Preface Here are my online notes for my Algebra course that I teach here at Lamar University, although Ihave to admit that it’s been years since I last taught this course. At this point in my career Imostly teach Calculus and Differential Equations.Despite the fact that these are my “class notes” they should be accessible to anyone wanting tolearn Algebra or needing a refresher in for algebra. I’ve tried to make the notes as self containedas possible and do not reference any book. However, they do assume that you’ve has someexposure to the basics of algebra at some point prior to this. While there is some review ofexponents, factoring and graphing it is assumed that not a lot of review will be needed to remindyou how these topics work.Here are a couple of warnings to my students who may be here to get a copy of what happened ona day that you missed.1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learnalgebra I have included some material that I do not usually have time to cover in classand because this changes from semester to semester it is not noted here. You will need tofind one of your fellow class mates to see if there is something in these notes that wasn’tcovered in class.2. Because I want these notes to provide some more examples for you to read through, Idon’t always work the same problems in class as those given in the notes. Likewise, evenif I do work some of the problems in here I may work fewer problems in class than arepresented here.3. Sometimes questions in class will lead down paths that are not covered here. I try toanticipate as many of the questions as possible in writing these up, but the reality is that Ican’t anticipate all the questions. Sometimes a very good question gets asked in classthat leads to insights that I’ve not included here. You should always talk to someone whowas in class on the day you missed and compare these notes to their notes and see whatthe differences are.4. This is somewhat related to the previous three items, but is important enough to merit itsown item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!Using these notes as a substitute for class is liable to get you in trouble. As already notednot everything in these notes is covered in class and often material or insights not in thesenotes is covered in class.
Equations With More Than One Variable In this section we are going to take a look at a topic that often doesn’t get the coverage that itdeserves in an Algebra class. This is probably because it isn’t used in more than a couple ofsections in an Algebra class. However, this is a topic that can, and often is, used extensively inother classes.What we’ll be doing here is solving equations that have more than one variable in them. Theprocess that we’ll be going through here is very similar to solving linear equations, which is oneof the reasons why this is being introduced at this point. There is however one exception to that.Sometimes, as we will see, the ordering of the process will be different for some problems. Hereis the process in the standard order.1. Multiply both sides by the LCD to clear out any fractions.2. Simplify both sides as much as possible. This will often mean clearing out parenthesisand the like.3. Move all terms containing the variable we’re solving for to one side and all terms thatdon’t contain the variable to the opposite side.4. Get a single instance of the variable we’re solving for in the equation. For the types ofproblems that we’ll be looking at here this will almost always be accomplished by simplyfactoring the variable out of each of the terms.5. Divide by the coefficient of the variable. This step will make sense as we workproblems. Note as well that in these problems the “coefficient” will probably containthings other than numbers.It is usually easiest to see just what we’re going to be working with and just how they work withan example. We will also give the basic process for solving these inside the first example.Example 1 Solve for r.(1A P rt= + )SolutionWhat we’re looking for here is an expression in the form,Equation involving numbers, , ,andr A= P tIn other words, the only place that we want to see an r is on the left side of the equal sign all byitself. There should be no other r’s anywhere in the equation. The process given above should dothat for us.Okay, let’s do this problem. We don’t have any fractions so we don’t need to worry about that.To simplify we will multiply the P through the parenthesis. Doing this gives,A P Prt= +Now, we need to get all the terms with an r on them on one side. This equation already has thatset up for us which is nice. Next, we need to get all terms that don’t have an r in them to theother side. This means subtracting a P from both sides.A P Prt− =As a final step we will divide both sides by the coefficient of r. Also, as noted in the process
