Linear equations with a parameter

A linear equation can contain a parameter, which we usually denote by p. Our aim is then to discuss what solution the linear equation has depending on the parameter p.

Motivation

Try to solve the following problem: you have written a novel, “How Brave Betty and the Wise Old Owl Chased the Goblin out of the Wood”, and you want to have it printed as a book and give copies to your friends. You have £100 to spend, which is 10,000 pence, so we will work in pence. The price is as follows: there is a basic fee of 100 pence per book, plus 1 penny for every page. Your novel has 300 pages. How many books can you afford to order?

This is a simple problem on linear equations. We set up the equation like this: the unknown x will be the number of books we can afford. So we get the equation (100 + 300)x = 10,000. The expression in brackets is the price of one book (100 is the basic fee, 300 is the cost of the pages). We solve the equation:

$$\begin{array}{rcl}(100+300)x& =& \mathrm{10,000}\\ 400x& =& \mathrm{10,000}\\ x& =& \frac{\mathrm{10,000}}{400}\\ x& =& 25\end{array}$$

We can afford to buy 25 books. However, the price per page is too high, so let’s reduce it to 0.8 pence. How many books do we get now? Again we set up an equation, (100 + 300 · 0.8)x = 10,000, and solve it:

$$\begin{array}{rcl}(100+300\cdot 0.8)x& =& \mathrm{10,000}\\ (100+240)x& =& \mathrm{10,000}\\ 340x& =& \mathrm{10,000}\\ x& =& \frac{\mathrm{10,000}}{340}\\ x& \approx & 29\end{array}$$

We can buy 29 books. To be honest, with such a large number of pages we might ask for a little discount on each page, mightn’t we? Perhaps just 0.6 pence? Why not. So let’s work out how many books we get.

You can probably guess that we could keep changing the price for ever. What is an elegant way to handle this? We use a parameter, another letter, which will stand for the price of one page of our book. We denote this parameter by the letter p. To keep things bounded, let’s say that the price per page will lie in the interval (0, 1] pence. It won’t be zero, because nobody will print it for free, and nobody will charge more than a penny per page. What will the equation look like then?

$$(100+300p)x=\mathrm{10,000}$$

This fits our idea of the price – 100 pence is the basic fee, which never changes. And we work out the cost of 300 pages by multiplying the price per page, that is, the parameter p, by 300. What is the solution of the equation? We need to get x on its own:

$$\begin{array}{r}(100+300p)x=\mathrm{10,000}\\ x=\frac{\mathrm{10,000}}{100+300p}\end{array}$$

That’s all. Now all we need is the price per page and we can work out how many books we get. If the price per page were 0.2 pence, then substituting for p we have:

$$x=\frac{\mathrm{10,000}}{100+300\cdot 0.2}=\frac{\mathrm{10,000}}{100+60}=\frac{\mathrm{10,000}}{160}\approx 62$$

We could buy 62 books. That’s quite nice.

The general equation with a parameter

With a general equation with a parameter, we try to find out what solution the equation has depending on the value of the parameter. An example of such an equation is:

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

First we find out when the unknown x disappears from the equation. That happens when the coefficient of the unknown is zero, that is, when 2p + 2 = 0. We solve this equation and get:

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

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

Now we still need to work out the solution for p ≠ − 1. We solve it in the usual way, except that we don’t get a “final” solution but a solution that depends on the parameter p. We get x on its own:

$$\begin{array}{rcl}(2p+2)x-6& =& 0(+6)\\ (2p+2)x& =& 6(÷(2p+2))\\ x& =& \frac{6}{2p+2}\\ x& =& \frac{3}{p+1}\end{array}$$

This is the final result. If p ≠ − 1, the solution of the equation is the fraction 3/(p + 1).