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 area of a square of side \(x\) is \(x^2\), so the sum of the areas of the enclosures is \(3x^2 + y^2\).
Hence, our first equation is \(3x^2 + y^2 = 147\).
Now a square of side \(x\) has perimeter \(4x\), and so the sum of the perimeter of the enclosures is \((3 \times 4x) + 4y\).
Our second equation therefore is \(12x + 4y = 96\).
In order to solve these, we’ll now use the second equation to express \(y\) in terms of \(x\). Rearranging, we get
\[\begin{align*} 3x + y &= 24 \\ \implies y &= 24 - 3x. \end{align*}\]If we substitute this into the area equation above, we have
\[\begin{align*} 3x^2 + (24 - 3x)^2 &= 147 \\ \implies 3x^2 + 576 - 144x + 9x^2 &= 147 \\ \implies 12x^2 - 144x + 429 &= 0\\ \implies 4x^2-48x+143=0\\ \implies (2x-13)(2x-11)=0\\ \implies x = 5.5, 6.5. \end{align*}\]Alternatively we could use the quadratic equation formula to solve here, to get the two solutions for \(x\) given by
\[\begin{align*} x & = \frac{144 \pm \sqrt{144^2 - 4 \times 12 \times 429}}{24}\\ & = \frac{144 \pm \sqrt{144}}{24}\\ & = \frac{144 \pm 12}{24}. \end{align*}\]So we again get \(x_1 = 6.5\) and \(x_2 = 5.5\).
Now by using the perimeter equation above, we find
\[\begin{align*} x_1 = 6.5, \quad \rightarrow \quad y_1 &= 24 - 3 \times 6.5 \\ &= 4.5 \\ x_2 = 5.5, \quad \rightarrow \quad y_2 &= 24 - 3 \times 5.5 \\ &= 7.5. \end{align*}\]Thus there are two solutions, namely \(x_1 = 6.5\) yd., \(y_1 = 4.5\) yd. and \(x_2 = 5.5\) yd., \(y_2 = 7.5\) yd.
How ‘likely’ is this, that we find two viable solutions for \((x,y)\) here?
If we sketch the graphs of \(3x^2 + y^2 = 147\) (an ellipse) and \(3x + y = 24\) (a straight line), we are asking the line to cut the ellipse in two places, both in the first quadrant.
![the ellipse and the line intersecting twice in the first quadrant](/thinking-about-algebra/r5505/images/img-5505.png)
If the perimeter was \(100\) yd. instead of \(96\) yd. do we still get two solutions?