Ch. 3 Equations and Inequalities

Ch. 3 Equations and Inequalities 3.1 Solving Linear Equations Graphically There are 2 methods presented in this section for solving linear equations graphically. Normally I would not cover solving linear equations in this manner, but this is as close as we will get to solving systems of linear equations (in this class), so I do not want to skip the section. Method #1: Intersection of Equations (the solution of a system) 1) Treat the left and right sides as two different functions (linear equations in 2 variables) 2) Graph each equation 3) The x-coordinate of the point of intersection is the solution. Since we are really only solving a linear equation in 1 variable and the y-coordinate is just the value of each expression (left side/right side) when they are evaluated at the solution. Solve the following linear equation in one variable graphically. 3 4x = 2 3x y Step 1: Left side is an equation & Graph y 1 = -4x + 3 Step 2: Right side is an equation & Graph y 2 = -3x + 2 Step 3: Find the intersection point & label The x-coordinate is the solution x Check & Note left & right value is y-coordinate Could you imagine what a hassle this would be if the answer was a fraction(decimal) or a huge(small) number? Well, that s one of the beautiful things about these graphing calculators of ours. Let s take those out and learn a few tricks to using them. Solve the same problem using the calculator that we solved by hand. Step 1: You still have to get the 2 equations as above. Step 2: Find the key that looks like Y= and push it. Step 3: Using the X,T,θ,n key and the ( ) and + key in the equations to Y 1 and Y 2 (you can move between those with the arrow keys) Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 1

Step 4: Find the ZOOM key and choose Standard (use arrows or enter 6) This graphs the 2 equations. Step 5: 2 nd TRACE will get you into the CALC menu, and you need the INTERSECT function (use arrows or enter 5). Once there, Y 1 should be in the upper left corner, if it isn t then down arrow until it is and ENTER Now, Y 2 should be in the upper left corner, press ENTER again. It will now say GUESS in the lower left corner, press ENTER again and your intersection will be shown. Remember you want the X-Value. Your Turn Using the same method, solve 1 / 2 (3x 4) = 3 / 4 (x 6) + 1 y x Try the graphing calculator with this exercise as well. Now the 2 nd method. This method finds the x-intercept of the equation by setting the function created by the equation equal to zero (put all terms on one side so that the other side equals zero; this is equivalent to letting y=0, the same way we did when we found x-intercepts in Ch. 2). Mehod #2: X-Intercept Method 1) Move all terms to either the left or right using addition property of equality 2) Graph the equation and locate the x-intercept. The x-coordinate is the solution. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 2

Redo the 1 st example using this method. 3 4x = 2 3x Step 1: Put all terms on one side 3 4x 2 + 3x = 0 -x + 1 = 0 Step 2: Graph as if this is: y = -x + 1 Step 3: Locate the x-intercept This is the solution (x-coord) y x Now the calculator method. Step 1: Repeat the process for putting in the equation as above for Method 1. This time you are using the equation from step 2 of our process (y = -x + 1). Step 2: Again graph the equation. Note: If there are other equations in your calculator you can prevent them from being graphed by moving your cursor to the equal sign (use the arrow keys) and pressing ENTER The equal signs will be unhighlighted, meaning that they won t be graphed. Step 3: Again go to the CALCULATE menu (see the instructions above) and this time choose the ZERO (use the arrow keys or press 2). The calculator will prompt LOWER BOUND? in the lower left corner and you should press ENTER Now it (use the up/down arrow key) to move the cursor along the line until it is on the other side of the x-intercept, and press ENTER again. Now it will prompt you with GUESS? in the lower left corner and you will press ENTER again and be rewarded with the x-intercept. Now try both the by hand method and the calculator method with the second example. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 3

