There are four separate square enclosures, three of side \(\quantity{x}{yd.}\) and one of side \(\quantity{y}{yd.}\) The sum of the areas of the enclosures is \(\quantity{147}{sq. yd.}\) and the sum of their perimeters is \(\quantity{96}{yd.}\) Obtain two equations containing \(x\) and \(y\) and solve them.

The three green squares here each have side length \(x\).

The side of the blue square is chosen to make the sum of the perimeters of the four squares \(96\).

Can you find the \(x\) value the question requires?

When is the area smallest?

Why is the width of the green and blue squares together constant?