# Complex numbers examples

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Equations and Inequalities Figure 1 ChAPTeR OUTl Ine 2.1 The Rectangular Coordinate Systems and graphs 2.2 l inear equations in One variable 2.3 models and Applications 2.4 Complex numbers 2.5 Quadratic equations 2.6 Other Types of equations 2.7 l inear Inequalities and Absolute value Inequalities Introduction For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some foreign location. What most Americans do not realize is that from September through December, territorial possession dominates our lifestyles while watching the NFL. In this area, territorial possession is governed by the referees who make their decisions based on what the chains reveal. If the ball is at point A (x , y ), then it is up 1 1 to the quarterback to decide which route to point B (x , y ), the end zone, is most feasible. 2 2 7374 CHAPTER 2 e aqu tio N as Nd iN qeu alitise l eARnIng Obje CTIveS In this section you will: • Plot ordered pairs in a Cartesian coordinate system. • Graph equations by plotting points. • Graph equations with a graphing utility. • Find x-intercepts and y-intercepts. • Use the distance formula. • Use the midpoint formula. 2.1 The Re CTAng Ul AR COORdInATe SySTemS And gRAPhS y Schiller Avenue Mannhelm Road Bertau Avenue McLean Street Wolf Road North Avenue x Figure 1 Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations. Plotting Ordered Pairs in the Cartesian Coordinate System An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis. While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis. e C Th artesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2.SECTION 2.1 t he r cet a Ngul ar c oor di Nte a s sty ems a Nd g ra Phs 75 . y-axis Quadrant II Quadrant I x-axis Quadrant III Quadrant IV Figure 2 The center of the plane is the point at which the two axes cross. It is known as the origin, or point (0, 0). From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 3. y 5 4 3 2 1 x –5 –4 –3–2 –1 14 2 3 5 –1 –2 –3 –4 –5 Figure 3 Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form (x, y). An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point (3, −1) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure 4. y 5 4 3 2 1 x –5 –4 –3–2 –1 14 2 5 –1 (3, –1) –2 –3 –4 –5 Figure 4 When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.76 CHAPTER 2 e aqu tio N as N d iN qeu alitise Cartesian coordinate system A two-dimensional plane where the • x-axis is the horizontal axis • y-axis is the vertical axis A point in the plane is defined as an ordered pair, ( x, y), such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin. Example 1 Plotting Points in a Rectangular Coordinate System Plot the points (−2, 4), (3, 3), and (0, −3) in the plane. Solution To plot the point (−2, 4), begin at the origin. The x-coordinate is −2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction. To plot the point (3, 3), begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction. To plot the point (0, −3), begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in Figure 5. y 5 (–2, 4) 4 (3, 3) 3 2 1 x 14 2 5 –5 –4 –3–2 –1 3 –1 –2 –3 (0, –3) –4 –5 Figure 5 Analysis Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis. graphing equations by Plotting Points We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two- dimensional plane is a graph in two variables. Suppose we want to graph the equation y = 2x − 1. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. Table 1 lists values of x from −3 to 3 and the resulting values for y. x y = 2x − 1 (x, y) −3 y = 2(−3) − 1 = −7 (−3, −7) −2 y = 2(−2) − 1 = −5 (−2, −5) −1 y = 2(−1) − 1 = −3 (−1, −3) 0 y = 2(0) − 1 = −1 (0, −1) 1 y = 2(1) − 1 = 1 (1, 1) 2 y = 2(2) − 1 = 3 (2, 3) 3 y = 2(3) − 1 = 5 (3, 5) Table 1SECTION 2.1 t he r cet a Ngul ar c oor di Nte a s sty ems a Nd g ra Phs 77 We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See Figure 6. This is not true for all equations. y 7 (4, 7) 6 5 (3, 5) 4 3 (2, 3) 2 1 (1, 1) x –5 –4 –3–2 –1 14 2 3 5 –1 (0, –1) –2 (–1, –3) –3 –4 (–2, –5) –5 –6 (–3, –7) –7 Figure 6 Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph. How To… Given an equation, graph by plotting points. 1. Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs. 2. E nter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make the graphing simpler. 