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 ?

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.

300

1. To find the value of x+y, we can solve the system of equations using the method of elimination. Subtracting the second equation from the first equation, we get:

(x+y+3z) - (x+y+z) = 600 - 400
2z = 200
z = 100

Substituting the value of z into the second equation, we get:

x + y + 100 = 400

Rearranging this equation, we have:

x + y = 300

Therefore, the value of x+y is 300.

2. Let's denote the number of trays containing both a cup and a plate as "a".

From the given information, we know that:
- The number of trays containing cups = 15
- The number of trays containing plates = 21
- The total number of trays = 25

Using the principle of inclusion-exclusion, we can determine the number of trays containing both a cup and a plate:

Number of trays containing both a cup and a plate = Number of trays containing cups + Number of trays containing plates - Total number of trays
a = 15 + 21 - 25
a = 36 - 25
a = 11

Therefore, there are 11 trays that contain both a cup and a plate.

3. Let's break down the given inequality:

6 < |x - 3| < 7

For the interval 6 < |x - 3| < 7 to hold true, we need two conditions to be satisfied simultaneously:

1) 6 < x - 3 < 7
2) 6 < - (x - 3) < 7

First, solve condition 1:

6 < x - 3 < 7
9 < x < 10

Next, solve condition 2:

6 < -(x - 3) < 7
6 < -x + 3 < 7
3 < -x < 4
-4 < x < -3

Since x < 0, we can consider the intersection of the two intervals:

-4 < x < -3

One possible value of |x| (absolute value of x) within this range is:

|x| = x when x < 0

Therefore, one possible value of |x| is |-3| = 3.