CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 1. The basic laws of Algebra Section 1 of these notes covers the same material as Sections 1.1, 1.2, and 1.6 of the text (in a somewhat very different way). If you need more details about a topic discussed here, ask me or look there. 1.1. The most important basic fact of algebra. This class is all about the manipulation of algebraic formulas. An algebraic formula is just a bunch of numbers, arithmetic symbols, and letters of the alphabet thrown together, such as 2x 2 3 or 3xy 2x. The letters of the alphabet that occur in an algebraic formula are called unknowns, because they stand for hidden numbers that we cannot see. Underneath it all, an algebraic formula stands for some hidden number. You can and should, therefore, treat it like a number in all of your dealings with it. For example, just like you can cancel the 4 s to show that you can also show that 4 3 4 = 3, (x 2 + 1) x (x 2 + 1) = x by canceling the (x 2 + 1) s. However, just because a cancellation rule works for one choice of numbers, doesn t mean it will work for an algebraic formula The Most Important Basic Fact of Algebra. When you simplify an algebraic expression, whatever you do to it needs to work for every number you can plug in for its unknowns. So, for example, if you changed (x+3) 2 into x 2 +3 2 on a quiz, I would mark you wrong, because if you plug in x = 1 into the equation (x + 3) 2 = x 2 + 3 2 you get (1 + 3) 2 = 1 2 + 3 2, which is false because the left side is 4 2 = 16, while the right side is 1 + 9 = 10. Usually, if you are not sure whether or not one of your algebraic moves is right, that means it is wrong. In situations where you are unsure, you should always check your formula by plugging in actual numbers for the unknowns and seeing if they work. Example 1.1. Plug some numbers into the following equations to see if they work. (x + y) 3 = x 3 + y 3 Solution: Plugging in x = 1 and y = 1 we get 1

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 2 (1.1) (1.2) (1 + 1) 3 = 1 3 + 1 3 2 3 = 1 + 1 8 = 1!!! This last equation is clearly false, so the formula (x + y) 3 = x 3 + y 3 is false as well. x2 +2x+1 = x + 1 x+1 Solution: With x = 3 we get: 3 2 + 2 3 + 1 = 3 + 1 3 + 1 9 + 6 + 1 = 4 4 16 4 = 4 Which checks out, and if you plug a few more numbers in yourself you ll see that it keeps on working. This indicates that the equation is in fact true, although to actually know for sure we need to use some laws of arithmetic. IMPORTANT: In this last example, plugging in x = 1 yields the following: ( 1) 2 + 2( 1) + 1 = ( 1) + 1 ( 1) + 1 (1.3) 1 2 + 1 = 0 0 0 0 = 0!!! This last equation is a problem, but not because it is false. Rather, it is senseless, because we divided by 0 on the left side. In this case, we don t say that the algebraic formula is wrong. Instead, we just say that it doesn t make sense for x = 1 and move on. 1.2. Laws of algebra from laws of numbers. You are expected to be familiar with the basic algebra discussed in Sections 1.2 of the text, here is a small, representative selection of problems for you to test yourself with. Homework 1, # 1 Compute the following expressions. (a) 7 + ( ) 4 10 5 (b) 4.3 ( 8.7) (c) ( 4)( 2)3( 2) (d) ( 3) 2 (e) 3 2 (f) ( ) 3 2 2 (g) ( ) 3 3 2

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 3 (h) (5 6)2 2(3 7) 89 3 5 2 (i) 7(2x + 3) (j) 7 4[3 (4y 5)] Look at problems 13-130 in Section 1.2 for more practice. The laws of arithmetic are facts and equations that are true for all numbers. You used a bunch of them when you were doing the problem set above. Since they are true for all numbers, we can also plug algebraic expressions into them and get out true statements. There are too many of them to list here, however here are three of the ones I see students have trouble with most often when using them on algebraic expressions. The Distributive Law: The distributive law says that, for any three numbers a, b, and c: (1.4) a(b + c) = ab + ac So, for example, equation 1.4 says that 5(2 + 3) = 5 2 + 5 3, which checks out: (1.5) 5(2 + 3) = 5 2 + 5 3 5 5 = 10 + 15 25 = 25 Since 1.4 is true for all numbers, we can also apply it to algebraic formulas, for example: (x 2 + 1)(3x 1 x ) = (x2 + 1)3x x2 + 1. x Here I just substituted x 2 + 1 for a, 3x for b, and 1 for c in 1.4. Notice the distributive x law can also be applied backwards to pull out a common factor like so: 3x 3 + 6x 2 = 3x 2 x + (3x 2 )2 = 3x 2 (x + 2). Here the distributive law 1.4 is applied in the second equality by setting a = 3x 2, b = x, and c = 2. However, notice that what makes the reverse distributive law difficult to use is the first step, where I had to recognize the common factor 3x 2. This just takes practice to get good at. Exercises 111-130 in Chapter 1.2 of the book provide good practice for the distributive property. Try these ones right now: Homework 1, # 2 (a) Use the distributive property to rewrite 7x(2x 2 + 1). (b) Use the reverse distributive property to factor 8x 2 4x 3. Exponents: Remember if n is a positive whole number, then by definition a n = n {}}{ a a a.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 4 We used this formula on the last list of problems. Recall at the beginning of Section 1.1, we showed that the equation (x + y) 3 = x 3 + y 3 is false. To find the correct simplification, we use the exponential law with a = x + y, then simplify by using the distributive law: (x + y) 3 = (x + y) (x + y) (x + y) = (x 2 + 2xy + y 2 )(x + y) = x 3 + 3x 2 y + 3xy 2 + y 3. Also remember that negative exponents are defined by the formula a n = 1 a n. This definition is made to ensure that that the rule a n a m = a n+m works even when n and m are not both positive whole numbers. The other rules you need are and (ab) n = a n b n (a n ) m = a nm. Remember these rules work for any numbers n, m, even negative ones. Here are some problems from section 1.6 that will check how good you are with exponents. Homework 1, # 3 Simplify the following: (a) (2xy 2 ) ( 3 (b) ) 4 x 3 y 4 30x 2 y 5 6x 8 y 3 (c) (d) ( 4a 2 ) 2 (2a 5 ) Look at problems 1-124 in Section 1.6 for more practice Fraction addition: The easy law of fraction addition tells us that, for any three numbers a, b, and c: a (1.6) c + b c = a + b c This allows us to perform the following kind of simplification: x 2 + 2x = x2 x x + 2x x = x + 2. Here the first equality is just Equation 1.6 in reverse with a = x 2, b = 2x, and c = x, and the second is basic cancellation. The problem with the easy law of fraction addition is that you need the bottom of both fractions to be the same in order to add them together. When your fractions are different on the bottom, you need to use the hard law of fraction addition: (1.7) a b + c d = ad + bc bd

