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

Figure5.1.3Amy and d'Marie's distances.

Each of the two lines in Figure 5.1.3 has an equation, as discussed in Chapter 4. The line reperesenting Amy 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 reperesenting d'Marie appears to have \(y\)-intercept \((0,0)\) and slope \(2\text{,}\) so its equation is \(y=2t\text{.}\) 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 {}={} \frac{4}{3}t \amp {}+{} \amp 4 \\ y \amp {}={} 2t \\ \end{alignedat} \right.\text{.} \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.