Introduction: types of equations

Equations arise all over in science, engineering, economics. You’ll be seeing them throughout your mathematical career. We want to learn to solve a few important types of equations. And in doing so, we’ll use everything we’ve learned. Let’s look at an example. We want to know where the graphs of these two affine functions intersect. Now as you know, the graph of an affine function is a straight line. So we’re looking for the intersection point of two straight lines in the Cartesian plane. Now each function is of the type mx plus b. That’s what affine means, m being the slope and b being the y-intercept. Now the slopes here are different you’ll note. 1 is 3 and the other is minus 2.

Therefore, the lines are not parallel and they will meet in a single point. And we label the coordinates of that point (x, y). Our goal then is to find x and y. How do we do it? First, we note that the fact that the point (x, y) lies on the graph of f means that y equals f of x, which is 3x minus 7. We note a similar fact for g. g of x is minus 2x plus 8, which also has to be y. So we deduce from this an equation between 3x minus 7 and minus 2x plus 8. Here it is, an equation. If we solve this equation for x, we will find the value of x.

How do we do it? Well, we could add 2x to each side of the equation. That would give us 5x minus 7 on the left and 8 on the right. And now we add– by the way, before I do the next step, let me mention that this is often summarized as saying the minus 2x that we had on the right hand side has gone over to the left side and has become a plus 2x. And now we add 7 to each side. And we get 5x equals 15. Then we divide each side by 5. We get x equals 3. Voila, we have x.

To find y, we can just go back and say, y is f of 3 which turns out to be 2. Thus, the coordinates of the point we were looking for are the numbers 3, 2. Now the equation that we solved here was this one, which is a linear equation in one variable. We say one variable because there is one unknown. What does linear mean? It means that whenever the variable appears, it’s only in an additive term multiplied by a constant coefficient. So terms like x squared, or x cubed, or 1 over x are said to be nonlinear. Let’s now look at a second example.

We want to know where the graphs of our affine function f, same as before, and a new function h, a quadratic polynomial, where these two graphs intersect in the Cartesian plane. As before, we let the intersection point be (x, y). We want to find x and y. We obtain a following equation for x. 3x minus 7 is x squared plus 4 x minus 6. We work on this equation a bit. We take some terms from right to left and vice versa. And we come out with the simple quadratic equation x squared plus x plus 1 equals 0. Now we are experts in such equations. We know that calculating the discriminant will tell us something.

We calculate it, it turns out to be minus 3. Because it’s strictly negative, this equation has no solution. The quadratic polynomial has no roots. Now is that a problem? Well, not really. It’s not a paradox. And it’s not a mystery because there’s nothing that guarantees, after all, that a straight line in the plane and a parabola in the plane– which is what the graph of h is– that they have to intersect. And, in fact, in this case, they don’t intersect. We’ve just proven that they don’t. So there are no solutions to this equation. And that can happen for an equation. It can have 0 solutions. Now our next example is a really ancient one.

Leonhard Euler was the greatest mathematician of all time. He basically invented modern mathematics. In 1770, he published an enormously influential algebra text. And in that text, we find the following exercise. This exercise is what we call a word problem, because you see, it’s full of words. Now originally, these were Latin words so this is a translation. And I’ve also put in the word kilos just to make it a little more understandable. But, you know, the kilogram had not yet been invented when this exercise was proposed. Now as you see, it involves a conversation between a horse and a mule. The horse says, “If I had 100 kilos of your load, mine would then be twice as heavy as yours.”

And the mule answers in somewhat similar terms. The idea is we need to figure out what their loads are from the two statements they made. How do we do it? Well, we first have to assign letters to the variables that we wish to evaluate. So I’m going to let m be the load of the mule, whatever it is, and h the load of the horse. Then you can translate that sentence pronounced by the horse as h plus 100 equals twice m minus 100. And similarly for what the mule answers. And we have here what is called a system of two simultaneous equations, as they say, and two unknowns. Now equations are wonderful machines for turning unknowns into knowns.

And we’re going to be solving this equation a little later on, but not right away. These examples that we’ve looked at show that the equations that arise can involve one or more variables, they can be linear, they can be non-linear. They may also involve, as we shall see, polynomials, radicals, absolute values, and even other elements like rational functions. We’re going to see that the solution methods to find the solutions will depend on the type of equations. There are, however, some general principles. And there are specific dangers to be aware of.