Class 1 -- Linear equations in one variable I. Manipulating Algebraic Expressions A. Plugging In B. Combining Like Terms II. Solving Basic Equations A. Addition Principle B. Multiplication Principle C. Addition and Multiplication Together D. Practice III. Word Problems A. Coin Problems B. Uniform Motion Problems C. Investment Problems D. Mixture Problems Introduction No single person is responsible for the invention of algebra. The basic building blocks of algebra as we know it today were invented in the Arab world from about 700 to 1400 A.D. The term algebra comes from an Arabic book "Hisab al-jabr almuqa-balah". "Al-jabr" literally means reunion. The author of the book is al-Khowarizmi, whose name is the basis for the word algorithm, meaning a set of instructions used to solve a problem. Arithmetic is a way of dealing with numbers of known values. Algebra gives us a way of manipulating and doing arithmetic on numbers whose values we don't know. The basic idea of algebra is the unknown, a symbol that stands for a number. We use can do everything to the unknown we could do to numbers: add to it, subtract from it, multiply it, divide it, square it, etc. This ability to maniuplate unknown or arbitrary numbers makes algebra a powerful method of solving problems, and of representing relationships in the real world. I. Manipulating Algebraic Expressions OK, now we know how to write down algebraic expressions. So what? What good does this do us? Well, part of the answer lies in our ability to create formulas into which we can plug arbitrary numbers, and to manipulate and simplify these formulas. A. Plugging In (5) An algebraic expression represents a statement about some number. Once we've got it, we often want to evaluate it for a particular number. This process is called "plugging in", because we plug a number into our algebraic formula and see what comes out. All we do is replace the variables by the given numbers and evaluate. Consider, for example, the expression -3d / |(ab - 4c) / (2b + c)| What does this expression give us when a = 2, b = 3, c = -1, and d = -4? Have class go through this. The answer is 6. It's just than simple. B. Combining Like Terms (10) Evaluating algebraic expressions for given numbers is fine, but it's not necessarily easier than just working with the numbers to begin with. However, one thing that makes algebraic expressions useful is that we can manipulate and simplify them. One way of simplifying algebraic expressions is combining like terms. By "like terms", we mean terms that have the same unknown, the same variable. Consider for example the expression: 3x + 4x What does this mean in English? (Get class to answer.) Notice that 3x just means x plus iteself 3 times, and 4x just means x plus itself four times. Well, if we write that out, it's just x + x + x + x + x + x + x Clearly, a simpler way to write this is just 7x. We can do similar things with any expressions where the variable is the same. However, we can't combine dissimilar variables. Consider, for example: 5x + 3y - 2x - 8y How could we combine like terms here? Get class to do this. Note that we can't combine the x and the y terms. They represent different unknowns, and there's no way to write them together as a single term. However, if we had: 3x + xy, could we combine like terms here? Sort of: we could write this as (3 + y) x. After all, y is just a number! Often we need to use the distributive property to combine like terms -- that's just a fancy way of saying we need to multiply out parentheses. Consider: 3 (x - 2(x + 2y)) The first thing we do with an expression like this is multiply it out: Get class to do this. The result is 3x - 6x + 4y. Now we can combine like terms. (Have class do this.) This gives -3x + 4y. II. Solving Basic Equations Manipulating expressions isn't very useful unless we can actually figure out what some of the unknowns are. To do that, we need to solve algebraic equations. The simplest type of equation to solve is one involving just a single variable multiplied by or added to constants. These are called linear equations. We can solve all linear equations using just two ideas. A. Addition Principle Let's consider a simple english statement: "Some number decreased by eight is thirteen" (on board) What's that in algebra? (have class answer) It's "x - 8 = 13" (on board) We can look at this and just guess the answer, but I want to show you a formal trick to deal with equations that look like this. The neat thing is that, with an equation, I can add or subtract anything I want to one side -- as long as I also do it to the other. What I mean by this is that if I have the equation "1 = 1" (on the board), I can also get a true equation by adding two to both sides: "1 + 2 = 1 + 2" (on board). Adding the same thing to both