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 5 For example: x + 1 x 1 + x x 2 1 = (x + 1)(x2 1) + x(x 1) (x 1)(x 2 1) Where here I just plugged a = x + 1, b = x 1, c = x, and d = x 2 1 into Equation 1.7. You should also be able to multiply the top and bottom out to get x 3 + 2x 2 2x 1 x 3 x 2 x + 1. By the end of this class, you will have even deeper skills which allow you to see that this last expression can be simplified down further to x 2 + 3x + 1 x 2 1 We won t go further into the nitty gritty of adding fractions of algebraic expressions until much later in the class, so there s no need to brush up further on them right now. But here s an example of the kind of thing you will need to be comfortable doing: Example 1.2. Rewrite the expression (x + 1) 1 + (x 1) 2 as a single fraction with no parentheses. Solution: This requires use to use all three of the major arithmetic laws I pointed out (and some minor ones that I didn t). (x+1) 1 +(x 1) 2 = 1 1)(x 1) +(x x + 1 1 = 1 1 + (x + 1)(x 1)(x 1) 1 (x + 1) = x3 x 2 x + 2. x + 1 2. Solving for x, using algebra in real life. This section discusses Chapters 1.4 and 1.5 of the book. 2.1. Solving for x. In Section 1 of these notes, we went over some ways to decide whether or not an equation f(x) = g(x) is true for all values of x, so that we could tell whether or not it was okay to change f(x) into g(x). This task is different from solving a given equation f(x) = g(x) for x. In this case, we don t expect f(x) to be equal to g(x) for all values of x. We just hope they are equal for some value of x, and want to find it. Example 2.1. Solve the equation 3x 1 = 2x + 1 for x. Solution: In addition to substituting equivalent algebraic substitutions for one another, we can do almost anything we want to one side of the equation, so long as we do the same

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 6 thing to the other side. 3x 1 = 2x + 1 3x 1 + 1 = 2x + 1 + 1 (2.1) 3x = 2x + 2 3x 2x = 2x + 2 2x x = 2 So we see that this equation does have exactly one solution, x = 2. Homework 2, #1: Solve the following equations for x. (a) 7x + 4 = x + 16 (b) 3(x 4) 4(x 3) = x + 3 (x 2) (c) 5 + x 2 = x+3 3 8 (d) 8x + 1 = x + 43 (e) 7(x + 1) = 4[x (3 x)] (f) x+1 = 5 x+2 3 7 For further practice, look at problems 1-38 of Chapter 1.4 of the book. We will learn more advanced x-solving techniques later in the course. 2.2. Algebra in real life: In my experience, word problems are just learned best by working examples, although the book breaks down the process into 4 steps in Chapter 1.5, which you may find helpful. Often, word problems in math classes are somewhat contrived, but I happen to like this one from the book (it is a slight rewording of Example 3 of Chapter 1.5): Example 2.2. A store offers you a choice of two texting plans. Plan A has a monthly fee of $20 with an additional charge of $0.05 per text. Plan B has a monthly fee of $10 with an additional charge of $0.10 per text. Which one should you get? Solution: As I ve worded the problem, it doesn t have a right or wrong answer, just like a real world problem will not have a right or wrong answer. But we can still use math to make a more informed decision. The store has offered you a very common kind of deal. Plan A requires you to pay more up front than Plan B, but you only have to pay half as much money per message with plan A. So Plan B will be cheaper if you don t plan on texting very much, and Plan A will be cheaper if you plan on texting a lot. Only you can guess how many text messages you will send in a month, but using math we can at least figure out which plan will be cheaper based on this guess. So, suppose x is the number of text messages you think you will send in a month. Then Plan A will charge you the $20 up front, plus an additional $0.05 x at the end of the month. So your total Plan A expenses will be 20 +.05x dollars. Meanwhile Plan B charges you only $10 up front, but then it tacks on $0.10 per message you send, which puts your Plan B expenses at 10 + 0.10x. As we observed earlier, Plan B will be cheaper initially, but once the number x of text messages gets big enough, Plan A will become cheaper. In order to figure out how big x

