Let's say for now that distance is d.įor the boy we have d = 200t and for his mother we have d = 500(t - 6). Since she covers 500 yards every minute, in (t - 6) minutes she covers 500 times (t - 6) yards, or 500(t - 6) yards.īy the time she catches up to him, they both have covered the same distance. His mother starts bicycling 6 minutes later, so she rides for (t - 6) minutes. Since the boy covers 200 yards every minute, in t minutes he will cover 200 times t yards, or 200t yards. We want to figure out how long it takes the mother to catch up to the boy, and how far she needs to ride to do so. She rides 500 yards every minute (She is an Olympian and a gold medalist). She gets on her own bicycle and starts following the boy. He rides 200 yards every minute.Ħ minutes later, his mother realizes her son forgot his lunch. Deriving a systemĪ boy gets on his bicycle and starts riding to school. But before we start, let's see how we end up with a particular system by looking at a real life example. In this article we will learn how to solve systems of linear equations using two fun methods. For example, d can stand for distance, and t for time. If equations describe some process, the letters can be chosen by the roles they play. Often they are designated by the letters x and y. As the name suggests, there are two unknown variables. To simplify the illustration, we will consider systems of two equations. A system of linear equations is when there are two or more linear equations grouped together. A linear equation is an equation that graphs a line.
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