The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1 in., the area would be increased by 176 in.squared. What are the length and width of the rectangle?
HELP HAS TO BE DONE BY 12
To solve this problem, we can begin by setting up equations based on the given information. Let's denote the width of the rectangle as "w" and the length as "l".
According to the given information, we know that the length is 3 inches more than the width, so we can write the equation:
l = w + 3
We are also given that if the length is doubled and the width is decreased by 1 inch, the area of the rectangle is increased by 176 square inches. The original area can be calculated as:
Area = length × width = l × w
If the length is doubled and the width is decreased by 1 inch, the new dimensions are (2l) and (w - 1). The new area can be calculated as:
New Area = (2l) × (w - 1)
Since the new area is increased by 176 square inches, we can write the equation:
New Area - Original Area = 176
Substituting the values for the area, length, and width, the equation becomes:
(2l × (w - 1)) - (l × w) = 176
Now, we can solve the system of equations.
Substitute l = w + 3 into the equation:
(2(w + 3) × (w - 1)) - ((w + 3) × w) = 176
Simplifying this equation, we have:
(2w + 6) × (w - 1) - (w^2 + 3w) = 176
Expanding and combining like terms, the equation becomes:
2w^2 - 2 + 6w - 6 - w^2 - 3w = 176
Combine the like terms:
w^2 + w - 2 = 176
Rearranging the equation:
w^2 + w - 178 = 0
Now, we can solve this quadratic equation. Using factoring, completing the square, or the quadratic formula, we find that the solutions are w = -18 and w = 9.
Since the width cannot be negative, we discard the negative solution.
Therefore, the width of the rectangle is 9 inches.
Using the previously defined equation, we can find the length:
l = w + 3
l = 9 + 3
l = 12
Therefore, the length of the rectangle is 12 inches and the width is 9 inches.