Example 5.1.2

Fabiana and David are running at constant speeds in parallel lanes on a track. David starts out ahead of Fabiana, but Fabiana is running faster. We want to determine when Fabiana will catch up with David. Let's start by looking at the graph of each runner's distance over time, in Figure 5.1.3.

Each of the two lines has an equation, as discussed in Chapter 4. The line representing David appears to have \(y\)-intercept \((0,4)\) and slope \(\frac{4}{3}\text{,}\) so its equation is \(y=\frac{4}{3}t+4\text{.}\) The line representing Fabiana appears to have \(y\)-intercept \((0,0)\) and slope \(2\text{,}\) so its equation is \(y=2t\text{.}\)

Figure 5.1.3 David and Fabiana's distances.

When these two equations are together as a package, we have what is called a system of linear equations:

\begin{equation*} \left\{ \begin{alignedat}{4} y \amp {}={} \tfrac{4}{3}t \amp {}+{} \amp 4 \\ y \amp {}={} 2t \end{alignedat} \right. \end{equation*}

The large left brace indicates that this is a collection of two distinct equations, not one equation that was somehow algebraically manipulated into an equivalent equation.

As we can see in Figure 5.1.3, the graphs of the two equations cross at the point \((6,12)\text{.}\) We refer to the point \((6,12)\) as the solution to this system of linear equations. To denote the solution set, we write \(\{(6,12)\}\text{.}\) But it's much more valuable to interpret these numbers in context whenever possible: it took \(6\) seconds for the two runners to meet up, and when they met they were \(12\) meters up the track.