Width and length

Question

what is the width and length of an community sports complex being built in Oak valley. The perimeter of the rectangle playing field is 345 yds. the length of the field is 3 yards less than triple the width?

Answer 1

L = length, W = width

2L + 2W = 345 (1)

L = 3W - 3 (2)

if we put (2) in (1)

2(3W - 3) + 2W = 345

6W - 6 + 2W = 345

8W = 345 + 6

W = 43.875 and L = 3*43.875 - 3 = 128.625

Answer 2

In this problem, we have to set up a system of 2 equations.

The perimeter, 345 m, is the entire distance around the playing field.

345 = l + l + w + w (l = length and w = width).

We can simplify this to:

345 = 2l + 2w

We also know that the length is 3 yards less than triple the width.

In other words, l = 3w - 3

We can put this expression into the equation we found earlier to put everything in terms of w.

345 = 2*(3w - 3) + 2w

345 = 6w - 6 + 2w

345 = 8w - 6

351 = 8w

w = 43.875 yards

If 345 = 2l + 2w and w = 43.875 yards, then:

345 = 2l + 2*43.875

345 = 2l + 87.75

257.25 = 2l

l = 128.625 yards

We can double-check that we're correct by putting these values into both our equations and making sure they work.

perimeter = (2*43.875) + (2*128.625) = 345. This is correct.

Also, l = 3w - 3 = (3*43.875) - 3 = 128.625. This is also correct.