Your Turn Using 2nd method, solve 1 / 2 (3x 4) = 3 / 4 (x 6) + 1 y x Try the graphing calculator with this exercise as well. Finally, we must discuss equations that have no solution or infinitely many solutions. An equation that has no solution yields an untrue statement when solving it using typical linear methods. When solving using method one we would see that the lines are parallel (have the same slope) and that there is no intersection. When we use method 2 we will see that there is no x-intercept (we have a horizontal line; not y = 0) This is a special type of a equation called a contradiction (an equation with no solution) and its solution set is written as a null set (symbolized with or { }). An equation with infinitely many solutions yields a true statement, when solving it using traditional linear methods. When using method one we will see only one line and that every point is the same. Using method 2 we will get the horizontal line y = 0 in which case every point is the x-itnercept. This is called an identity (an equation with infinite solutions and its solution set is all real numbers). I will allow you to write R rather than {x x is a real numbers}, but note that the book indicates the set builder notation as the correct answer. Here is the same information, in list form. Three Types of Equations Contradiction Instead of variable = #, a number will equal a different number, making a false statement. This means that there is no solution. or {} is the solution set. Identity Instead of variable = #, a number will equal itself or variable will equal itself, making a true statement. This means that there are infinite solutions. R is the solution set. Conditional An equation with only one solution, variable = #, is considered conditional. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 4

State conditional, contradiction or identity and give the solution. a) c + 5 = c 5 b) 2(a + 1) = 2(a 5) + 12 c) 2(a + 1) = 2(a 5) + 12 + a Note: It is a common error when working with contradictions, identities and conditional equations to give the answer of {} when the equation is truly conditional with a solution set of {0}. I m going to leave it up to you to investigate these further. Remember our linear equations from 1.8? These come back into play in this chapter. First, we will write two linear equations for two scenarios using the same independent variable, and then we will set them equal to find our solution the break-even point for the scenarios. The cost for producing a CD is $550 plus $2 per CD and the CD s are sold 10 in a pack. If the sale price is $45 per pack, how many sales must be made to break even? Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 5

3.2 Linear Inequalities and Problem Solving Review of Inequality Symbols < > The OLD way of graphing simple inequalities (contain only one inequality) < -- Graph with an open circle and a line to the left, if in standard form > -- Graph with an open circle and a line to the right, if in standard form -- Graph with a closed circle and line to the right, if in standard form -- Graph with a closed circle and a line to the left, if in standard form Summary of the NEW way < -- ) and a line to the left > -- ( and a line to the right -- [ and a line to the right -- ] and a line to the left Let s practice standard form 1 st. Write the variable of the left and the number on the right. Rewrite the inequality so that the arrow points to the same thing it did originally. Examples: Put the following in standard form a) 5 x b) 2 y c) 5002 > m Now some graphing. Examples: Graph each of the following (on a number line) a) x > 3 b) 2 > y c) -2 v d) t 5 Parts b and c in the example are examples of inequalities written in nonstandard form. To read an inequality written in nonstandard form, we must read right to left, instead of our usual left to right. Thus it is easiest to always put inequalities in standard form before reading and trying to graph them. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 6

Graphing of compound inequalities (contains 2 inequalities): A compound inequality is read as an and which is an intersection (overlap). 15 < x < 20 is read as: x is greater than 15 and x is less than 20 Graph the following compound inequalities a) 8 x < 9 b) 2 < x 5 c) 5 x > -2 d) 5 x -2 Note: The authors chose to discuss compound inequalities in 3.3, but I am going to at least cover the graphing and interval notation for both in this section. Now let s extend this to another new way of writing a solution set. Up to this point you have been taught roster form and set builder notation. We will learn the last method here. It is called interval notation. It is essentially the shortcut of writing what you just saw on the number line. It tells about the inclusion or exclusion of left endpoint ( Endpoints are the beginning or end of the solution set) and the right endpoint and slams them together with a comma in the middle. When graphing simple inequalities, we can get a solution set that travels out to negative or positive infinity (-, ). Infinity is an elusive point since you can never reach it, and therefore in interval notation we always use a parenthesis around infinity. Summary Endpoint included [ or ] Endpoint not included ( or ) Negative Infinity (- Positive Infinity ) Visually Relating to the Number Graph of an Inequality x > 5 Will be graphed with an parenthesis (open circle) on 5 and a line to the right with an arrow on the end showing that it continues on to infinity. ( 5 Interval Notation (5, ) Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 7

