The sum and product of two numbers, \(x\) and \(y\), are \(10\) and \(12\) respectively. Form appropriate equations, plot [on graph paper] the corresponding graphs and hence obtain approximate values for \(x\) and \(y\).

The information given in the question tells us that \[\begin{align*} x + y &= 10, \\ xy &= 12. \end{align*}\] By rearranging these equations to make \(y\) the subject, we get \[\begin{align*} y &= 10 - x, \\ y &= \frac{12}{x}. \end{align*}\]

The following graph shows both the line (corresponding to the first equation) and the hyperbola (corresponding to the second equation).

A plot of the line $y = 10 - x$ in blue and the hyperbola $xy = 12$ in red. The axes range from $-15 \le x, y \le 15$, with major gridlines at every multiple of $5$ and minor gridlines at every integer.

The graph shows that there is an approximate solution with \(x \approx 1.4\) and \(y \approx 8.6\) and another one with \(x \approx 8.6\) and \(y \approx 1.4\). These give essentially the same solution, as we are looking for two numbers with a given sum and given product.

The equations can be solved exactly, and without much difficulty. By substituting the equation \(y = 10 - x\) into the equation \(xy = 12\), we have \[x(10 - x) = 12\] so that \[x^2 - 10x + 12 = 0.\] By the quadratic formula, we have that \[\begin{align*} x &= \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \\ &= \frac{10 \pm \sqrt{100 - 48}}{2} \\ &= \frac{10 \pm \sqrt{52}}{2}. \end{align*}\]

The smaller of these values is \(1.394448\!\ldots\), which is close to our visual estimate.