1. Three times the width of a certain rectangle exceeds twice its length by two inches. Four times its length is twelve more than its perimeter. Write a system of equations that could be used to solve this problem. (hint: P = 2L + 2W)
2. Your mother has left you in charge of the annual family yard sale. Before she leaves you to your entrepreneurial abilities, she explains that she has made the job easy for you: everything costs either $1.50 or $3.50. She asks you to keep track of how many of each type of item is sold, and you make a list, but it gets lost sometime throughout the day. Just before she’s supposed to get home, you realize that all you know is that there were 150 items to start with (your mom counted) and you have 41 items left. Also, you know that you made $227.50. Write a system of equations that you could solve to figure out how many of each type of item you sold.
#1. What's the trouble? Just write in math what they have said in words.
3W = 2L+2
4L = 12+2(W+L)
You can rearrange these as you see fit.
#2. If you sold x items at $1.50 and y at $3.50, then
x+y = 150-41
1.50x + 3.50y = 227.50
1. Let's assume the width of the rectangle is W and the length is L.
According to the given information:
- Three times the width of the rectangle exceeds twice its length by two inches: 3W = 2L + 2
- Four times its length is twelve more than its perimeter: 4L = 2L + 2W + 12
So the system of equations can be written as:
1st equation: 3W = 2L + 2
2nd equation: 4L = 2L + 2W + 12
2. Let's assume the number of items costing $1.50 is x, and the number of items costing $3.50 is y.
According to the given information:
- The total number of items at the start was 150: x + y = 150
- The number of items remaining is 41: x + y - (number of items sold) = 41
- The sales made $227.50: 1.50x + 3.50y = 227.50
So the system of equations can be written as:
1st equation: x + y = 150
2nd equation: x + y - (number of items sold) = 41
3rd equation: 1.50x + 3.50y = 227.50
1. Let's define the variables for this problem:
- Let W represent the width of the rectangle
- Let L represent the length of the rectangle
Given the information in the problem, we can set up the following equations:
1) "Three times the width of a certain rectangle exceeds twice its length by two inches":
3W = 2L + 2
2) "Four times its length is twelve more than its perimeter":
4L = 2L + 2W + 12
These two equations form a system that can be used to solve for the width (W) and length (L) of the rectangle.
2. Let's define the variables for this problem:
- Let x represent the number of items that cost $1.50
- Let y represent the number of items that cost $3.50
Given the information in the problem, we can set up the following equations:
1) "there were 150 items to start with":
x + y = 150
2) "you have 41 items left":
x + y = 41
3) "you made $227.50":
1.50x + 3.50y = 227.50
These three equations form a system that can be used to solve for the number of items that cost $1.50 (x) and the number of items that cost $3.50 (y).