( ] -2 ] -1 0 5-2 x Will be graphed with an bracket (closed circle) on 5 and a line to the right with an arrow on the end showing that it continues on to infinity. Interval Notation (-, -2] -1 < x 5 Will be graphed with an parenthesis (open circle) on -1 and a bracket (closed circle) on 5 and a line in between. Interval Notation (-1, 5] Now, go back to the examples that we graphed and write their interval notation. You will find it a lot easier at first if you have the graph right in front of you. Solving Linear Inequalities If all we must do to solve a linear inequality is to add or subtract then they are solved in the exact same manner as a linear equation. Recall the steps to solving a linear equation are: 1) Simplify (distribute, clear, combine like terms) 2) Use the Addition Property of Equality to move terms across equal sign 3) Use the Multiplication Prop. of Eq. to remove the numeric coefficient & produce the solution var = #. [Mult. by a neg. reverses the sense of the inequality (flips the inequality)] Solve, graph and give the interval notation for following the linear inequalities a) 2x + 7 < 9 + x b) 2x 15 + x 15 + 2(x 1) If we must multiply or divide by a positive number to get the solution, then the steps for finding the solution are still the same! Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 8

We can use the graphing methods of the 1 st chapter here as well, but we will be looking for the x-coordinates before or after (depending upon the inequality) the intersection. To help further this we can change the way the calculator graphs one of the equations so we can read our solution more easily. Let s use the next example to show this. Graphing Solve, graph and give the interval notation for following the linear inequality 5x 15 + 2 15 + 3x Step 1: Enter the left side as Y 1 and change the line type by moving the cursor to the left of Y 1 and pressing the ENTER key so you see a bold line. Step 2: Enter the right side as Y 2 leaving the line type alone. Step 3: See the instructions above for finding the intersection of the lines. Step 4: From the intersection to the left (because it is ) are the solutions to this inequality. You are reading the x-values of the coordinates where the y-values of the left side s equation (Y 1 ) are less than the y-values of the right s equation (Y 2 ). The problem occurs when I must multiply or divide by a negative number! Whenever I multiply or divide by a negative number, the sense of the inequality must reverse. Solve, graph and give the interval notation for following the linear inequalities a) 3 / 4 x 1 2x + ¼ b) -2x + 5 > 9 To see why it is true that the sense of the inequality must reverse, let s rework these problems and keep the numeric coefficient of the variable positive. Notice that you must read the inequality in nonstandard form (right to left) and the answer is the same as above. It can also be seen on a graph of the lines. Graph the first of the examples below and you will see that the y-values of the left are greater than the y-values of the right when the x s are less than the intersection s x- coordinate. Solve, graph and give the interval notation for following the linear inequalities a) 3 / 4 x 1 2x + ¼ b) -2x + 5 > 9 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 9

We may also encounter problems that have no solution or infinite solutions. Just as with linear equations in 1 variable or systems, we will find a true statement when the solution is infinite solutions and a false statement when the solution is a null set. Solve. a) 4(x 1) 3(x 2) + x b) 2(x + 3) + 5 < 2x 4 Word Problems & Inequalities at least more than > minimum maximum between compound in standard form small < x < large #86 p. 251 Intermediate Algebra by Blitzer The bank charges $8/month plus $0.05 per check. The credit union charges $2/month plus $0.08 per check. How many checks could be written each month to make the credit union a better deal (another way of asking this question is what is the maximum number of checks that can be written each month to make the credit union a better deal)? #88 p. 252 Intermediate Algebra by Blitzer A company manufactures and sells custom stationery. The weekly cost is $3000 plus $3 to produce each package of stationery. The selling price for a package is $5.50. How many packages must be sold each week for the company to make a profit (another way of asking this question is how many packages must be sold so that the profit is more than the cost)? Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 10

