Harder Geometric Word Problems

The perimeter P of this rectangular area with (as-yet unknown) length L and width w is given by:

The area A is given by:

I will divide my perimeter equation above by 2, so I am dealing with smaller numbers. This gives me the following system (or set or collection) of equations:

I can solve either one of these equations for either one of the variables, and then plug the result into the other equation for the other variable. I think I'll solve the addition equation for L in terms of w; then I'll plug the result into the multiplication equation:

Once again, I've come up with two valid solutions. Neither is negative, so I can't discard either of them. I'll check each.

If w = 30, then:

If w = 44, then:

So, in this particular (and unusual) case, each of the solutions is valid; and taking one solution for the width will give the other solution as the length, and vice versa.

The important point is that the shorter side (whether I refer to it as the "width" or the "length") is the "narrow" side that is across the front of the lot.

The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width w. I'll start by doing things orderly, with clear and complete labelling:

The second statement, "four times its length is twelve more than its perimeter", compares the length L and the perimeter P. I will again be complete with my labelling:

I'll replace the perimeter variable with the expression (above) for the perimeter:

So now I have my two equations:

There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I'll take the first equation and solve for w:

Now I'll simplify the second equation, and then plug in this above expression for w:

Then:

The question didn't ask me to "Find the values of the variables L and w". It asked me to "Find the dimensions of the rectangle," so my hand-in answer is: