10.8

There are so many unknowns in life, that an equation can easily have more than one. We’re going to look at such cases. Let’s begin by this simple example. I want to find the affine function whose graph goes through the two given points in the Cartesian plane. As you know the, graph of an affine function is a straight line. So we’re looking for the equation of the straight line that goes through these two points. Let’s call the function that we don’t know, let’s call it phi. Since it’s affine, it’s going to be of the form mx plus b. For certain parameters, m and b, and the goal is to identify those parameters. How do we do it?

54

We first note that phi of 2 has to equal 8 because that’s what it means for the point 2,8 to be in the graph of phi. That gives us the equation 2m plus b equals 8. Similarly, phi must map the number 3 to the value 13. That corresponds to 3m plus b equals 13. Putting these together then, we have what is called a system of simultaneous equations, as they say, and two unknowns– m and b.

87.8

We refuse, by the way, resolutely refuse, to be bothered by the fact that our unknowns are called m and b. True, they’re usually called x and y, but we are not the slaves of our notation, and if we wish to call them m and b, then we certainly shall do so. How are we going to find m and b from these two equations? Well, let’s number the equations. That’s useful for referring to certain operations that we might do to them, and operations such as the following. I’m going to subtract the first equation from the second. When you look on the left-hand side, you’ll see that the b will cancel out, and I’ll get m, and on the right-hand side, 5.

127.2

So m equals 5. Armed with that fact, I go back in the first equation, and I deduce that b is minus 2. We can check our answer. The couple, the pair, 5 minus 2 satisfies the two equations that we had, and we check. Here’s a problem that we saw earlier. Euler’s problem from 1770. We didn’t solve it at the time. You may recall it involved a conversation between a horse and a mule. The clues that were contained in the conversation gave rise to two equations, and the two unknowns, h and m. Let’s look at those two equations now, and see how we can solve them simultaneously.

171.8

The first thing we do is number them, and the next thing I’m going to do is rewrite them in standard form. This first new equation I’ve generated is simply my original first one in which I’ve brought some terms over to the left and to the right. The 2m came over to the left as minus 2m. The plus 100 went over to the right as a minus 100. And the second equation, similarly, is handled, and we now have what is called the standard form of the system. The constants are on the right, the unknowns, in the same order, are on the left. Now, the next thing I’m going to do to this system is I’m going to change the second equation.

211.7

Do you see what was done? We took the second previous equation and multiplied it by 2. So now we have 6h minus 2m equals 800. What’s the advantage of having done that? Answer, we now have the same coefficient in front of the m, minus 2. Therefore, if we now subtract the first equation from the second, the minus 2m will cancel out, and we’ll get 5h on the left, and 1,100 on the right, and we’ll deduce that h is 220. Given that value of h, we go back in the first equation to find m, and we find that m is 260. We can check our answer. See that it satisfies the equation.

258.4

Incidentally, it did turn out then that the poor mule was carrying more weight than the horse. Notice that the solution is unique. There’s no other solution to the system but the one we found. There’s a name for this method that we use. It’s called elimination. Why? Because we play with the equations, we add, and subtract, and multiply by them by constants in order to eliminate some variables, and come to the conclusion. It’s a very good technique. There’ll be one other that we’ll see later. Now, here’s the general form of a generic linear system of two equations and two unknowns. I’ve given the name star to such a system.

301.1

In this notation, the unknowns are called x and y, and the parameters are a, b, c, d, p, and q. The a, b, c, d are the coefficients of the unknowns, and the p and q are the data on the right-hand side. Now, in the two earlier examples that we’ve solved, the particular cases of this star system, there was a unique solution to the system. That is not necessarily the case at all times. It is a fact that a system like star can admit no solutions whatever. We say it is inconsistent. Here’s a simple example.

340.6

If you’re looking for two numbers whose sum is equal 1, and whose sum is also equal 2, well, you won’t be able to find such numbers. There are no solutions simultaneously to these two equations. It’s also a fact that a system such as star can admit infinitely many solutions. Here’s an example. The system x minus y equals 0, and 2x minus 2y equals 0 is always satisfied by any pair of the form x x, and there are infinitely many such things. Now, it turns out there’s an easy way to know ahead of time, just by looking at the system, whether or not it has a unique solution, and we’ll see what this method is next time.