3.3 Compound Inequalities Intersection and Union These are two ideas that you may not be familiar with. They come from set theory and are important in our study of inequalities. Intersection is a mathematical and. It means contained by all sets. It is the overlap when visualized. It is abbreviated. You may be able to remember this better if you think of it looking like an A without the cross in the middle. Find A B when A = {2,4,6} & B = {0,2,4,6,8} In the last example, A is a subset of B (written A B) and therefore the intersection is A. Find C D when C = {1,3,5,7} & D = {2,4,6,8} The above example s solution is called a null set or an empty set. The empty set is shown with empty braces {} or with the symbol. It is not wise to use crossed zeros in math because of this symbol!! Union is a mathematical or and it means either one or the other so it joins the sets. Its abbreviation is. Find A B when A = {1,2,3} & B = {4,5,6} Note: When there is no overlap the union does not care. This is unlike the intersection which would be a null set. Find C D when C = {2,4,6} & D = {2,4,6,8, } Note: When one is a subset of the other, the union is the larger set. Whereas, when one is a subset of the other the intersection is the smaller set. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 11

Graph the following and then give the solution set in interval notation a) c 2 or c > -3 b) c 2 and c > -3 c) d > 0 and d 5 d) d > 0 or d 5 e) 2k + 5 > -1 and 7 3k 7 f) 2k + 5 > -1 or 7 3k 7 g) 2a + 3 7 and -3a + 4-17 h) 2a + 3 7 or -3a + 4-17 i) 2x 1 3 and -x > 2 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 12

Solving Compound Linear Inequalities Step 1: See the problem as having 3 parts a left, a middle and a right. Step 2: Simplify the three parts if necessary Step 3: Remove all constants from the middle by adding their opposite to all three part, the left and right sides of the inequality and the middle Step 4: Remove the numeric coefficient of the variable term (middle) by multiplying each part by the reciprocal (don t forget to reverse the sense of both inequalities if your numeric coeffiecient is negative!) Step 5: Check your answer Solve, graph and give the interval notation for following the linear inequalities a) 2 x 1 < 5 b) 2 3 / 4 x 1 2 ¼ Don t forget to switch the sense of the inequality if you must multiply or divide by a negative number! Solve, graph and give the interval notation for following the linear inequality 7 -(x + 5) 2 An alternate way of doing any compound inequality is to treat it as an intersection (and) of the middle to left and the middle to right, which is helpful in solving problems like the following. #92 p. 238 x + 3 < 2x + 1 < 4x + 6 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 13

3.4 Absolute Value Equations Recall the meaning of the absolute value. The distance from zero regardless of direction, so for every positive real number there are two possible solutions for an absolute value equation. It is an entirely different story if we are talking about zero or a negative real number! Why? Well, because the absolute value is always a positive number since it is the distance from zero, so there is no solution when the absolute value equals a negative number. And when the absolute value is zero, there can only be one solution because there is only one possibility for solutions that are zero units from zero! x = 5 so x can be either +5 or 5 since the absolute value of a number and its opposite are the same. x = 0 x can only be zero x = -9 The definition of an absolute value says that we get a positive number, so this can t be possible and thus there is no solution. Now, it isn t always just a single variable within the absolute value symbols, but the idea doesn t change. Set the inside equal to the endpoint or its opposite and solve the resulting equations. The 2 exceptions do not change. Solve, graph and give the solution set. a) x + 3 = 7 b) 2x + 3 = 9 c) 7x 5 = 0 d) 2 / 3 x + 3 / 5 = -5 On to the next level of difficulty; if there is a value added to or subtracted from the absolute value expression, you want to isolate the absolute value expression using the addition property of equality. Solve and give the solution set. a) x 5 4 = 0 b) 2 / 3 x + 3 / 5 + 5 = -5 c) 2x 9 + 14 = 10 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 14

