Parametric linear equations

A linear equation may contain a parameter, which we usually denote by p. Our goal is then to conduct a discussion of what solution a linear equation has depending on the parameter p.

Motivation

Try solving the following problem: You have written a novel, "How Rumcajs and Snow White drove the Nutcracker out of the forest", and you want to have it printed as a book to give to a friend. You have 10 000 CZK at your disposal. The price is as follows. Your novel is 300 pages long. How many books can you afford to buy?

This is a simple linear equation problem. We will set up the equation as follows: the unknown x will represent the number of books we can buy. So we get the equation: (100 + 300)x = 10 000. The expression in brackets represents the price of one book (100 is the base, 300 is the price per page). We solve the equation:

$$\begin{eqnarray} (100+300)x&=&10 000\\ 400x&=&10 000\\ x&=&\frac{10 000}{400}\\ x&=&25 \end{eqnarray}$$

We can afford to buy 25 books. However, the price per page is too high, so we reduce it to 0,8. How many books can we get now? Let's set up the equation again: (100 + 300 · 0,8)x = 10 000 and solve:

$$\begin{eqnarray} (100+300\cdot0{,}8)x&=&10 000\\ (100+240)x&=&10 000\\ 340x&=&10 000\\ x&=&\frac{10 000}{340}\\ x&\thickapprox&29 \end{eqnarray}$$

We can buy 29 books. In fact, with such a relatively large number of pages, perhaps we could ask for some discount on each page, right? Maybe just 60 pennies, 0,6 crowns? But yeah. So let's calculate how many books we get.

You can probably guess that we could change the price endlessly like this. What's the elegant way to solve this? We'll solve it by using a parameter, by using another letter to represent the price per page of our book. We will denote this parameter by the letter p. To restrict it, let's say that the price per page will be in the interval (0, 1> crowns. It will not be zero, no one will print it for free, and no one will make it more expensive than a krona per page. What will the equation look like then?

$$(100+300p)x=10 000$$

It fits with our idea of price - a hundred crowns is the base, it doesn't change in any way. And we calculate the price for 300 pages by multiplying the price per page, i.e. the parameter p, by three hundred. What will be the solution of the equation? We have to isolate x:

$$\begin{eqnarray} (100+300p)x=10 000\\ x=\frac{10 000}{100+300p} \end{eqnarray}$$

That's all. At this point, we just need to know the price per page and calculate how many books we get. If the price per page were twenty pennies, we have p:

$$x=\frac{10 000}{100+300\cdot0{,}2}=\frac{10 000}{100+60}=\frac{10 000}{160}\thickapprox62$$

We would have bought 62 books. That's pretty nice.

The general parametric equation

For a general parametric equation, we are trying to find out what the solution to the equation is, depending on the value of the parameter. An example of such an equation might be:

$$(2p+2)x-6=0$$

The first thing we find is when the unknown disappears from such an equation x. This occurs when the coefficient in front of the unknown is zero, i.e., when 2p + 2 = 0. We solve this equation and get:

$$\begin{eqnarray} 2p+2&=&0\\ 2p&=&-2\\ p&=&-1 \end{eqnarray}$$

If the parameter p equals minus one, then the unknown x drops out of the equation and we get the form of the equation: −6 = 0. This equation has no solution.

Now we will calculate the solution in the case that p≠ − 1. We will calculate the solution classically, only we will not get a "final" solution, but a solution depending on the parameter p. We will isolate x:

$$\begin{eqnarray} (2p+2)x-6&=&0\quad/+6\\ (2p+2)x&=&6\quad/:(2p+2)\\ x&=&\frac{6}{2p+2}\\ x&=&\frac{3}{p+1} \end{eqnarray}$$

This is the final result. In the case that p≠ − 1, the root of the equation is equal to the fraction 3/(p + 1).