A librarian bought 10 library books for a total of $138.62; how many
at $12.95 and how many at $15.99?
There are two ways to work it.
FIRST WAY (using one unknown)
Let x = the number bought at $12.95 and 10-x be the number bought at $15.99
Then
Total money spent for the x books at $12.95 each, [or 12.95x dollars]
PLUS
Total money spent for the 10-x at $15.99 each, [or 15.99(10-x) dollars]
EQUALS
Total money spent for all the books, [or 138.62 dollars]
12.95x + 15.99(10-x) = 138.62
Remove the decimals by multiplying through by 100
1295x + 1599(10-x) = 13862
Remove the parentheses using the distributive principle
1295x + 15990 - 1599x = 13862
Combine the two x-terms on the left
-304x + 15990 = 13862
Subtract 15990 from both sides
-304x = 13862 - 15990
-304x = -2128
Divide both sides by -304
x = (-2128)/(-304)
x = 7
So x = 7 is the number bought at $12.95 each.
Then substitute 7 for x in 10-x to find that 10-7 or 3 books were
bought ar $15.99.
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SECOND WAY (using two unknowns)
Let x = the number bought at $12.95 and y be the number bought at $15.99
Then the first equation coms from:
Number of $12.95 books [or s]
PLUS
Number of $15.99 books [or y]
EQUALS
Total number of books [or 10]
So the first equation is
X + y = 10
Then,
Total money spent for the x books at $12.95 each, [or 12.95x dollars]
PLUS
Total money spent for the 10-x at $15.99 each, [or 15.99y dollars]
EQUALS
Total money spent for all the books, [or 138.62 dollars]
12.95x + 15.99y = 138.62
So we have the system of two equations in two unknowns
x + y = 10
12.95x + 15.99y = 138.62
Solve that system and get x=7, y=3.
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Play tickets were $6 per adult and $4 per child. If $960 total
sales were for 180 tickets, how many were adult?
That's the same problem except put $6 and $4 in place of
$12.95 and $15.99, 180 in place of 10 and $960 in place of
$138.62
6x + 4(180-x) = 960
Solve that and get 120 adult tickets and 60 children tickets.
Edwin
AnlytcPhil@aol.com