Finally, we investigate what happens when an absolute expression equals an absolute expression. To solve these equations, set the expression in the left absolute value equal to the expression in the right side absolute value and then set the left equal to the opposite of the right side (make sure to distribute the negative!). Solve and give the solution set. a) 5x + 1 = 4x 7 b) x + 4 = 7 x c) 4y 5 = - -5 Although we will not be taking class time to do this, it is worth noting how we would visualize the solutions to all of these equations. Recall in the first section when we discussed the method of solving 2 equations simultaneously by finding their intersection point. Well, we can again visualize the solutions in this way. The only additional calculator knowledge that you need is where to find the absolute value function. This is located in your MATH menu. While in the Y= menu, press the MATH menu, scroll across to NUM and then down to abs( [it is the 1 key if you prefer to use that instead of the arrows]. You are graphing an absolute value equals a horizontal line in equations in the first 2 groups of examples. In this last type you are graphing 2 absolute value functions. The x-intercept method can be used as well. In the first 2 sets set you are finding the x- intercepts of absolute value functions with horizontal translation. The last type are a little trickier to conceptualize as they are the x-intercepts of a function that is the difference of 2 absolute value functions. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 15

3.5 Absolute Value Inequalities Let's do some thinking about absolute values for a bit. a means a without its sign, therefore the absolute value is always greater than zero if a is either larger or smaller than zero and if a = 0 then the absolute value is still zero. So, this leads us to the question: Can a ever be negative (less than zero)? x < 0 has no R solution since there is no number that will ever make this a true statement since the absolute is never less than zero. Now we need to give some consideration to x > #. There is one special case and that is when x is greater than a negative number. What values will make x greater than any negative number? you x > -2 this is all real numbers since no matter what number put in the value is always zero Now the x > the real numbers 0. This requires you to give a little consideration to the numbers on the number line and the fact that as you go out in each direction the number are *getting larger in terms of their absolute value. Also considered that in the negative direction as the numbers get larger in terms of their absolute value, they also get smaller in terms of their actual value, just as the positive numbers get larger. This gives us a hint as to how to solve the equations x > #. x > 5 since all positive numbers greater than 5 will make this true and all negative numbers less than 5 will make this true this is how we go about solving these problems. Our last consideration is x < #, when that number is greater than zero. Again let's consider the positive numbers and the negative numbers that will make it a true statement. Positive numbers between zero and the number at hand will make a true statement. Negative numbers between zero and the opposite of the number at hand will also make the statement true, with much the same reasoning that we had above (*). x < 5 since all positive numbers less than 5 will make this a true and all negative numbers greater than 5 will make this a true statement, this is how we go about solving these problems. Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 16

Very generally spoken!! For more detail see above. Summary of Solutions When a is a positive number x = a x = a or x = -a x < a -a < x < a (x is between the + and values of a) x > a x > a or x < -a (you may have seen this expressed in the following way, but you will not see me express it this way as it is not a correct statement, although it gets the job done. -a > x > a) When a is a negative number x = a x < a x > a R If a is zero we must be cautious of the endpoints! x > 0 {x x R, x 0}* x 0 R x < 0 x 0 x = 0* *x can be any algebraic expression and therefore you may have to solve to find the value of the variable!! Solving an Absolute Value Equation 1) Get into x > a or x < a form by using addition/ multiplication properties 2) Set up appropriately to remove the absolute values a) If x < a -a < x < a b) If x > a x > a or x < -a 3) Solve the resulting linear inequalities as usual 4) Check your solution Find the solution set of the following a) x = 9 b) x = -7 c) c + 1 < 2 d) a 2 + 3 < 4 e) a 2 + 3 < 2 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 17

Note: The absolute value can't be less than zero, so once you have completed step 1 of the process you know the answer. f) a + 3 0 2 Note: The absolute value can't be less than zero so you are only finding a single number that makes this statement equal to zero! g) c + 3 > 9 h) d 3 2 > 4 i) e + 2 0 j) 4 2y 0 Note for j: The absolute value is always greater than zero so you don't have to do anything to answer that the solution is all real numbers. k) 16a 4 > 0 Note: The endpoint is not included, so every real number except where 16a 4 = 0 will be the solution! Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 18

3.6 Graphing Linear Inequalities in 2 Variables See Ch. 2 notes for discussion of 3.6 Y. Butterworth Notes for Ch. 3 Martin/Gay-Greene 19