<PRE><B>Question 1184380</B>
&lt;br&gt;
Informally, using logical reasoning and common sense....&lt;br&gt;
Since the cost of each cup is half the cost of each bowl, the cost of 2 cups is the same as the cost of 1 bowl.  So buying 2 cups and 3 bowls costs the same as buying 1+3=4 bowls.&lt;br&gt;
Since she paid $112, the cost of each bowl was $112/4 = $28.&lt;br&gt;
ANSWER: $28 for each bowl&lt;br&gt;
With formal algebra....&lt;br&gt;
b = # of bowls
c = # of cups&lt;br&gt;
3b+2c=112  [1]  the cost of 3 bowls and 2 cups was $112
c=(1/2)b  [2]  each cup costs half as much as each bowl&lt;br&gt;
Substitute [2] into [1]:&lt;br&gt;
3b+2((1/2)b) = 112
3b+b = 112
4b = 112
b = 112/4 = 28&lt;br&gt;
As you see, the formal algebraic solution follows exactly the same path as the informal one.&lt;br&gt;
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