A rectangle's length is 6 less than twice its width. The perimeter of the rectangle is 36 units. Set up and solve a system of equations to determine the length and width.
P = 2L + 2W
36 = 2(2W - 6) + 2W
36 = 6W - 12
24 = 6W
? = W
To solve this problem, we can set up a system of equations based on the given information.
Let's consider the width of the rectangle as "w" units.
According to the problem, the length of the rectangle is 6 less than twice its width. This can be expressed as:
Length = 2w - 6
The perimeter of a rectangle can be calculated as the sum of the lengths of all four sides. In this case, the perimeter is given as 36 units. Since a rectangle has two sides of length "w" and two sides of length "2w - 6", we can set up an equation to represent the perimeter:
Perimeter = 2w + 2(2w - 6) = 36
Now, we can solve the system of equations by substituting the value of the perimeter equation into the length equation:
2w + 2(2w - 6) = 36
Simplifying:
2w + 4w - 12 = 36
6w - 12 = 36
Adding 12 to both sides:
6w = 48
Dividing by 6:
w = 8
Now that we have the value for "w," we can substitute it back into the length equation to find the length:
Length = 2w - 6
Length = 2(8) - 6
Length = 16 - 6
Length = 10
Therefore, the width of the rectangle is 8 units, and the length is 10 units.