3. Select x-values that will yield y-values with little eo ff rt, preferably ones that can be calculated mentally. 4. Plot the ordered pairs. 5. Connect the points if they form a line. Example 2 Graphing an Equation in Two Variables by Plotting Points Graph the equation y = −x + 2 by plotting points. Solution First, we construct a table similar to Table 2. Choose x values and calculate y. x y = − x + 2 (x, y) −5 y = − (−5) + 2 = 7 (−5, 7) −3 y = − (−3) + 2 = 5 (−3, 5) −1 y = − (−1) + 2 = 3 (−1, 3) 0 y = − (0) + 2 = 2 (0, 2) 1 y = − (1) + 2 = 1 (1, 1) 3 y = − (3) + 2 = −1 (3, −1) 5 y = − (5) + 2 = −3 (5, −3) Table 278 CHAPTER 2 e aqu tio Ns a Nd iN qeu alitise Now, plot the points. Connect them if they form a line. See Figure 7. y 7 (–5, 7) 6 5 (–3, 5) 4 3 (–1, 3) (0, 2) 2 1 (1, 1) x –7 –6–5 –4 –3–2 –1 14 2 3 56 7 –1 (3, –1) –2 –3 (5, –3) Figure 7 Try It 1 1 _ Construct a table and graph the equation by plotting points: y = x + 2. 2 graphing equations with a graphing Utility Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style y = . The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen. For example, the equation y = 2x − 20 has been entered in the TI-84 Plus shown in Figure 8a. In Figure 8b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows −10 ≤ x ≤ 10, and −10 ≤ y ≤ 10. See Figure 8c. Plot1 Plot2 Plot3 WINDOW \Y = 2X20 Xmin = −10 1 \Y = Xmax = 10 2 Xscl = 1 \Y = 3 \Y = Ymin = −10 4 \Y = Ymax = 10 5 \Y = Yscl = 1 6 x \Y = Xres = 1 7 (a) (b) (c) Figure 8 (a) enter the equation. (b) This is the graph in the original window. (c) These are the original settings. By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See Figure 9a and Figure 9b. WINDOW y Xmin = −5 Xmax = 15 Xscl = 1 Ymin = −30 Ymax = 10 Yscl = 1 x Xres = 1 (a) (b) Figure 9 (a) This screen shows the new window settings. (b) We can clearly view the intercepts in the new window. Example 3 Using a Graphing Utility to Graph an Equation 2 4 _ _ Use a graphing utility to graph the equation: y = − x − . 3 3 Solution Enter the equation in the y = function of the calculator. Set the window settings so that both the x- and y-intercepts are showing in the window. See Figure 10. y 2 4 y = − x − 3 3 3 2 1 x –5 –4 –3–2 –1 14 2 3 5 –1 –2 Figure 10SECTION 2.1 t he r cet a Ngul ar c oor di Nte a s sty ems a Nd g ra Phs 79 Finding x-intercepts and y-intercepts The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation y = 3x − 1. To fi nd the x-intercept, set y = 0. y = 3x − 1 0 = 3x − 1 1 = 3x 1 _ = x 3 1 _ , 0 x-intercept   3 To find the y-intercept, set x = 0. y = 3x − 1 y = 3(0) − 1 y = −1 (0, −1) y-intercept We can confirm that our results make sense by observing a graph of the equation as in Figure 11. Notice that the graph crosses the axes where we predicted it would. y y = 3x − 1 4 3 2 1 x –5 –4 –3–2 –1 14 2 3 5 –1 –2 –3 –4 Figure 11 given an equation, find the intercepts. • Find the x-intercept by setting y = 0 and solving for x. • Find the y-intercept by setting x = 0 and solving for y. Example 4 Finding the Intercepts of the Given Equation Find the intercepts of the equation y = −3x − 4. Th en sketch the graph using only the intercepts. Solution Set y = 0 to find the x-intercept. y = −3x − 4 0 = −3x − 4 4 = −3x 4 _ − = x 3 4 _ − , 0 x−intercept   3 Set x = 0 to fi nd the y-intercept. y = −3x − 4 y = −3(0) − 4 y = −4 (0, −4) y−intercept80 CHAPTER 2 e aqu tio N as Nd iN qeu alitise Plot both points, and draw a line passing through them as in Figure 12. y 3 4 2 _ − , 0 1 3 x –5 –4 –3 –2 –1 14 2 3 5 –1 –2 –3 –4 (0, –4) –5 –6 Figure 12 Try It 2 3 _ Find the intercepts of the equation and sketch the graph: y = − x + 3. 