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 7 needs to get in order to make Plan A cheaper than Plan B, we must find the value of x where Plan A and Plan B cost exactly the same. But given the formulas above, this just means solving the equation 20 +.05x = 10 +.1x for x. We get: (2.2) 20 +.05x = 10 +.1x 10 +.05x =.1x 10 =.05x 200 = x Therefore, even if you are tempted to go with Plan B because it s cheaper at first, Plan A will be the better choice if you send more than 200 texts in a month. And since 200 text messages comes out to just 200 6.67 text messages a day during the average month, 30 I m willing to bet that most of us here would be better off with Plan A. Incidentally, this is usually the way things work in the world: those who choose to pay more up front pay less in the long run. Homework 2 #2: (a) If the temperature is C degrees Celsius, then it is F = 9 C + 32 degrees Farenheit. 5 Suppose it is a 98 degrees Farenheit in Salinas. What s the Celsius temperature of Salinas? (b) An Inn charges you $250 to stay the night, and this price includes an 8% sales tax. How much money goes to the Inn, and how much to the government? (c) The toll for a bridge is $5.00 every time you cross it, unless you purchase a discount pass for $30, which lowers the price to only $3.50 every time you cross it, for a month. How many times a month would you have to cross the bridge in order to make the discount pass worth it? Please show your work using algebra. (d) After a 20% reduction, you purchase a television for $336. What was the television s cost before the reduction? Problems 17-40 in Chapter 1.5 of the book are all great for further study. 3. Functions and Graphs In this part of the notes we cover Section 2.1, then jump back to 1.3, and end with a little bit of 2.2. 3.1. A conceptual shift. The word set in the definition below is just a short mathematical word for a collection of objects. Definition 3.1. If A and B are sets, then a function f from A to B is a rule which assigns a single member of B to every member of A. This definition is quite general, and Chapter 2.1 of the book gives many examples of functions. The only kind of functions we will be concerned with in this class are functions between sets of numbers. In fact, the real point of this section is to teach you a new way to think about algebraic formulas.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 8 Instead of thinking of a formula like 2x 2 as something which stands for some unknown number, we will think of it as defining a function f which assigns, to every number x, the new number f(x) = 2x 2. It is like a machine that takes numbers as input and outputs new numbers. So, in this case, f(x) eats 1 and turns it into 2(1) 2 = 2, it eats 1 and turns it into 2( 1) 2 = 2, it eats 2 and transforms it into 2(2) 2 = 8. We no longer think of x as some fixed hidden number, but instead think of it as a slot to put numbers into. Homework 3, # 1 Compute f(0), f(1), f(π), f( 2.2), and f( 10) for the following functions. (a) f(x) = 1 1 x (b) f(x) = x (c) f(x) = x2 1 x+10 For more practice, try problems 9-19 of Section 2.1. There are two important definitions to remember for functions. Definition 3.2. If f is a function from A to B, then A is called the domain of f. In particular, if f(x) is a function of numbers defined by an algebraic formula, then we always assume the domain of f is the set of numbers for which the algebraic formula makes sense. So, for example, the domain of f(x) = 1 is the set of all numbers except 1. The domain of f(x) = x 2 is all numbers because it always at least makes sense. Meanwhile the x 1 domain of f(x) = x is the set of all non-negative numbers. To see why, let s remember what x is in words: x is the non-negative number y such that y 2 = x. So 0 = 0, since 0 2 = 0, and 4 = 2, since 2 2 = 4. However, as it stands, 1 is undefined, because there is no real number y such that y 2 = 1. Definition 3.3. If f is a function from A to B, then B is said to be the range of f. In particular, if f(x) is a function of numbers defined by an algebraic formula, then we always assume the range of f is the set of numbers y such that f(x) = y for some x in the domain of f(x). If we think of f(x) as something which shoots at numbers, then the range of f(x) is the set of numbers which it actually hits. So, for example, if f(x) = x 2, then when f(x) loads x = 2, it hits x = 4, and we conclude that 4 is in the range of f(x). Similarly, since f(3) = 3 2 = 9, when f(x) loads 3, it hits 9. So 9 is in the range of f(x). However, notice that no negative number is in the range of x 2, because even if you plug in negative numbers, only positive numbers pop out. Homework 3,# 2 Find the domain and range of the following functions. (a) f(x) = 2x + 1 (b) f(x) = 78x 125 (c) f(x) = 25 x 2 (d) f(x) = x 1 x

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 9 (e) f(x) = x 2 3.2. Graphs of Equations. Recall that Cartesian coordinates are a way to assign a pair of numbers to every point in the plane. The way they work is simple. You draw a horizontal line with notches in it corresponding to numbers, and call it the x-axis. You then draw a vertical line with numbered notches in it and call it the y-axis (it is also customary to make the x-axis and y-axis meet at 0 on both). Do this on a separate piece of paper. Once these axes have been drawn, you have a way to associate a pair of numbers (a, b) to every point on the plane. The first number a is the vertical shadow cast by the point on the x-axis, and the second number b is the horizontal shadow cast by the point on the y-axis. You can also go in the reverse direction: starting with a pair of numbers (a, b) you can find the point on the plane which corresponds to it. Practice this with a couple of points yourself. This is probably one of the best ideas of all time, because it allows us to make a connection between algebraic equations and shapes in space. Definition 3.4. Suppose f(x, y) and g(x, y) are algebraic expressions in two unknowns, x and y. Then the graph of the equation f(x, y) = g(x, y) is the set of points (a, b) in the Cartesian plane which make the equation true when you substitute a for x and b for y. So, for a simple example, the graph of the equation (x + y) 2 = x 2 + 2xy + y 2 is the entire plane, because all pairs of numbers (a, b) make the equation true when you plug in a for x and b for y. Another simple example is the equation x 2 + y 2 = 1 whose graph consists of no points at all! However, most graphs will be lines or curves of some kind. For example, the graph of the equation (3.1) x 2 + y 2 = 4 is a circle of radius 2. To see why, notice that if (a, b) is a point on the graph, then a 2 + b 2 = 4 = 2 2. But if we set c = 2, then this becomes a 2 + b 2 = c 2, the formula for the Pythagorean theorem. Homework 3, # 3: Draw the graphs of the following equations. (a) x 2 + y 2 = 9 (b) x + y = 1 (c) xy = 1 (d) 2x y = 1 (e) x 1 4 y = 2 (f) 2x 2 + y 2 = 4

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 10 For more practice, try problems 11-26 of Section 1.3 (some of them are harder than you need to know, but worth trying). 3.3. Graphs of Functions. Definition 3.5. Suppose f(x) is a function. Then the graph of f(x) is just the graph of the equation y = f(x). In other words, the graph of f(x) is the set of points of the form (x, f(x)). It s easy to use the vertical line test to determine whether or not a curve in the plane is the graph of a function, and it s also easy to figure out the domain and range of a function by looking at its graph. Homework 3, # 4 (a) Explain in your own words why a vertical line can only intersect the graph of a function in one point at most. (b) Suppose G is the graph of a function f(x). Explain why the domain of f(x) is the shadow of G on the x-axis, and the range of f(x) is the shadow of G on the y-axis. 4. Linear Functions Today we finish up Section 2.2, then move on to Section 2.4. 4.1. A bit more about the graphs of general functions. Last class period I ended by introducing graphs of functions, and had you think about three basic facts: (1) A vertical line can only intersect the graph of a function in at most one point (Vertical Line Test). (2) The domain of a function is the shadow that its graph casts on the x-axis. (3) The range of a function is the shadow that its graph casts on the y-axis. Let s put these ideas to use now. Homework 4, # 1 I m going to draw six graphs on the board. For each of them, do the following: (a) Use the vertical line test to determine whether it is a function. (b) If it is a function, find its domain. (c) If it is a function, find its range. For more practice, try problems 11-18 and 31-40 of Chapter 2.2 of the book. 4.2. Linear functions and the slope-intercept form. Suppose A, B, and C are constants. Then the graph of the equation (4.1) Ax + By = C

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 11 will always be a straight line. Homework 4, # 2 Verify this by plotting the graphs of the following equations. (a) 2x y = 1 (b) 3x = 6 (c) 8x + 4y = 0 For more practice, try problems 1-14 in Section 2.4 of the book. We can almost 1 always rewrite Equation 4.1 so that it defines y as a function of x: (4.2) Ax + By = C By = Ax + C y = A B x + C B Those of you who have seen this before will recognize that this is just the slopeintercept equation of a line, which is normally written like this: (4.3) y = mx + b Where m and b are constants. Notice that if b = 0, then this equation just becomes y = mx. Homework 4, # 3 Draw the graph of the following functions. (a) y = 2x (b) y = 1 2 x (c) y = 2x Notice that these lines all pass through the point (0, 0), but they point in different directions. In general, the direction a line points in is determined by the constant m in Equation 4.3, which is called the slope. Slope Rule: If a line has slope m, then for every unit you move right in the x-direction, you move m units vertically in the y direction. If the slope m is written as a fraction c, where d is a positive number, then we can say d that the line rises 2 by c units for every d units you move right. In any event, for the functions in Problem 3 above, you could always use the following procedure to find the graph: (1) Draw a dot at (0, 0) 1 The only time the algebra won t work is if B = 0, but in that case Equation 4.1 becomes Ax = C, whose graph is just the vertical line of points (x, y) with x = C A. 2 Although if c is a negative number it will actually fall by c units as you move right by d units.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 12 (2) Move right 1 unit, and vertically m units, then draw another dot. (3) Draw the straight line through these two dots. Now it is a general fact about functions that the graph of f(x) + b is what you get by moving f(x) up by b units (or down if b is negative). So, to draw the graph of a linear function of the form y = mx + b, we only need to change the first step of the previous procedure: (1) Draw a dot at (0, b) (2) Move right 1 unit, and vertically m units, then draw another dot. (3) Draw the straight line through these two dots. Use this procedure to solve the following homework problem. Homework 4, #4 (a) y = 2x + 1 (b) y = 1 3 x + 3 (c) f(x) = 3 4 x 2 For more practice on problems like these, try problems 29-40 in Section 2.4 of the book. Since it s so easy to graph an equation in slope-intercept form, it usually saves a lot of time to simplify a linear equation down to slope-intercept form first before graphing it. Homework 4, #5 Graph these equations by putting them into slope-intercept form first (and if you can t for some reason improvise!). (a) 2x 1 3 y = 2 3 (b) 7x = 3 + y (c) 1 4 y + 2x = 1 x (d) 4x = 12 For more practice, try problems 41-62 of Section 2.4. 5. The point-slope form, finding the equation of the line. In this part of the notes we discuss Sections 2.4 (a bit more) and Section 2.5 of the book. 5.1. Going from line to equation, finding the slope. In the previous class period we saw how to change a linear equation into slope-intercept form. (5.1) y = mx + b Where m tells us the slope and b is the y-intercept of the line. Once we know m and b, it is easy to find the line. The slope-intercept form is also nice when you want to solve the reverse problem: starting with the graph of a line function, find an equation for it. All you need to be able to do is read off the line s y-intercept and determine its slope. Recall how finding the slope

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 13 just means finding the rise over run of the graph. HW 5, # 1: I am going to draw three graphs on the board. Find the slope-intercept form for each of them. Exercises 27 and 28 of Section 2.4 each have three lines drawn on them. For more practice, find their slope-intercept form as well. Suppose you know the coordinates (x 0, y 0 ) and (x 1, y 1 ) of two points on the graph of a line, where x 1 > x 0. Then, the run of the line between the two points is x 1 x 0 and; the rise of the line between the two points is y 1 y 0. Since the slope is just the rise over run, this leads to a nice formula for the slope. Slope Formula: Suppose that (x 0, y 0 ) and (x 1, y 2 ) are both points on the line l. Then the slope m of l is given by: (5.2) m = y 1 y 0 x 1 x 0 Notice that it doesn t matter which two points (x 0, y 0 ) and (x 1, y 1 ) that we choose on the line l, the answer will always come out the same anyway. Homework 5, # 2: I am going to draw three lines on the board, and label some points that lie on them. Use these points and the slope formula of equation 5.2 to find their slope. For more practice with the slope formula, try exercises 15-28 of Section 2.4 (most of these exercises don t draw a graph for you, they just list two points on the line for you to plug into the slope formula). 5.2. The Point-Slope formula. In the future, when doing calculus you will often only know the following two things about a line: The slope of the line; and some arbitrary point on the line. Using this information, you will then need to find an equation describing the line. The following simple formula will always do the trick: Point-Slope formula: Suppose that a line l has slope m and passes through the point (a, b). Then the following is an equation of the line: (5.3) y b = m(x a) The parentheses in this formula are important! If we simplify the right side of Equation 5.3, we get mx ma, not mx a!

