the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were double and if the width were decreased by 1 inch, the area would be increased by 66inches ^2. What are the length and width of the rectangle?
Let w = width and w+3 = length.
2(w+3)(w-1) - w(w+3) = 66
Solve for w, then w+3.
Can someone please explain what the -w(w+3) in the solution stands for?
To solve this problem, we can start by assigning variables to the length and width of the rectangle. Let's use "L" for length and "W" for width.
We are given that the length is 3 inches more than the width. Therefore, we can write the equation: L = W + 3.
We are also given that if the length were double and the width were decreased by 1 inch, the area would be increased by 66 square inches. This means that (2L)(W - 1) = LW + 66.
Now we can substitute the value of L from the first equation into the second equation to solve for W.
(2(W + 3))(W - 1) = (W + 3)W + 66.
Simplifying this equation gives us:
(2W + 6)(W - 1) = W^2 + 3W + 66.
Expanding the equation further gives us:
2W^2 - 2 + 6W - 6 = W^2 + 3W + 66.
Combining the like terms, we have:
2W^2 + 4W - 8 = W^2 + 3W + 66.
Subtracting W^2 and 3W from both sides gives us:
W^2 + W - 74 = 0.
Now we have a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. In this case, the equation can be factored as:
(W - 7)(W + 11) = 0.
Setting each factor equal to zero gives us two possible values for W:
W - 7 = 0 --> W = 7
W + 11 = 0 --> W = -11
Since the width cannot be negative, we discard the value W = -11.
Therefore, the width of the rectangle is W = 7.
Substituting this value back into the first equation, we find the length:
L = W + 3 = 7 + 3 = 10.
So, the length of the rectangle is L = 10 inches and the width is W = 7 inches.