Question

Pls help! Solving for dimenions

Answer 1

The dimensions are 17 inches (length) by 4 inches (width).

Explanation:

W = Width

L = Length

Since the problem says that the length, L, equals 9 more inches than 2 times its width, the equation would be:

L = 9+2*W

This would be the same as:

L = 2W + 9

The formula for the area of a rectangle is:

L*W = Area

In the problem, we are given that the area equals 68 inches, so after plugging in the variables for the equation, we get:

(2W+9) * (W) = 68

Then we distribute:

2W^2 + 9W = 68

Then we set it equal to zero:

2W^2 + 9W - 68 = 0

Then we factor it out:

(2W+17) (W-4) = 0

We set each part equal to zero:

2W +17 = 0

2W = -17

W = -17/2

W-4 = 0

W = 4

Since we know that the lengths can only be positives, we disregard the negative solution. Therefore, W, the width, is equal to 4.

We then plug it into the equation to solve for length.

L = 2(4) + 9

L = 17

Then we plug in the lengths and widths to the solution. (FYI: it is typically written as length x width.)

We get:

The dimensions are 17 inches by 4 inches.

Answer 2

17 inches, 4 inches

Explanation:

Let the width = x.

Then the length is 2x + 9.

area = length × width

area = (2x + 9)x

area = 2x² + 9x

area = 68

2x² + 9x = 68

2x² + 9x - 68 = 0

2 × 68 = 136

136 = 2³ × 17

8 × 17 = 136

17 - 8 = 9

2x² + 17x - 8x - 68 = 0

x(2x + 17) - 4(2x + 17) = 0

(x - 4)(2x + 17) = 0

x = 4 or x = -17/2

2x + 9 = 8 + 9 = 17

The length is 17 inches, and the width is 4 inches.