sides leaves a true equation true. This is called the "addition principle". So, with "x - 8 = 13", suppose I add eight to both sides: (on board) x - 8 = 13 (on board) +8 +8 ---------- x = 21 Notice that the plus eight cancels out the minus eight! I'm left with an equation that is just the variable by itself -- so it's obvious here that x is 21. We can see a principle here: if we want to figure out what the variable is, we want to get the variable by itself on one side of the equation. To do that, we can add or subtract to cancel out whatever is with the variable. Thus, if I had "x + 3 = -9" (on board), what would I do to both sides to get the x by itself (have class answer). I would just subtract 3 from both sides: x + 3 = -9 (on board) -3 -3 ----------- x = -12 Again, I immediately know what x is. B. Multiplication Principle The addition principle allows us to "undo" addition to or subtraction from the variable, so that we can get the variable by itself. Is there a similar principle for multiplication/division? Yes! If we take the equation "2 = 2", note that we can multiply both sides by, for example, 1/3. 2 . 1/3 = 2 . 1/3 (on board) 2/3 = 2/3 Thus, multiplying both sides of a true equation by the same number generates another true equation. This is called the multiplication principle. OK, now let's try applying it to variables. Consider "4x = 32" (on board). Suppose I multiply both sides by 1/4: 1/4 . 4x = 1/4 . 32 (on board) x = 8 Just as we did with addition, we've cancelled out the 4 that was multiplying the x and gotten x by itself. To get rid of a multipication by four, I divide by four, or, equivalently, multiply by one fourth. Suppose I had "-6x = 15" (on board) Here my x is multiplied by -6. What should I do to get x by itself? (let class answer) Right: I multiply by -1/6, or divide by -6. Notice that I need to multiply or divide the -6 by another negative to leave behind a positive x. (-1/6) . -6x = (-1/6) . 15 x = -15/6 = -5/2 So we've got x again. C. Addition and Multiplication Together The final step is to combine these two principles. Consider the equation "3x + 8 = 17" (on board). We want to figure out x, so we need to get x by itself. In this equation, x is both muliplied by 3 and added to 8. We need to undo both, but which first? We can be guided by the order of operations. If we look at "3x + 8", the order of operations tells us that we would do the multiply first, then the addition. To undo these, we need to work in reverse order, so we undo the addition first, and then the multiplication. So how do we undo the addition? (let class answer) We subtract 8 from both sides: 3x + 8 = 17 - 8 -8 ----------- 3x = 9 OK, now we've got just multiplication. So let's undo that. What should we do to both sides? (let class answer) We divide both sides by 3, or multiply them by 1/3: 1/3 . 3x = 1/3 . 9 x = 3 And thus we have the answer. We can check to make sure this is right by plugging in to the original equation: "3x + 8 = 17". If we plug in a 3 for x, we get: 3x + 8 = 17 3(3) + 8 = 17 9 + 8 = 17 17 = 17 So this checks out, which means that 3 really is the right value for x. III. Solving Word Problems A. OK, now we're done with the less useful part of the class: pushing symbols around on a piece of paper. Now we're going to get to the more useful part: applying your ability to push around symbols on a piece of paper to real problems posed in English. We're going to go over a few of the more common types of algebraic problems you'll encounter, and develop strategies for solving them. A. Coin Problems The first type of problem you'll encounter is coin problems. In a coin problem, you are given objects of different values, such as different coins. You're told something about the number of such objects, and about their total values. This happens all the time if you're working in a bank. Here's an example of a coin problem: In my pocket there are five more dimes than there are nickels. The total value of the dimes and nickels in my pocket is $1.55. How many of each coin do I have? (on board) The strategy we take for this type of problem is simple: 1. Make a table with the number and the value of each coin. (on board) Coin | Number | Value (on board) ------------------------- Dimes | | Nickels | | 2. Pick one of the unknowns to be the variable, and fill in that line of the table. (on board) What are the unknowns in our problem? (let class answer) In this case, we have two unknowns: the number of dimes and the number of nickels. Which should we pick to be the variable? (let class choose) Actually it doesn't matter which we pick, since both choices are equally easy in this problem. In later problems, it may be easier to pick on or the other. OK, now that we've picked one, let's fill in that line of the table. (Assuming the class