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 14 Homework 5, # 3 Find the slope-intercept form of the following lines. Hint: Start by finding the point-slope form, then use algebra to switch it to slope-intercept. (a) l passes through ( 1, 2), has slope m = 1 2 (b) l passes through (1, 0), has slope m = 4 (c) l passes through (2, 2) and (4, 1) (d) l passes through (1, 1), has slope m = 3.14 (e) l passes through (1, 2) and (0, 1) (f) l has x-intercept 3, and y-intercept 4. Notice that on parts (c), (e), and (f), you first need to use the Slope Formula from Equation 5.2. For more practice, try problems 1-28 of Section 2.5. You might be wondering, how do we know that the point-slope formula works? Well, once we know the slope m of the line, and a single point (a, b) on the line, then the slope formula given in Equation 5.2 tells us that, for any other point (x, y) on the line, we must have: (5.4) m = y b x a Where here we have substituted (x, y) for (x 1, y 1 ), and (a, b) for (x 0, y 0 ) in the slope formula. After we multiply both sides of this equation by (x a), we get (5.5) (x a)m = y b And this is just the point-slope formula written in reverse order. 5.3. One more little thing about parallel and perpendicular lines. A diagram I draw on the board (together with a little bit of deductive logic), tells us the following fact: Perpendicular Line Slope Formula: Suppose l has slope m and that the line p is perpendicular to l. Then the slope of p is 1 m. On the other hand, it s pretty easy to see that parallel lines will have the same slope. These facts, together with the point-slope formula, allows us to solve the following kind of problem. Homework 5, # 4 Find equations for the following lines: (a) l is parallel to y = 2x but passes through (4, 2). (b) l passes through the point ( 2, 7) and is perpendicular to the line y = 5x + 4. (c) l passes through ( 2, 2) and is parallel to the line with equation 2x 3y = 7. (d) l passes through ( 2, 2) but is perpendicular to the line with equation 2x 3y = 7. For more practice, try problems 45-56 in Section 2.5 of the book.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 15 6. Solving small systems of linear equations In this part of the notes we cover Chapter 3.1 of the book. 6.1. Solutions of Systems of Equations. A system of equations in 2 variables is just a collection of equations in which the same 2 variables occur (and we will almost always choose these variables to be x and y). For example: (6.1) y = x 2 1 = xy A solution of a system of equations in 2 variables is a pair of numbers (a, b) which make every equation in the system come out true when you substitute a for x and b for y. For the system above, it is easy to check by plugging in that the pair ( 1, 1) is a solution, while (1, 1) is not. Homework 6, #1 Verify that ( 1, 1) is a solution to the system above by actually plugging in x = 1, y = 1. Also, show that (2, 3), (1, 1), and ( 1, 1) are not solutions by plugging them in. Then plug these same pairs of numbers into the system (6.2) x 2 = 2 y 2 x = y Which pairs are solutions, and which are not? Usually, there will only be a handful of solutions to a system of equations, if any. 6.2. Finding solutions: the substitution method. It is often difficult or even impossible to find the solutions to a system of equations, if it is too complicated. One way to try, however, is to use the substitution method. The substitution method: Suppose you have a system of two equations in two unknowns, x and y. Then the following steps will lead to a solution: (1) Use algebra to reduce one equation to the form y = f(x),. (2) Plug in f(x) for y in the other equation, then solve for x. (3) Plug each value of x back into y = f(x) to find corresponding y values. Rather than explain with more words, it is easier to explain the process by example: Example 6.1. Find all solutions of the original example system y = x 2 1 = xy Solution: The first equation can be rewritten as y = x 2. Plugging in x 2 for y in the second equation gives us 1 = x 3.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 16 Multiplying both sides by 1 then turns it into 1 = x 3 and by taking the cube root of both sides we find x = 1. Finally, plugging x = 1 back into the first equation gives y = ( 1) 2 = 1. Therefore there is only one solution, and it is given by the pair ( 1, 1). Homework 6, #2 Use the substitution method to find all solutions to the system x 2 = 2 y 2 x = y For more complicated systems of equations the substitution method won t work, because you won t even be able to do step (1),or maybe step (2). 6.3. Using graphs to understand systems of equations. As we discussed in earlier sections, any equation in two unknowns x and y has a graph of points in the plane. This graph consists of all points (a, b) on the plane which make the equation come out true. If we have a system of two equations E 1 and E 2 in two unknowns, then we will have two graphs, call them G 1 and G 2. If the point (a, b) lies in G 1 and G 2, then the pair of numbers x = a, y = b solves E 1 and E 2. And on the other hand, if the pair of numbers x = a, y = b solves both E 1 and E 2, then the point (a, b) lies on both graphs G 1 and G 2. Therefore we have the following basic fact: FACT: If G 1 and G 2 are the graphs corresponding to a system of two equations in two unknowns, then the solutions of this system correspond to intersection points of G 1 and G 2. Homework 6, # 3 Draw the graphs corresponding to both systems of equations occurring earlier in the notes. Verify that the intersections do indeed correspond to the solutions we found earlier. 6.4. Linear systems of equations. For more complicated systems of equations the substitution method won t work because you just can t do step (1), or maybe you get stuck on step (2). Using graphs to find solutions is also very hard. However, for the time being you will only need to be concerned with a very special kind of systems: Linear Systems. For linear systems, it is easy to find the graphs, and the substitution method always works. Definition 6.2. A system of equations in two variables is said to be linear if every equation in the system can be algebraically reduced to the form: (6.3) Ax + By = C Where, A, B, and C are constants.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 17 Neither of the two systems we have been working with so far have been linear. However, all of the systems in the following homework problem are. Homework 6, #4 Find all solutions (if any) to the following systems of equations using the substitution method. (a) (b) (c) x + y = 6 y = 2x x 3y = 3 3x + 5y = 19 y = 2 5 x 2 2x 5y = 10 For more practice, try problems 25-42 in Chapter 3.1 of the book. Homework 6, #5 Use the graphing method to find all solutions (if any) to the following systems of equations. (a) (b) (c) x + y = 4 x y = 2 x y = 2 y = 1 2x 3y = 6 4x 6y = 24 Homework 6, #6 In your own words, explain why there are only three possibilities for a system of linear equations in two unknowns: (1) There are no solutions. (2) There is exactly one solution. (3) There are infinitely many solutions. Hint: Use the big fact from Section 6.3 of these notes. In case (1), the system is called inconsistent, and in the third case the system is said to have a dependency, although the reasons for this latter terminology will not be clear to you for awhile. 7. Three variables. In this part of the notes we finish up Section 3.1 and cover Section 3.3 as well. 7.1. The addition method for two variables. Besides the substitution and graphing methods, there is yet another way to solve systems of linear equations. This method is usually faster than the other two methods, and it extends easily to larger systems of linear equations in more variables.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 18 Suppose we have a system of two linear equations in two variables as in the last section: (7.1) A 1 x + B 1 y = C 1 A 2 x + B 2 y = C 2 Recall that a pair of numbers (a, b) is a solution to this system if both equations come out true when a is substituted for x and b for y. In addition to the usual rules for simplifying equations that you already know, there is one new move that we can perform on systems of equations which does not change its set of solutions. Addition Rule: In the system of equations 7.1 above, the first equation tells us that A 1 x+b 1 y is interchangeable with C 1 (that s what it means for them to be equal!). Therefore, we can add any multiple n of A 1 x + B 1 y to the left side of the bottom equation, while adding the same multiple n of C 1 to its right side without changing the solutions to the system: (7.2) And this simplifies to: A 1 x + B 1 y = C 1 n(a 1 x + B 1 y) + A 2 x + B 2 y = C 2 + nc 1 (7.3) A 1 x + B 1 y = C 1 (na 1 + A 2 )x + (nb 1 + B 2 )y = nc 1 + C 2 In other words, you can always add a multiple of the top row to the bottom row and still get the same solutions. Of course, we can also add or subtract the bottom row from the top as well. This new rule, when used with other basic algebraic rules, is very powerful. Example 7.1. Consider the system (7.4) x + y = 4 x y = 2 from your previous homework. We use the addition rule to add the top equation to the bottom one, and get: (7.5) x + y = 4 2x = 6 If we divide both sides of the bottom equation by 2 (which also does not change the set of solutions) we get: (7.6) x + y = 4 x = 3 Finally, we can subtract the bottom equation from the top, and get: (7.7) y = 1 x = 3

