Final answer:
To find the original dimensions of a rectangle given the length is twice the width minus 7 meters, and with changes in dimensions, the new perimeter is 66 meters; we use algebraic expressions and solve for the original width and length.
Step-by-step explanation:
Understanding Rectangle Dimensions
To solve for the original dimensions of the rectangle, let's let the width be represented by w and the length as L. According to the problem, the original length of the rectangle is 7m less than twice its width. This can be written as:
L = 2w - 7
After decreasing the length by 1m and the width by Am, the new dimensions become L - 1 for the length and w - A for the width. The problem then states that the perimeter is 66m, which can be expressed as:
2(L - 1) + 2(w - A) = 66
Now we can substitute the expression for L into the perimeter equation:
2((2w - 7) - 1) + 2(w - A) = 66
Simplifying, we get:
4w - 16 + 2w - 2A = 66
And further simplifying:
6w - 2A = 82
However, we are not given the value of A, we might assume A is a typo and should also be 1. This would yield:
6w - 2 = 82
Solving for w gives:
w = 14 m
Substituting back into the length equation:
L = 2(14) - 7 = 21 m
Therefore, the original dimensions of the rectangle are 14 meters in width and 21 meters in length.