Chapter 37
Notes for the twelfth week

(Still in progress)

Notes also available as PDF. For now, graphs are only in the PDF version.

37.1 Covered So Far

37.2 What Will Be Covered

Topic: Algebra and graphs for mathematical modeling.

A mathematical model is a mechanism used for predicting responses from data.

Consider these real life “word problems.” The models often can be expressed algebraically; we will be covering a few of these forms.

Sometimes an algebraic view, working with symbols, is the most useful, and sometimes a graphical view, working with plots, is the most useful. Different people and different problems may require different views for fully understanding them. Regardless, each view can serve as a good check.

The algebraic models and relationships we will cover:

 7.3 An Algebraic Example

Consider the following problem:

Some proposition loses by 4 votes out of 100 votes cast. How many voted yes and how many voted no?

One of the first questions to answer is if there is enough information to find a solution. The answer here is yes, although it may not be obvious from the description above.

For now, we approach this problem algebraically using symbols. This is the approach you remember for general word problems. I won’t explain everything here; this is an example we can use throughout. Afterwards, I’ll describe a graphical approach we can use to quickly see if there is any hope of solution.

The first step in an algebraic approach is to assign variables to the unknowns. Here, let Y be the number of yes votes, and let N be the number of no votes. For now, we do not worry about requiring these to be integers or non-negative numbers. One of the techniques in modeling and algebra is knowing what to ignore and when to ignore it. Often, you ignore some properties of the data. Once you have a final result or solution, you can re-apply those properties to see if it makes sense.

Following Pólya’s principles, we next rephrase the problem by relating the variables with equations.

loses by 4 votes:N = Y + 4
total of 100 votes:Y + N = 100

The general algebraic method for solving systems of equations is to simplify them into forms that lead to a result. This is a general plan for solving algebraic problems, but it needs broken into sub-plans that may not be obvious when you start.

Here, we can reduce the problem over two variables, Y and N, into a problem over one variable, N, by substituting the first equation into the second’s left-hand side:

\eqalignno{ Y + N & = Y + (Y + 4) & &\text{by substitution} & & & & \cr & = (Y + Y ) + 4 & &\text{by the associative property} & & & & \cr & = 2Y + 4 & &\text{by evaluating $Y + Y $}. & & & & }

Now we substitute 2Y + 4 for Y + N and transform both sides of 2Y + 4 = 100:

\eqalignno{ 2Y + 4 & = 100, & &\text{inital equations} & & & & \cr (2Y + 4) - 4 & = 100 - 4, & &\text{subtract the same from equal quantities} & & & & \cr 2Y + (4 - 4) & = 96, & &\text{associative property} & & & & \cr 2Y + 0 & = 96, & &\text{additive inverse} & & & & \cr 2Y & = 96, & &\text{additive identity} & & & & \cr {1\over 2} ⋅ 2Y & = {1\over 2} ⋅ 96, & &\text{mult. equal quantities by the same} & & & & \cr 1 ⋅ Y & = 48, & &\text{mult. inverse and evaluation} & & & & \cr Y & = 48, & &\text{mult. identity}. & & & & }

Then Y = 48 and N = 100 - 48 = 52. So the final path by which we solved the problem:

  1. Rephrase the problem algebraically.
  2. Substitute to eliminate one variable, simplifying the problem.
  3. Solve a linear equation in one variable.
  4. Then substitute back to obtain the other variable.

One of the primary topics we will cover is when to skip all the intermediate steps. Many of the algebra rules we and the text present are designed to allow skipping from the two equations directly to 2Y + 4 = 100 and then more quickly to Y = 48. The reasons given above are purely general, and algebraic notation provides the means of applying those reasons generally.

Note that the answer makes sense according to what we know of the problem domain. You cannot have fractional or negative votes. The results are positive integers, so they make sense.

For an example where the number theory we covered helps identify a incorrect data, consider a slight variation where N = Y + 3. From the solution above, we know that the difference will satisfy 2Y + 3 = 100. If Y is an integer, we know that 2\mathrel{∣}2Y and 2\mathrel{∣}100. But 2 ∤ 3, so we know this equality cannot have an integer solution. So if our initial method is carried out correctly, i.e. we substituted N = Y + 3 into N + Y = 100 correctly, we know that the problem’s initial data must be incorrect. This is what I mean by number sense.

37.4 The Example’s Graphical Side

Thinking of the equations above as relationships, we can plot them on an N-Y graph. I will go into lines and linear forms later. For now, it suffices to remember Euclid’s axiom that two points define a line.

For N = Y + 4, consider the points where Y = 0 and Y = 20. These give the values N = 4 and N = 24. Drawing a line between these points

See PDF version.


 ow consider N + Y = 100. Two points suggest themselves immediately, one at N = 0 and one at Y = 0. Adding a line through these points:

See PDF version.


 ven just sketching these without being two precise shows us that the solution may make sense. One line slopes up and one slopes down, so they will intersect somewhere. And a quick sketch shows they intersect with both variables taking positive numbers. This style of graphical reasoning often helps show if a solution is possible or impossible. A quick sketch does not show that the variables are positive integers, but the sketch does justify carrying out the algebraic work.

37.5 Definitions

Now for the painful part. We need a common set of definitions.

37.6 Algebraic Rules for Transformations Between Equivalent Equations

Each of these keeps the solution set invariant or unchanging. Invariance is a very powerful property. (See the story of Emmy Noether, who used the properties of invariants to fundamentally change not only abstract algebra but also mathematical physics.)

We won’t prove these, but we will provide general examples to justify them.

37.7 Transformation Examples

See the text’s Examples 1, 2, and 3 in Section 7.1. For example 3, you can add the fractions directly without multiplying by the common denominator as well. Multiplying the second fraction by {3\over 3} gives

{(x + 7) + 3(2x - 8)\over 6} = {7x - 17\over 6} = -4.

37.8 Manipulating Formulæ by Transformations

Consider an equation for the perimeter of a rectangle, P = 2L + 2W. When you need to compute one variable from the others, treat the variables you know as numbers. To compute the length L given the perimeter P and width W, classify

To find a formula for L, treat P and W as if they were numbers and solve for L.

\eqalignno{ 2L + 2W & = P, & & & & \cr 2L & = P - 2W & &\text{ by subtracting $2W$ from both sides, and} & & & & \cr L & = {P - 2W\over 2} & &\text{ by dividing by the non-zero constant two}. & & & & }