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 19 Homework 7, #1 Solve the same systems of equations as in Homework 6, #4 and #5, but use the addition method (it s probably a good idea to start with the problems of #5 first, because they are easier). For more practice with the addition method in two variables, try exercises 43-58 of Chapter 3.1 in the book. 7.2. Systems of Linear Equations in 3 variables. Now let s consider systems of three linear equations in three unknowns x, y, and z, which have the following form: (7.8) A 1 x + B 1 y + C 1 z = D 1 A 2 x + B 2 y + C 2 z = D 2 A 3 x + B 3 y + C 3 z = D 3 Here, all the capital letters are assumed to be constants. A solution to this system is a triple of numbers (a, b, c) such that all three equations come out true when you substitute a for x, b for y, and c for z. All of the reasoning is the same for these systems as it is for the systems in two variables. In particular, you can still use the substitution method and the addition method (and there is even something like a graphing method ) to find solutions to the equations. However, the substitution method is now usually much harder than the addition method. Example 7.2. Let us solve the following system of equations: (7.9) x + y + 2z = 11 x + y + 3z = 14 x + 2y z = 5 Just for the sake of teaching you a lesson, I m going to solve this system in class using the substitution method. However, I m not going to torment myself by writing it all out here, especially since you should probably never use the substitution method on a system this big. Instead, I will use the addition method to quickly solve this equation. Notice first that if we subtract the top equation from the middle equation we get: (7.10) x + y + 2z = 11 z = 3 x + 2y z = 5 If we add the middle equation to the bottom equation, and then also subtract 2 times the middle equation from the top equation, we get: (7.11) x + y = 5 z = 3 x + 2y = 8 Now if we subtract the top equation from the bottom, we get:

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 20 (7.12) x + y = 5 z = 3 y = 3 Finally, if we subtract the bottom from the top, we end with: (7.13) x = 2 z = 3 y = 3 This tells us that the original equation had only one solution, the triple (2, 3, 3). We can check that this is right by actually plugging it into the original system! Homework 7, #2 Solve the following systems of equations: 2x + y 2z = 1 (a) 3x 3y z = 5 x 2y + 3z = 6 2x + y = 2 (b) x + y z = 4 3x + 2y + z = 0 2x + y + 2z = 1 (c) 3x y + z = 2 x 2y z = 0 7z 3 = 2(x 3y) (d) 5y + 3z 7 = 4x 4 + 5z = 3(2x y) For more practice do problems 5-22 of Chapter 3.3 in the text. 7.3. The geometric significance of linear equations in 3 variables. we can use Cartesian coordinates to associate a pair of numbers with every point on the plane, we can use Cartesian coordinates to associate a triple of numbers to every point in space. The graph of an equation in three variables f(x, y, z) = g(x, y, z) is then just the set of points (a, b, c) in space which make the equation come out true when a = x, b = y, and c = z. When an equation has the form Ax + By + Cz = D its graph is guaranteed to be a plane. The set of solutions to a system of three such equations as in 7.8 above is then the set of all points in space which lie on all three planes. It follows that there are only three possibilities for the number of solutions to a system of three equations in three unknowns. (1) There is exactly one solution. (This is the most common).

