Question

The length of a rectangle is 5yd more then doubled the width and the area of the rectangle is 63. what is the length and width?

Answer 1

Length: 14 yd

Width: 4.5 yd

Explanation:

Let's use the variable w to represent the width of the rectangle.

We are given that the length of the rectangle is 5yd more than doubled the width, so we can represent the length as 2w+5.

We are also given that the area of the rectangle is 63.

We know that the area of a rectangle is equal to the length multiplied by the width, so we can write the following equation:

(2w+5) * w = 63

We can then solve for w:

2w²+ 5w - 63 = 0

We can factor this equation as follows:

2w²+14w-9w-63=0

2w(w+7)-9(w+7)=0

(w+7)(2w-9) = 0

This means that w=-7 or w=9/2.

However, the width cannot be negative, so w=9/2.

Now that we know the width, we can find the length by substituting w=3 into the equation for the length:

l = 2w+5 = 2*9/2+5 = 14

Therefore, the width of the rectangle is 9/2 or 4.5 yd and the length is 14 yd.

Answer 2

• 4.5 yd and 14 yd

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Let the width of the rectangle be x.

Given that, the length is 5 yd more than doubled the width, therefore, the length is 2x + 5 yd.

Area of the rectangle is 63 square units.

Set up an equation:

• 63 = (2x + 5) × x
• 63 = 2x² + 5x
• 2x² + 5x - 63 = 0

Solve it by factorizing:

• 2x² + 14x - 9x - 63 = 0
• 2x(x + 7) - 9(x + 7) = 0
• (x + 7)(2x - 9) = 0

There are two solutions;

• x = - 7 and x = 4.5

But distance cannot be negative, so x = 4.5.

Length of the rectangle is:

• 2(4.5) + 5 = 9+5 = 14 yd

Therefore, the width of the rectangle is 4.5 yd and the length is 14 yd.