math

x+y+3z=600
x+y+z=400

1. In the system of equations above, what is the value of x+y?

2. There are 25 trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If 15 of the trays contain cups and 21 of the trays contain plates, how many contain both a cup and a plate?

3. If 6<lx-3l<7 and x<0, what is one possible value of lxl ?

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asked by Julia
  1. 1.
    For
    (x+y)+3z=600
    (x+y)+z=400
    solve for p in
    p+3z=600
    p+z =400
    From which you'll get z=100, and therefore p=(x+y)=?

    2.
    Hint:
    |C|=15 (cardinality of C = number of elements in the set)
    |P|=21
    We also know that the cardinality of the union of the two sets is 25, i.e.
    |C∪P|=25

    To find |C∩P| (number of trays containing both), we use the principle of inclusion and exclusion:
    |C∪P|=|C|+|P|-|C∩P| ...(1)
    Substitute the above values into (1) and solve for |C∩P|.

    3.
    Hint: for
    6<|x-3|<7
    to be satisfied, x cannot be an integer, so there are many possible answers within a certain range.
    Also, x<0 means x takes on a negative value.
    First find out what |x-3| should be, and then solve for x.

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  2. 300

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    posted by Louis

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