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 21 (2) There are infinitely many solutions. (3) There are no solutions. 8. Applications of Systems of Equations In this part of the notes we cover Section 3.2 and a little more of 3.3. 8.1. The Fruit Stand. The ability to solve systems of linear equations is often useful when you are trying to isolate the individual effects of two or more quantities which always work together. What I mean is best shown by the following contrived, but easily understood, fruit stand word problem. Example 8.1. John goes to a fruit stand and pays a total of $7 for two apples and three oranges. Jane goes to the same fruit stand and pays a total of $6 for one apple and four oranges. What was the price of apples, and what was the price of oranges? A good first step with word problems is to identify important unknown quantities. In this problem, the important unknown quantities are the price of apples and the price of oranges. Let x denote the price of apples, and let y stand for the price of oranges, both in dollars. The next step is to convert the facts we have into equations. In this case, the first sentence tells us that 2x + 3y = 7 The second sentence tells us that x + 4y = 6. The price of apples x and the price of oranges y will have to make both of these equations come out true. Thus, we have now finally converted our word problem into a math problem: we must solve the system { 2x + 3y = 7 (8.1) x + 4y = 6 If we subtract twice the bottom row from the first, we get the new system (8.2) { 5y = 5 x + 4y = 6 From the top equation we see y = 1, and plugging this into the bottom equation tells us x = 2. It follows that apples must have cost $2, and oranges $1. Homework 8, #1 John pays $12 for three bananas, two mangos, and two boxes of strawberries. Jane pays $10 for five bananas and two boxes of strawberries. Javier pays $20 for five mangos and four boxes of strawberries. What is the price of bananas? There are many problems which are just fruit stand problems in disguise. Here s one for you to think about over the weekend:

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 22 Homework 8, #2 Rex the dog likes two brands of dog food: bacon-flavored and steakflavored. Bacon-flavored costs $10 a bag, and steak-flavored costs $15 a bag. He ended up eating 10 total bags of dog food last month, on which you spent a total of $120. How many of the bags that he ate were bacon-flavored? Finally, I would like you to know how to do this kind of problem: Homework 8, #3 You invested $7000 in two accounts paying 6% and 8% annual interest. If the total interest you earned for the year was $520, how much was invested at each rate? For full homework credit on this I need to see some work, not just an answer. For more practice, try problems 12-18, and 25-28 in Section 3.2 of the book. 8.2. Finding parameters of geometric shapes. Any parabola with a vertical axis of symmetry will be the graph of a function of the form f(x) = ax 2 + bx + c. Suppose now that someone just draws the graph of such a parabola, and gives you the coordinates of three points on it. Can you use these to determine the full of equation of the parabola? The answer is yes, using systems of linear equations. Example 8.2. Suppose a parabola with vertical axis of symmetry passes through the points ( 1, 2), (0, 4) and (1, 1). Find the equation of this parabola. We know that, whatever the equation is, it has the form y = ax 2 + bx + c. The unknowns for us in this situation are the numbers a, b, and c. Once we find them, we find the equation. Knowing that the parabola passes through ( 1, 2) tells us that the equation 2 = a( 1) 2 + b( 1) + c is true, which is the same as 2 = a b + c. Similarly, the fact that (0, 4) lies on the parabola tells us that 4 = a 0 2 + b 0 + c. So in fact c = 4. Finally, since (1, 1) lies on the graph, we see that 1 = a 1 2 + b 1 + c. All told then, we see that we want a solution of the system a b + c = 2 (8.3) c = 4 a + b + c = 1 If we subtract the second equation from the top and bottom, we get (8.4) a b = 2 c = 4 a + b = 3

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 23 If we add the bottom to the top we get (8.5) 2a = 5 c = 4 a + b = 3 From the first equation we see that a = 5 2, plugging this into the bottom we see b = 1 2, and from the middle equation we immediately see that c = 4. Therefore the equation of the parabola that we are after is y = 5 2 x2 + 1 2 x + 4. Homework 8, #4 Find the equation y = ax 2 + bx + c of the parabolas which pass through the following triples of points. If you cannot find them, explain why. (a) ( 1, 4), (0, 1), (2, 9). (b) ( 2, 1), (1, 1), (2, 4). (c) (0, 2), (0, 3), (1, 4) For more practice, try problems 31 and 32 in Section 3.3 of the book. Similarly, any equation of the form ax 2 + by 2 = 1 gives us the equation of an ellipse centered at zero with vertical and horizontal axes of symmetry. It makes intuitive sense that any such ellipse will be determined by two points, and using the same method as above (except easier) we can find its equation. Homework 8, #5 Find the equation of the ellipse ax 2 + by 2 = 1 which passes through the following pairs of points. If you cannot find such an ellipse, explain why. If there are many, explain why. (a) (4, 0), (0, 1) (b) (1, 1), (2, 0) (c) (1, 1), (1, 1) (d) (1, 1), (1, 1.1) 9. Inequalities This section of the notes covers 4.1-4.3 of the text. 9.1. The meaning of an inequality. We say that a < b if there is some positive number c such that a + c = b. We say that a b if a < b or a = b. Finally, a > b and a b mean the same thing as b < a and b a, respectively. Example 9.1. The following inequalities are true: 2 < 4

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 24 π 4 2 > 4 Definition 9.2. A number a is a solution of the inequality f(x) < g(x) if the inequality comes out true when you substitute a for x. Similar definitions apply for the symbols >,, and. When you solve an equation of the form f(x) = g(x), there are usually only a handful of solutions at most. However there are usually an infinite number of solutions to inequalities. Example 9.3. The inequality 2x < x 2 is true whenever x > 2 or x < 0. You might be wondering how I figured that out. The basic idea is to use algebra to simplify the inequality like you would with an equation. For example, if a < b, then for any number c, a + c < b + c. Therefore, given any inequality f(x) < g(x) we can add a function h(x) to both sides and get a new inequality f(x) + h(x) < g(x) + h(x) which has the same solutions. For example, using this fact we can simplify the inequality like so 2x < x 2 (9.1) ( 2x) + 2x < x 2 2x 0 < x 2 2x 0 < x(x 2) And in this last simple form it is easy to see that the inequality is true when x < 0 or 2 < x. Notice, however, that if 0 < x < 2, then x is positive while (x 2) 0 is negative, which forces x(x 2) to be negative. Homework 9.1 Solve the following inequalities: (a) 2x 11 < 3 + x (b) 4(x + 2) 3 2x + 8 (c) x 2 > 2x+1 3 4 (d) 1 x 2x + 1 3 Notice that you can partially verify whether your answer is correct by plugging in a few values of x and seeing if they work. For more practice try problems 1-38 of Section 4.1.

