<PRE><B>Question 1187739</B>
&lt;br&gt;
Let c be the rental cost for each chair
Let t be the rental cost for each table&lt;br&gt;
The total cost to rent 5 chairs and 2 tables is $18:  5c + 2t = 18&lt;br&gt;
The total cost to rent 3 chairs and 8 tables is $55.  3c + 8t = 55&lt;br&gt;
(1) Solving the problem informally, using logical reasoning....&lt;br&gt;
Since the cost of 5 chairs and 2 tables is $18, the cost of 20 chairs and 8 tables is 4*$18 = $72.
But the cost of 3 chairs and 8 tables is $55; the difference between the two cases is 20-3 = 17 more chairs for an additional $72-$55 = $17.  So the rental cost for each chair is $17/17 = $1.
And then since 5 chairs at $1 each and 2 tables cost $18 to rent, the cost for the 2 tables is $13, so the cost per table is $6.50.&lt;br&gt;
ANSWER: $1 per chair and $6.50 per table&lt;br&gt;
CHECK:
5c+2t = 5(1)+2(6.50) = 5+13 = 18
3c+8t = 3(1)+8(6.50) = 3+52 = 55&lt;br&gt;
(2) Using formal algebra....&lt;br&gt;
5c + 2t = 18
3c + 8t = 55&lt;br&gt;
Multiply the first equation by 4:
20c + 8t = 72&lt;br&gt;
Compare that equation to the second original equation:
17c = 17&lt;br&gt;
Solve to find the cost for each chair:
c = 17/17 = 1&lt;br&gt;
Use that cost per chair in the first original equation to find the cost per table:
5(1)+2t = 18
5+2t = 18
2t = 18-5 = 13
t = 13/2 = 6.5&lt;br&gt;
ANSWERS: c=1; t=6.5  --  i.e.  $1 per chair and $6.50 per table&lt;br&gt;
Observe that the solution using formal algebra uses EXACTLY the same calculations as the informal solution.&lt;br&gt;
But we need to understand the formal algebraic solution, because an informal solution will not be possible when the problems get a lot more complicated.&lt;br&gt;
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