The total cost is $51:<br>
2x+5(x+1)+3(14-2x)=51
2x+5x+5+42-6x=51
x=4<br>
ANSWER: the number of meat to purchase is x+1 = 5<br>