Question 1172900
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The first given rental is for 2 chairs and 5 tables, for a cost of $33.  The second is for 8 chairs and 3 tables, for a cost of $30.<br>
Observe that the second rental has 4 times as many chairs as the first.<br>
So imagine a third rental that is 4 times the first; it has 8 chairs and 20 tables, for a total of $132.<br>
Now the second and third rentals have the same numbers of chairs; the second had 3 tables and the third has 20 tables.  The difference in the number of tables is 17; and the difference in cost is $132-$30 = $102.  That means the cost of each table is $102/17 = $6.<br>
So the first rental of 2 chairs and 5 tables, for a cost of $33, includes 5($6) = $30 for the tables.  So the two chairs cost $3, and then each chair costs $1.50.<br>
ANSWERS: $6 for each table; $1.50 for each chair.<br>
A solution using formal algebra uses exactly the same calculations as the informal solution above.<br>
2c+5t = 33 (1)
8c+3t = 30 (2)<br>
8c+20t = 132 (3) [(1), multiplied by 4]<br>
17t = 102  [the difference between (2) and (3)
t = 6<br>
2c+5(6) = 33
2c = 3
c = 1.5<br>