picked dimes) Coin | Number | Value (on board) ------------------------- Dimes | d | Nickels | | So we have "d" dimes. What is the value of those dimes, in terms of dollars or cents? (let class answer) Right: it's just 10d cents, or 0.1d dollars. So we fill that in. Coin | Number | Value (on board) ------------------------- Dimes | d | 10d Nickels | | 3. Fill in the other lines of the table in terms of the variable. (on board) We have "d" dimes. The problem says we have five more dimes than nickels. So how many nickels do we have? (let class answer) We have "d - 5" nickels, so we fill that in on the table. Coin | Number | Value (on board) ------------------------- Dimes | d | 10d Nickels | d-5 | What is the value of those "d-5" nickels? (let class answer) Each one is worth 5 cents, so the value is just 5 times d-5, and we write that on the table. Coin | Number | Value (on board) ------------------------- Dimes | d | 10d Nickels | d-5 | 5(d-5) 4. Write an equation based on the total value of the coins. (on board) OK, now we can write down our equation. We can write down an expression for the total value of the dimes and nickels based on what we have written on the table. What is this expression? (let class answer) This is just simple addition: "10d + 5(d-5)" (on board) Aha! The problem told us that the total value of the dimes and nickels is $1.55. There's that magic word "is". That means we can write an equation. What is the equation? (let class answer) It's "10d + 5(d-5) = 155" (on board) -- note that we're working in cents here, so $1.55 is 155 cents. 5. Solve the equation. (on board) At this point we're down to symbol manipulation. We just need to solve the equation. To do this, first we simplify and combine like terms: (ask class to do this) 10d + 5d - 25 = 155 15d - 25 = 155 Now we solve: (have class tell you how to do this) 15d - 25 = 155 +25 +25 -------------- 15d = 180 1/15 . 15d = 1/15 . 180 d = 12 Now the last step. 6. Write the answer in English. (on board) So, in English, how many of each coin do I have? (let class answer) "I have 12 dimes and 7 nickels." (write on board) That's it. Speech: Now: notice that you start out with _two_ things you don't know: the number of dimes and the number of nickels. But you have _two_ independent pieces of information: the difference between the number of dimes and the number of nickels (an equivalent piece of information would have been the total number of coins), and the total value of the coins. These two independent pieces of information are needed to _solve for_ the two numbers. Note also that the problem does not give you all of the information you need to solve it - it does not tell you the monetary value (how many cents each is worth) of a dime or a nickel. You know this of course so it's no problem, but it illustrates some important things you need to do to solve these problems. First of all, write down each stage of your work _clearly_ with a new line for each step you take. Second, carry your units along with you. In this case we are talking dollars and cents, but in other cases it might be miles per hour (for speed), or miles (for distance) or degrees centigrade (for temperature), and any number of other units. Doing this will help you keep your work clear. Third, make sure you have all the information you need to solve the problem, and recognize when you don't. Let's take a very simple example: there are 2 million married people in country X. How many of them are women? To answer this, you'd have to know in addition the customs of country X. In the USA, a marriage is one man/one woman so the answer is 1 million; but other countries have other customs and laws, and without this addtional information you can't answer the question. And note, too, the importance of _independent_ pieces of information. For example, a dime is worth 10 cents. It's also worth 0.1 dollars. From the point of view of solving the equation, these are not independent in the sense that you can use either to solve the equation, and knowing both doesn't increase your ability to solve the equation. We'll see more examples. And finally: if you need two pieces of information to calculate/solve for two quantities, how many do you think you need to solve for _three_ quantities? ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ B. Uniform Motion Problems Another type of problem you'll encounted is uniform motion problems. In a uniform motion problem, you're given objects travelling at different speeds and asked about when they meet, or similar questions about their positions. Here's an example: A car leaves Los Angeles heading for San Francisco, travelling at 60 miles per hour. One hour later, a car leaves San Francisco heading for LA, travelling at 75 miles per hour. The two cities are 465 miles apart. How long after the first driver leaves do they pass each other, and where are they when they pass? (on board) For this type of problem, we also have a strategy. The first step is the most important. 