Question 772284: There are eight chess sets. The least expensive is $5 and the most is$196. The price of the third is equal to the price of the first and second sets combined. The price of the fourth set equals the price of the second and third sets combined. The price of the fifth, sixth, seventh, and eighth sets also follow this rule. Each one equals the sum of the prices of the two previous.
I think it is 5+x=y?
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
Let $x = the cost of 2nd chess set.
The 1st chess set costs $5
The 2nd chess set costs $x
The 3rd chess set costs $(5+x) <-- add the previous two, $5+$x
The 4th chess set costs $(5+2x) "
The 5th chess set costs $(10+3x) "
The 6th chess set costs $(15+5x) "
The 7th chess set costs $(25+8x) "
The 8th chess set costs $(40+13x) "
And since the 8th chess set is the
most expensive and is given to be
$196, we have the equation
$(40+13x) = $196
40+13x = 196
13x = 156
x = 12
So their prices are
$5, $12, $17, $29, $46, $75, $121, $196
Edwin
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