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## Precalculus

Skip to 0 minutes and 10 seconds We now continue our discussion of linear systems, two equations and two unknowns. We have given our generic system of this type the name star, and we’re now going to identify a constant, which tells us a lot about the solutions of star. It’s called the determinant of star. It’s simply the number that you get by calculating ad minus bc. It turns out that this determinant is the key to knowing whether you have a unique solution. We’re going to prove the following proposition. It makes two statements. If the determinant of the system is different from 0, then the system star has a unique solution. It has a solution, and it has exactly one.

Skip to 0 minutes and 59 seconds On the other hand, if the determinant is 0, then the system star will have either no solutions at all– it will be inconsistent– or else it will have an infinite number of solutions. Let’s consider the example of the horse and the mule system. The determinant for that system, if one works it out, turns out to be 5. Therefore, we know that there should be a unique solution to the system, as indeed there was. We saw it when we solved the system. Now, we’re going to prove this proposition for two reasons. First of all, it’s fun to do proofs.

Skip to 1 minute and 37 seconds And secondly, the nature of the proof will yield a systematic method for solving systems of equations, which is very useful sometimes even for bigger systems than just two variables. So here’s the proof. To begin with, I remark that we can take either a or c to be non-zero, or else the variable x will be completely absent. Just to be definite, let’s say that a is different from 0. Other cases are similar. In that case, we can use the first equation in order to express x in terms of y. x turns out to be p minus by over a. Given that conclusion, we then plug this expression for x into the second equation.

Skip to 2 minutes and 20 seconds Or, in more formal language, we substitute this for x in the second equation. There results a simple linear equation in the sole variable y. We can solve that for y. We arrive at this point, and you see at this point that the coefficient of y is precisely the determinant. And now the proof of our proposition is at hand. Because, consider first the case where the determinant is non-zero. Then we can divide across by it. We get a unique y from this last equation, and then we can go back and find the x that corresponds to it.

Skip to 2 minutes and 58 seconds On the other hand, if the determinant happens to be 0, then the left-hand side of our equation here is 0, and either a q minus c p will be 0 and agree with it or not. In the first case, then any y will work because 0 equals 0. And in the second case, we get an inconsistent system. It’s an impossible one. And that’s the end of the proof. QED, as they used to say. Now, the systematic method I was speaking of suggested by this proof is called the method of substitution. It applies to systems of, let’s say, two variable equations like this, but they don’t necessarily have to be linear in general for us to apply the method.

Skip to 3 minutes and 42 seconds Here’s how it works. We first look at one equation, let’s say the first, and we use it to express x as a function of y. That’s usually said by saying we solve the equation f equals 0 for x. We take the expression we have for x, and we substitute into the second equation. There results a single equation involving just y. We solve that equation, and we find y or possibly a set of possible values for y, and then we go back to the first equation and find the x’s that correspond to those values. And of course, always a good idea to check. Let’s apply this to an example.

Skip to 4 minutes and 24 seconds You see here a system of two simultaneous equations and two variables, x and y. It is a nonlinear system. Why is it nonlinear? Because you see the second equation has an x squared, also a y squared. In fact, the xy term is also a nonlinear term. Nonetheless, we can try and use substitution to solve this system. And we use the first equation because it allows us to express x directly as a function of y. We get x equals 2y minus 5 over 3. We plug that into the second equation. That means, in the second equation, everywhere we see an x we replace it by that expression.

Skip to 5 minutes and 6 seconds Now, we get something that’s a little unwieldy here, but our mastery of algebra allows us to simplify it. And when the dust settles, it turns out we have a quadratic equation for y. By factoring or else by the quadratic formula, we find the roots. We find that y is either 4 or 7. Given those two values, we go back and find the corresponding x values. They turn out to be either 1 or 3. So we have two possible solutions, the pair 1,4 and the pair 3,7.

Skip to 5 minutes and 41 seconds Let us note that there were exactly two solutions to this system, which is another way of showing that it’s nonlinear because we have seen that a linear system has either 0, 1, or an infinite number of solutions, whereas a nonlinear system can have two solutions or indeed any other number. Now, to be honest, I must say that nonlinear systems can’t always be solved explicitly. They are often impossible to do so. But all we can ask is that you do your best. We’ve come a long way on equations, and our next big topic is going to be inequalities.

# The determinant and substitution methods

The determinant, substitution, a nonlinear example