Help with word problem

Question

Mesa is building a sandbox. The length is 2 feet more than the width. Mikes sandbox is 4 feet less than 5 times the width.  If they need the same amount of material, whst would be the dimensions of the sandbox?

Answer 1

Here's what we know:

P = 2l + 2w (the perimeter of a rectangle is the two lengths and 2 widths added together)

Mesa:

l = w + 2 (the length is 2 feet longer than the width)

Mike:

P = 5w - 4 (the perimeter is four less than 5 times the width)

 

Since we have a value for l, and P, let's put them into the equation:

5w - 4 = 2(w + 2) + 2w

Distribute:

5w - 4 = 2w + 4 + 2w

Combine like terms:

5w - 4 = 4w + 4

add 4 to both sides:

5w = 4w + 8

subtract 4w from both sides:

w = 8

 

Now that we know w, we can solve for l:

l = 8 + 2

l = 10

 

Find the perimeter:

P = 5 * 8 - 4

P = 40 - 4

P = 36

 

Check our work with the rectangle formula:

P = 2 * 10 + 2 * 8

P = 20 + 16

P = 36