The dimensions of a rectangle are such that it's length is 9 in. more than its width. If the length were doubled and if the width were decreased by 4 in., the area would be increased by 110 in^2. What are the length and width of the rectangle?

L=9+W

Area=LW
Area+110=(2L(W-4))

so to solve for L, W substiture...

LW+110= 2(L)(W-4)
and L=9+W
(9+W)W+110=2(9+W)(W-4)

so multiply that out, you will have a quadratic in W. You can solve it with factoring, or the quadratic equation. Once you have W, calculate L

Let's suppose the width of the rectangle is 'x' inches.

According to the problem, the length is 9 inches more than the width, so the length would be 'x + 9' inches.

The area of a rectangle is given by the formula length multiplied by width, so the area of the rectangle in its original state would be 'x(x + 9)' square inches.

Now, if the length is doubled, it becomes '2(x + 9)' inches, and if the width is decreased by 4 inches, it becomes '(x - 4)' inches. The area of this new rectangle is given by the formula length multiplied by width, and it is increased by 110 square inches. So, we can set up the equation:

2(x + 9)(x - 4) = x(x + 9) + 110

Now, let's solve this equation step by step:

Expanding the left side of the equation:
2(x^2 + 5x - 36) = x^2 + 9x + 110

Simplifying the left side:
2x^2 + 10x - 72 = x^2 + 9x + 110

Rearranging the equation to bring all terms to one side:
2x^2 + 10x - 72 - x^2 - 9x - 110 = 0

Combining like terms:
x^2 + x - 182 = 0

Now, let's factorize this quadratic equation:

The factors of -182 that add up to +1 are +14 and -13. So, we can rewrite the equation as:

(x + 14)(x - 13) = 0

Setting each factor equal to zero and solving for x:

x + 14 = 0 or x - 13 = 0

x = -14 or x = 13

Since the width cannot be negative, we ignore the solution x = -14.

Therefore, the width of the rectangle is 13 inches.

Now, we can calculate the length by adding 9 inches to the width:

Length = 13 + 9 = 22 inches

Thus, the length and width of the rectangle are 22 inches and 13 inches, respectively.

To solve this problem, we need to set up and solve a system of equations based on the given information.

Let's assume the width of the rectangle is W inches.

According to the problem, the length is 9 inches more than the width, so the length would be (W + 9) inches.

The area of the rectangle is given by the formula: Area = Length × Width

So, the original area of the rectangle would be: Original Area = W × (W + 9)

Now, let's consider the second scenario where the length is doubled and the width is decreased by 4 inches. In this case, the new length would be 2(W + 9) inches, and the new width would be (W - 4) inches. The new area of the rectangle can be calculated as: New Area = 2(W + 9) × (W - 4)

According to the problem, the new area is 110 square inches more than the original area. So we can set up an equation:

New Area - Original Area = 110

[2(W + 9) × (W - 4)] - [W × (W + 9)] = 110

Now, let's solve this equation step by step:

2(W + 9)(W - 4) - W(W + 9) = 110
2(W^2 + 5W - 36) - (W^2 + 9W) = 110
2W^2 + 10W - 72 - W^2 - 9W = 110
W^2 + W - 182 = 0

Now we have a quadratic equation. Let's solve it by factoring or using the quadratic formula.

Factoring the quadratic equation: (W + 14)(W - 13) = 0

So, either (W + 14) = 0 or (W - 13) = 0

If (W + 14) = 0, then W = -14 (which is not possible for the width of a rectangle)

If (W - 13) = 0, then W = 13

Therefore, the width of the rectangle is 13 inches.

Using the information we obtained earlier, the length would be (W + 9) = (13 + 9) = 22 inches.

So, the dimensions of the rectangle are 13 inches (width) and 22 inches (length).