listed above the “coefficient” is not a number. In this case it is Pt. At this stage the coefficient ofa variable is simply all the stuff that multiplies the variable.A P A Pr rPt Pt− −= ⇒ =To get a final answer we went ahead and flipped the order to get the answer into a more“standard” form.We will work more examples in a bit. However, let’s note a couple things first. These problemstend to seem fairly difficult at first, but if you think about it all we really did was use exactly thesame process we used to solve linear equations. The main difference of course, is that there ismore “mess” in this process. That brings us to the second point. Do not get excited about themess in these problems. The problems will, on occasion, be a little messy, but the steps involvedare steps that you can do! Finally, the answer will not be a simple number, but again it will be alittle messy, often messier than the original equation. That is okay and expected.Let’s work some more examples.Example 2 Solve1 5aRV mb m⎛ ⎞= −⎜ ⎟⎝ ⎠for R.SolutionThis one is fairly similar to the first example. However, it does work a little differently. Recallfrom the first example that we made the comment that sometimes the ordering of the steps in theprocess needs to be changed? Well, that’s what we’re going to do here.The first step in the process tells us to clear fractions. However, since the fraction is inside a setof parenthesis let’s first multiply the m through the parenthesis. Notice as well that if we multiplythe m through first we will in fact clear one of the fractions out automatically. This will make ourwork a little easier when we do clear the fractions out.5mV ab= − RNow, clear fractions by multiplying both sides by b. We’ll also go ahead move all terms thatdon’t have an R in them to the other side.55Vb m abRVb m abR= −− = −Be careful to not lose the minus sign in front of the 5! It’s very easy to lose track of that. Thefinal step is to then divide both sides by the coefficient of the R, in this case -5ab.( )5 5 5 5 5Vb mVb m Vb m Vb m m VbRab ab ab ab ab− −− − − + −= = − = = =−Notice as well that we did some manipulation of the minus sign that was in the denominator sothat we could simplify the answer somewhat.In the previous example we solved for R, but there is no reason for not solving for one of theother variables in the problems. For instance, consider the following example.
Example 3 Solve1 5aRV mb m⎛ ⎞= −⎜ ⎟⎝ ⎠for b.SolutionThe first couple of steps are identical here. First, we will multiply the m through the parenthesisand then we will multiply both sides by b to clear the fractions. We’ve already done this work sofrom the previous example we have,5Vb m abR− = −In this case we’ve got b’s on both sides of the equal sign and we need all terms with b’s in themon one side of the equation and all other terms on the other side of the equation. In this case wecan eliminate the minus signs if we collect the b’s on the left side and the other terms on the rightside. Doing this gives,5Vb abR m+ =Now, both terms on the right side have a b in them so if we factor that out of both terms we arriveat,( )5b V aR m+ =Finally, divide by the coefficient of b. Recall as well that the “coefficient” is all the stuff thatmultiplies the b. Doing this gives,5mbV aR=+Example 4 Solve1 1 1a b c= + for c.SolutionFirst, multiply by the LCD, which is abc for this problem.( ) ( )1 1 1abc abca b cbc ac ab⎛ ⎞= +⎜ ⎟⎝ ⎠= +Next, collect all the c’s on one side (the right will probably be easiest here), factor a c out of theterms and divide by the coefficient.( )bc ac abc b a ababcb a− =− ==−
Example 5 Solve45 9yx=−for x.SolutionFirst, we’ll need to clear the denominator. To do this we will multiply both sides by .We’ll also clear out any parenthesis in the problem after we do the multiplication.5 9x −( )5 95 9y xxy y44− =− =Now, we want to solve for x so that means that we need to get all terms without a y in them to theother side. So add 9y to both sides and the divide by the coefficient of x.5 99 454xy yyxy= ++=Example 6 Solve4 31 8xyx−=+for x.SolutionThis one is very similar to the previous example. Here is the work for this problem.( )( )1 8 4 38 4 38 3 48 3 448 3y x xy xy xxy x yx y yyxy+ = −+ = −+ = −+ = −−=+As mentioned at the start of this section we won’t be seeing this kind of problem all that often inthis class. However, outside of this class (a Calculus class for example) this kind of problemshows up with surprising regularity.