Class 5 -- Equations I. Single-Variable Equations A. Addition Principle B. Multiplication Principle C. Addition and Multiplication Together D. Raising to Powers E. Fractions II. Linear Inequalities III. Absolute value problems A. Meaning B. Equations C. Inequalities I. Single-Variable 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. C. Raising to Powers The idea of solving for an unknown in an equation by doing something to get that unknown by itself extends to more complicated equations as well. We just need to know the opposite of a given operation. One example is raising to a power. Suppose we have an equation: sqrt(x) = 4 We want to get x by itself, so we must do something to both sides. The solution is to square both sides, because square root means the opposite of square. sqrt(x)^2 = 4^2 x = 16 Checking, you can immediately see that, indeed, sqrt(16) = 4. So we have solved the equation. The same applies to other powers: x^(1/3) = 2 Raising to the 1/3 power, or taking the cube root, is the opposite of cubing. Thus to solve we do [x^(1/3)]^3 = 2^3 x = 8 Checking, we see that 8^(1/3) = 2, so we have solved the equation. We do need to be a little careful with negatives, however: sqrt(x) = -3 sqrt(x)^2 = (-3)^2 x = 9 However, if we plug this in, it doesn't work: sqrt(9) = 3 != -3 This is potentially an issue whenver we are raising to a positive power, because raising to a positive power changes the sign. E. Fractions Another type of equation we can solve is ones where variables appear in fractions. We can do this using our experience from last week in cancelling fractions. Suppose we have 4/(x+1) = 2/(x-1) To solve this, we can multiply by factors to cancel the x+1 and the x-1 in the denominator. In this case, we'll multiply by (x+1)(x-1): [(x+1)(x-1)] 4/(x+1) = [(x+1)(x-1)] 2/(x-1) The x+1 cancels on the left, and the x-1 on the right. Thus we have 4(x-1) = 2(x+1) This is an equation of the sort we can already solve: 4x - 4 = 2x + 2 2x = 6 x = 3 It's important to check this: 4/(3+1) = 2/(3-1) 1 = 1 This checks out. The reason it is important to check is to make sure we don't end up with a zero in the denominator. If we'd gotten x = 1 instead of x = 3, then the x-1 would have been zero, and we would have been in trouble. II. Linear inequalities Something very closely related to equations is inequalities: statements that two quantities are unequal in some particular way. An inequality is just like an equation, except that we replace the = sign with an inequality sign: <, <=, >, >= Just to remind you what these signs mean: (on board) < Less than <= Less than or equal to > Greater than >= Greater than or equal to A useful mnemonic for this is that the sign is the mouth of a fish, and the fish opens its mouth for the bigger morsel. (Draw picture) I. Simple Inequalities Let's consider a simple equation we know how to solve, and replace the = sign with an inequality: (on board) 3x - 9 < 15 The goal is to get the variable by itself, just as we do with an equation. OK, how would we solve this if it were an equation? (Let class do this.) In the end we get: x < 8 (on board) That's all there is to it: just like an equation. Let's consider another example: 5 - 2x >= 7 How could we solve this? (Let class do it -- they'll pick one of the two ways below) Note that we can approach this two different ways. One way would be to move the x to the right side: (on board) 5 - 2x >= 7 +2x +2x 5 >= 7 + 2x -7 -7 -2 >= 2x -1 >= x So, in the end we get x <= -1. Now we could also solve the problem this way: (on board) 5 - 2x >= 7 -5 -5 -2x >= 2 OK, so now we divide by -2. Note that if we do this and leave the inequality as it is, we'll get x >= -1. But something is wrong here! That's exactly the opposite of what we got when we did the problem the other way. The problem is that when we multiply or divide by a negative, we have to flip the inequality: > becomes <, >= becomes <=, and vice-versa. Thus, when we divide by -2, we should write (on board) -2x / -2 <= 2 / -2 x <= -1 Now we get agreement. That's the only difference between solving inequalities and equations: with inequalities, you have to flip the inequality when you multiply or divide both sides by a negative. III. Absolute value problems A. Meaning A third closely related idea is equations and inequalities involving absolute values. First off, what is an absolute value? It just means the size of a quantity without regard to its sign. Absolute value is very simple: it's just like a parenthesis, but once you've evaluated everything inside it, take the positive of that number. Thus for example: |100| = 100 |-10| = 10 |3 - 5| = 2 |5 - 3| = 2 We can think of absolute value as a method of measuring distance. The distance between 3 and 5 is 2, and it doesn't matter is we ask about the distnace between 3 and 5 or the distance between 5 and 3. B. Equations Absolute values can be used in equations that describe distances. For example suppose we wanted to describe the points on a rule that are 3 inches from the 6 inch mark. (Draw picture.) We could write this algebraically as |x - 6| = 3 Clearly there are two possible answers: the 3" and 9" marks are both 3" from the 6" mark. Plugging these two possibilities in confirms that they work: |9 - 6| = |3| = 3 |3 - 6| = |-3| = 3 Algebraically, we can solve an absolute value equation by noticing that the quantity in absolute value signs could be equal to either the positive or the negative of whatever is on the other side of the equality. Thus |x - 6| = 3 ===> x - 6 = 3 or x - 6 = -3 We can then solve the resulting pair of equations to find the answer: x-6 = 3 ==> x = 9 x-6 = -3 ==> x = 3 If the equation doesn't come with the absolute value sign by itself on one side, we just have to manipulate it to put it into that form: |x - 6| + 4 = 6 |x - 6| = 2 x - 6 = 2 or x - 6 = -2 x = 8 or x = 4 That's all there is to solving an absolute value equation. C. Inequalities Finally we come to absolute value inequalities. We can imagine one just by extending our previous example: what are all the distances on the ruler that are less than 3" from 6"? The answer is obviously all the distances bewteen 3" and 9". We can solve this algebraically using exactly the same trick as before of dividing the equation into two pieces. We start with |x - 6| < 3 In order for this to be true, x - 6 has to less than 3 or more than -3. Thus we write x - 6 > -3 and x - 6 < 3 x > 3 and x < 9 We can combine this into a single shorthand: 3 < x < 9. That's the answer. Suppose that, instead, we wanted to find all the points on the rule that are *more* than 2" from the 5" mark. In this case we have |x - 5| > 2 We could make this true either by having x-5 be more than 2, or by having it be less than -2. Thus we have x - 5 < -2 or x - 5 > 2 x < 3 or x > 7 And this makes sense: the marks below 3" or above 7" are indeed those that are more than 2" from the 5" mark.