You are shopping for single-use cameras to hand out at a party. The daylight cameras cost $2.75 and the flash cameras cost $4.25. you must buy exactly 20 cameras and you want to spend between $65 and $75, inclusive. Write and solve a compound inequality for this situation. Then list all the solutions that involve whole numbers of cameras
To find a compound inequality that represents this situation, we can set two conditions: one for the total cost and another for the number of cameras.
Let's start with the condition for the total cost. The cost of each daylight camera is $2.75, and the cost of each flash camera is $4.25. To ensure that the total cost falls between $65 and $75, we can write the following inequality:
65 ≤ (2.75x + 4.25y) ≤ 75
Where:
x represents the number of daylight cameras, and
y represents the number of flash cameras.
Now let's establish the condition for the number of cameras. You want to buy exactly 20 cameras, so we can write the equation:
x + y = 20
To list all the solutions involving whole numbers of cameras, we can substitute values for either x or y and solve for the other variable. Here, we'll solve for y as a function of x:
y = 20 - x
Now we can substitute this expression for y in the compound inequality:
65 ≤ (2.75x + 4.25(20 - x)) ≤ 75
Let's simplify the inequality:
65 ≤ (2.75x + 85 - 4.25x) ≤ 75
65 ≤ (-1.5x + 85) ≤ 75
65 - 85 ≤ -1.5x ≤ 75 - 85
-20 ≤ -1.5x ≤ -10
To solve for x, divide all parts of the inequality by -1.5 and reverse the direction of the inequalities:
10/1.5 ≥ x ≥ 20/1.5
6.67 ≥ x ≥ 13.33
Since we're looking for whole numbers, the solutions for x are 7, 8, 9, 10, 11, 12, and 13.
By substituting these values back into the equation y = 20 - x, we can find the corresponding values for y:
For x = 7, y = 13
For x = 8, y = 12
For x = 9, y = 11
For x = 10, y = 10
For x = 11, y = 9
For x = 12, y = 8
For x = 13, y = 7
Therefore, the solutions that involve whole numbers of cameras are:
7 daylight cameras and 13 flash cameras,
8 daylight cameras and 12 flash cameras,
9 daylight cameras and 11 flash cameras,
10 daylight cameras and 10 flash cameras,
11 daylight cameras and 9 flash cameras,
12 daylight cameras and 8 flash cameras, and
13 daylight cameras and 7 flash cameras.
To solve this problem, we need to set up a compound inequality.
Let x represent the number of daylight cameras, and y represent the number of flash cameras.
The cost of daylight cameras is $2.75, so the total cost of daylight cameras is 2.75x.
The cost of flash cameras is $4.25, so the total cost of flash cameras is 4.25y.
We know that the total number of cameras (x + y) must be exactly 20.
The total cost of the cameras must be between $65 and $75, inclusive.
Combining all the information, we can write the compound inequality as:
65 ≤ 2.75x + 4.25y ≤ 75
To determine the whole number solutions, we can consider the possible values for x and y that satisfy the compound inequality.
Let's calculate the possible solutions step-by-step:
Step 1: Substitute different values of x and solve for y.
For x = 0:
65 ≤ 4.25y ≤ 75
15.29 ≤ y ≤ 17.64
Since we require whole numbers of cameras, y cannot have values between 15.29 and 17.64.
For x = 1:
65 - 2.75 + 4.25 ≤ 4.25y ≤ 75 - 2.75 + 4.25
66.25 ≤ 4.25y ≤ 76.25
15.59 ≤ y ≤ 17.94
Since we require whole numbers of cameras, y cannot have values between 15.59 and 17.94.
For x = 2:
65 - 2.75(2) + 4.25 ≤ 4.25y ≤ 75 - 2.75(2) + 4.25
67.5 ≤ 4.25y ≤ 77.5
15.88 ≤ y ≤ 18.24
Since we require whole numbers of cameras, y cannot have values between 15.88 and 18.24.
For x = 3:
65 - 2.75(3) + 4.25 ≤ 4.25y ≤ 75 - 2.75(3) + 4.25
68.75 ≤ 4.25y ≤ 78.75
16.18 ≤ y ≤ 18.54
Since we require whole numbers of cameras, y cannot have values between 16.18 and 18.54.
For x = 4:
65 - 2.75(4) + 4.25 ≤ 4.25y ≤ 75 - 2.75(4) + 4.25
70 ≤ 4.25y ≤ 80
16.47 ≤ y ≤ 18.83
For y to be a whole number, the solutions within this range are y = 17 and y = 18.
Hence, the solutions involving whole numbers of cameras are:
(4 daylight cameras, 17 flash cameras)
(4 daylight cameras, 18 flash cameras)