1. Draw a picture! (on board) Car 1 60 mph 75 mph Car 2 x ------> <------ x LA <------------------------------------> SF 465 miles 2. Make a table with the speed, time, and distance travelled for each object. (on board) Object | Speed | Time | Distance (on board) -------------------------------------- Car 1 | 60 | | Car 2 | 75 | | In this table, we fill in the speeds, since those are given. In other problems, other things might be given. Just fill in whatever you have. 3. Pick one of the unknowns to be the variable, and fill in that line of the table. (on board) From this point on, the strategy is basically the same as for coin and stamp problems. So, what should we choose as our variable? (let class answer) In problems like this, it is generally easier to pick time as the variable. That's not always true, but it is for this one. Let t be the time since car 1 left. We can put that in the table. Object | Speed | Time | Distance (on board) -------------------------------------- Car 1 | 60 | t | Car 2 | 75 | | How far does car 1 go in those t hours? (let class answer) If car 1 has been going for t hours, and it is going at 60 miles per hour, its total distance must be 60t miles. We write that in the table. Object | Speed | Time | Distance (on board) -------------------------------------- Car 1 | 60 | t | 60t Car 2 | 75 | | 4. Fill in the other lines of the table in terms of the variable. (on board) How long has car 2 been on the road -- in other words, what is its time, in terms of t? (let class answer) Since car 2 left 1 hour later, it has been going for 1 hour less. Thus, its time is t-1. What is the distance it has travelled? (let class answer) That's just 75 (t-1). Thus: Object | Speed | Time | Distance (on board) -------------------------------------- Car 1 | 60 | t | 60t Car 2 | 75 | t-1 | 75(t-1) 5. Write an equation based on the total distance travelled or the total time, whichever is given. (on board) In this case we know the total distance travelled, because we know how far apart the cities are. How would we write an equation based on this information? (let class answer) The total distance travelled is just "60t + 75(t-1)" (on board) We know that when the cars meet, they have travelled at total of 465 miles. That gives us an equation! 60t + 75(t-1) = 465 (on board) 6. Solve the equation. (on board) Again, we're down to symbol manipulation. How do we solve this equation? (let class answer) 60t + 75t - 75 = 465 135t - 75 = 465 +75 +75 --------------- 135t = 540 1/135 . 135t = 1/135 . 540 t = 4 7. Write the answer in English (on board) Here we were asksed two questions: when to the cars meet, and where are they when the meet? Let's answer each one. First, when do the cars meet? (let class answer) "The cars pass 4 hours after the first car left." (on board) Where are they when they meet? (let class answer) Here we need to plug in. The distance the first car has travelled is 60t miles, and we know t is 4. Plugging in, the car has travelled 240 miles. Thus: "The cars are 240 miles from Los Angeles when they pass." And we're done. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ C. Investment Problems Let's take a look at a couple of other common types of problems. Another common problem type is what I'll call an investment problem. In this type of problem, you have some money to divide up between investments that grow at different rates. Here's an example: You have $5000 to invest. You put some of it in a savings account that earns 2% interest per year, and some of it stocks that have a 6% return. At the end of the year, you have $5200. How much did you put in each investment? (on board) And, surprisingly enough, we've got a strategy for this one too. Even more surprising, our strategy involves a table. 1. Make a table with the principle invested, the interest rate, and the final value. (on board) Investment | Principle | Rate | Value (on board) ------------------------------------- Savings | | 2% | Stocks | | 6% | We will in the interest rates, because those are given. 