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 25 9.2. A warning about multiplication and division. It is tempting to try to solve an inequality like 2x < x 2 by dividing both sides of the equation by x, which turns it into 2 < x. But notice that this simplification actually throws out all the negative numbers that we found to be solutions earlier! This happened because we divided both sides of the inequality by the variable x. WARNING: You are not allowed to multiply or divide both sides of an equality by x! If a < b then it is only true that ac < bc if c is a positive number. If c is negative, then a < b implies ac > bc. For example, 1 < 2, and if we multiply both sides of this inequality by 2 we get 2 < 4, which is fine. However, if we multiply both sides of this inequality by c = 2 we have to reverse the inequality to get 2 > 4. Special Rules for Multiplication and Division: When solving an inequality f(x) < g(x), we can perform the following moves: (1) Changing it into Cf(x) < Cg(x) or f(x) C (2) Changing it into Cf(x) > Cg(x) or f(x) C < g(x) C > g(x) C if C is a positive constant. when C is a negative constant. We cannot multiply or divide both sides of an inequality by a variable x because it might stand for a positive C or a negative C, and the rules are different for positive and negative C. Homework 9.2: Solve the following inequalities. (a) 3x < 8 (b) 2x+1 > 1 2 (c) 2x 3 1 2x 3 3 (d) 3x 2 > 2x (e) 2x 1 > 1 x For more practice you can try the same problems that I listed after HW 9.2 of these notes. They will be easier than some of the problems above, but they are as hard as you need to know for the quizzes. 9.3. How to deal with the words and and or with inequalities. Solving a compound inequality just means solving a couple of inequalities at the same time. For example: Example 9.4. Find the set of all x such that 2x 1 > 3 AND x 1 8. Solution: The inequality 2x 1 > 3 reduces to x > 2, and the inequality x 1 8 reduces to x 9. Therefore, the set of all x which make both true is the set 2 < x 9, or (2, 9],

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 26 in interval notation. When solving compound inequalities, it is useful to draw a number line to keep track of things. When solving a compound inequality involving the word AND as above, the set of solutions will be the overlap of the sets of each equation you solve. If you are asked to find all solutions to g(x) < f(x) < h(x), this is the same as the set of x which solve g(x) < f(x) AND f(x) < h(x). Homework 9.3: Find the set of all x such that: (a) 1 4x < 1 and 3x 5 6 2 (b) 6 < x 4 1 (c) 2 x 4 2 3 (d) 1 < 1 x 7. Express your answers in interval notation. For more practice try problems 15-32 of Section 4.2. Here s a somewhat different kind of compound inequality. Example 9.5. Find the set of all x which solve 2x 3 > 2 x OR 1 x x 2. Solution: The first inequality reduces to x > 5. The second inequality reduces to 1 3, 3 x which is solved by all x 1 such that x 0. In interval notation, then, our solution is 3 (, 0) (0, 1) ( 5, ). 3 3 Again, it is useful to use number lines to keep track of everything. The difference is that, with compound inequalities which use the word OR, the set of solutions will consist of all the solutions that occur in either inequality, not just their overlap. Homework 9.4 Find the set of all x such that: (a) x < 3 or x 1 (b) 3x + 2 5 or 5x 7 8 (c) x 2 9 or x 2 < 1 (d) 1 1 x > 0 or x 1. Express your answers in interval notation. For more practice try problems 39-58. 10. The absolute value This section of the notes covers Section 4.3 of the notes. 10.1. Solving equations involving the absolute value. The absolute value function x is defined to be x when x > 0 and x when x < 0. Therefore, x = 3 when x = 3 OR when x = 3. Similarly, more complicated equations like might be true when 2x + 1 = 3 2x + 1 = 3

OR CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 27 (2x + 1) = 3, but you need to plug in at the end to make sure the solutions you find really do work. Here we find the two solutions x = 1 and x = 2 both work in the original. Example 10.1. Find the solutions of 2x 1 2x + x = 2x 3x 2. Solution: It usually saves time to first reduce the equation to the form f(x) = g(x) first. (10.1) 2x 1 2x + x = 2x 3x 2 2x 1 2x = x 3x 2 1 2x = 1 2 3 2 x Now that it is in this form, we break it up into two cases. When 1 2x = 1 2x, the equation becomes (10.2) 1 2x = 1 2 3 2 x 1 2 = 1 2 x 1 = x However, notice that the solution x = 1 does not actually work in the original equation! This is because 1 2x = 1 2x is not true when x = 1, like we assumed at first. So in fact the equation has no solution when 1 2x = 1 2x. The second case to consider is 1 2x = (1 2x) = 2x 1. In this case our equation becomes (10.3) 2x 1 = 1 2 3 2 x 7 2 x = 3 2 x = 3 7 You can now verify that x = 3 is a solution. By the way, x = 0 is also a solution, one 7 which we forgot the moment we divided both sides of the original equation by 2x. Think about that. Homework 10.1: Solve for x. (a) x = 1 (b) x 1 = 4 (c) 3x 2 = 5 x (d) 2 x + 1 3 = 4 2x (e) 3 2x + 1 = 3 2x