Please refrain from submitting entire homework assignments here. We are happy to help you with concepts and get you started, though.

Problems like this are solved with systems of equations. All you need to get started are the basic formulas and a little understanding of "translating" words to math.

Let L = the length of the rectangle and let W = the width.

If the length is 3 yds longer than the width, then we could say that L = W+3. OK so far? We'll call that equation (1)

We are also given that the perimeter is 46 yds.

The formula for perimeter is P = 2L + 2W, so in this case 2L + 2W = 46. That's equation (2)

Equation (1) gives us an expression for L, so let's substitute that expression into (2)

2L + 2W = 46

2(W+3) + 2W = 46 (Substituting W+3 for L)

2W + 6 + 2W = 46 (Distributive prop)

4W+6 = 46 (combining like terms)

4W=40 (Subtract 6 from both sides)

W=10 (Dividing by 4)

Now, we substitute this value of W into equation (1)

L=W+3

L=10+3

L=13

Now we know both length and width. Since Area = L x W,

the area of the rectangle is 10 x 13 = 130 yds^{2}

Best wishes!