2. Pick one of the unknowns to be the variable, and fill in that line of the table. (one board) OK, what should our variable be. (let class answer) We could choose either the amount invested in savings, or the amount in stocks. Both are equally good choices. (Assume we pick x = amount in savings.) OK, so let's fill in the table. If we invest x in the savings account, what will the value of the savings account be at the end of the year? (let class answer) Investment | Principle | Rate | Value (on board) ------------------------------------- Savings | x | 2% | 1.02x Stocks | | 6% | Note that the answer is 1.02x, not 0.02x. The amount the account increased, the interest, is 0.02x, but we still have the principle. That doesn't disappear! Thus, the total value is thus x + 0.02 = 1.02x. 3. Fill in the other lines of the table in terms of the variable. (on board) Notice a theme here? All these problems have basically the same strategy. OK, so we put x in savings. How much did we put in stocks then? (let class answer) Since we had a total of $5000, if x is in savings, $5000-x must be in stocks. Investment | Principle | Rate | Value (on board) ------------------------------------- Savings | x | 2% | 1.02x Stocks | 5000-x | 6% | OK, so what's the final value of the money we put in stocks? (let class answer) It's just like the savings amount: just multiply principle times rate. Investment | Principle | Rate | Value (on board) ------------------------------------- Savings | x | 2% | 1.02x Stocks | 5000-x | 6% | 1.06(5000-x) 4. Write an equation based on the values. (on board) In terms of our variable, what is the total value at the end of the year? (let class answer) 1.02x + 1.06 (5000-x) Here we are given the total value at the end of the year, so we can write an equation using that. (get class to write equation) 1.02x + 1.06 (5000-x) = 5200 5. Solve the equation. (on board) At this point we're down to algebra again. How do we solve this equation? (let class answer) 1.02x + 1.06 (5000-x) = 5200 1.02x + 5300 - 1.06x = 5200 -0.04x + 5300 = 5200 -5300 -5300 -------------------- -0.04x = -100 -0.04x / (-0.04) = -100 / (-0.04) x = 2500 6. Write the answer in English. (on board) So, what's the answer in English? (let class answer) "You invested $2500 in each investment." (on board) D. Mixture Problems OK, one last type of problem. In a mixture problem, we have two or more mixtures, each containing a certain percentage of the quantity we're interested in. We want to combine the subtances to get a mixture which has a third, intermediate percentage. Here's an example: A hospital technician has two saline solutions, one containing 2% salt and another containing 10% salt. A doctor asks for 200 milliliters of 4% saline solution. How much of each solution should the technician use? What do you think we should do first here? (let class answer) Make a table! What should the table contain? (let class answer) 1. Make table showing the content of each mixture, the amount used, and the total content of the result. (on board) Solution | Salinity | Amount | Total Salt ----------------------------------------- 1 | 2% | | 2 | 10% | | Again, we fill in what we know. In this type of problem, it's also useful to add a line to the table for the final result: (have class fill this in) Solution | Salinity | Amount | Total Salt ----------------------------------------- 1 | 2% | | 2 | 10% | | Result | 4% | 200 | 8 Note that we get the amount of salt just by multiplying the percent salt content by the amount. 2. Pick a variable, and fill in that line of the table. (on board) So, what should we use as our variable? (let class choose) We could pick either the amount of solution 1, or the amount of solution 2. Either choice is equally good. (Assuming the class picks x = solution 1.) OK, so how much total salt is in solution 1? (let class answer) So we fill in that table line: Solution | Salinity | Amount | Total Salt ----------------------------------------- 1 | 2% | x | 0.02x 2 | 10% | | Result | 4% | 200 | 8 Note that in this case, it is 0.02x, not 1.02x. This is a bit different than the investment problem. 3. Fill in the other lines of the table in terms of the variable. (on board) OK, so how much of solution 2 do we have to use if we use x milliliters of solution 1? (let class answer) It must be 200 - x, to give a total of 200. How much salt does this amount of solution 2 contain? (let class answer) It's just amount times salinity: 0.1 (200 - x) So we fill this in: Solution | Salinity | Amount | Total Salt ----------------------------------------- 1 | 2% | x | 0.02x 2 | 10% | 200-x | 0.1(200-x) Result | 4% | 200 | 8 4. Write an equation using the total amount. (on board) OK, so what's our equation here? (let class answer) It's 0.02x + 0.1(200-x) = 8 (on board) This is just because we know how much total salt must be in the resulting mixture, so we can set that equal to the amounts in the two parts that are going into it. 5. Solve the equation. (on board) By now this should be easy enough for you. How do we solve? (let class answer) 0.02x + 20 - 0.1x = 8 (on board) -0.08x + 20 = 8 -20 -20 ---------------- -0.08x = -12 -0.08x / -0.08 = -12 / -0.08 x = 150 6. Write the answer in English. (on board) What's the answer in English? "The technician should use 150 milliliters of solution 1, and 50 milliliters of solution 2." (on board)