4 Using the distance Formula Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the 2 2 2 plane. e Th Pythagorean Theorem, a + b = c , is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See Figure 13. y 7 (x , y ) 2 2 6 5 d = c 4 y − y = b 2 1 3 2 (x , y ) 2 1 (x , y ) 1 1 x − x = a 2 1 x 0 14 23 56 7 Figure 13 e r Th elationship of sides x − x and y − y to side d is the same as that of sides a and b to side c. We use the absolute 2 1 2 1 value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, −3 = 3. ) Th e symbols x − x and y − y indicate that the lengths of the sides of the triangle are 2 1 2 1 positive. To find the length c, take the square root of both sides of the Pythagorean Theorem. — 2 2 2 2 2 c = a + b → c = √a + b It follows that the distance formula is given as —— 2 2 2 2 2 d = (x − x ) + (y − y ) → d = √(x − x ) + (y − y ) 2 1 2 1 2 1 2 1 We do not have to use the absolute value symbols in this definition because any number squared is positive. the distance formula Given endpoints (x , y ) and (x , y ), the distance between two points is given by 1 1 2 2 —— 2 2 d = √(x − x ) + (y − y ) 2 1 2 1 Example 5 Finding the Distance between Two Points Find the distance between the points (−3, −1) and (2, 3). Solution Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 14.SECTION 2.1 t he r cet a Ngul ar c oor di Nte a s sty ems a Nd g ra Phs 81 y 4 (2, 3) 3 2 1 x –5 –4 –3–2 –1 14 2 3 5 –1 (2, −1) (−3, −1) –2 –3 –4 Figure 14 e Th n, calculate the length of d using the distance formula. —— 2 2 d = √(x − x ) + (y − y ) 2 1 2 1 —— 2 2 d = √(2 − (−3)) + (3 − (−1)) — 2 2 = √(5) + (4) — = √ 25 + 16 — = √41 Try It 3 Find the distance between two points: (1, 4) and (11, 9). Example 6 Finding the Distance between Two Locations Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions. Solution e fi Th rst thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at (1, 1). e Th next stop is 5 blocks to the east, so it is at (5, 1). After that, she traveled 3 blocks east and 2 blocks north to (8, 3). Lastly, she traveled 4 blocks north to (8, 7). We can label these points on the grid as in Figure 15. y Schiller Avenue (8, 7) Mannhelm Road Bertau Avenue McLean Street (8, 3) Wolf Road North Avenue (1, 1) (5, 1) (0, 0) x Figure 1582 CHAPTER 2 e aqu tio N as Nd iN qeu alitise Next, we can calculate the distance. Note that each grid unit represents 1,000 feet. • From her starting location to her first stop at (1, 1), Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop. • Her second stop is at (5, 1). So from (1, 1) to (5, 1), Tracie drove east 4,000 feet. • Her third stop is at (8, 3). There are a number of routes from (5, 1) to (8, 3). Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet. • Tracie’s final stop is at (8, 7). This is a straight drive north from (8, 3) for a total of 4,000 feet. Next, we will add the distances listed in Table 3. From/To Number of Feet Driven (0, 0) to (1, 1) 2,000 (1, 1) to (5, 1) 4,000 (5, 1) to (8, 3) 5,000 (8, 3) to (8, 7) 4,000 Total 15,000 Table 3 The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points (0, 0) and (8, 7). — 2 2 d = √(8 − 0) + (7 − 0) — = √64 + 49 — = √ 113 ≈ 10.63 units At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point (8, 7). Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways. Using the midpoint Formula When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, (x , y ) 1 1 and (x , y ), the midpoint formula states how to find the coordinates of the midpoint M. 2 2 x + x y + y 1 2 1 2 _______ _______ M = ,   2 2 A graphical view of a midpoint is shown in Figure 16. Notice that the line segments on either side of the midpoint are congruent. y x + x y + y 1 2 1 2 _______ _______ , 2 2 x 0 Figure 16SECTION 2.1 t he r cet a Ngul ar c oor di Nte a s sty ems a Nd g ra Phs 83 Example 7 Finding the Midpoint of the Line Segment Find the midpoint of the line segment with the endpoints (7, −2) and (9, 5). Solution Use the formula to find the midpoint of the line segment. x + x y + y 7 + 9 −2 + 5 1 2 1 2 _______ _______ _____ _______ , = ,     2 2 2 2 3 __ = 8,   2 Try It 4 Find the midpoint of the line segment with endpoints (−2, −1) and (−8, 6). Example 8 Finding the Center of a Circle e d Th iameter of a circle has endpoints ( −1, −4) and (5, −4). Find the center of the circle. Solution e Th center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point. x + x y + y 1 2 1 2 _______ _______ ,   2 2 −1 + 5 −4 −4 4 8 _______ ______ __ __ , = , − = (2, −4)     2 2 2 2 Access these online resources for additional instruction and practice with the Cartesian coordinate system. • Plotting Points on the Coordinate Plane (http://openstaxcollege.org/l/coordplotpnts) • Find x- and y-intercepts based on the graph of a l ine (http://openstaxcollege.org/l/xyintsgraph)84 CHAPTER 2 e aqu tio Ns a Nd iN qeu alitise 2.1 SeCTIOn exeRCISeS veRbAl 1. Is it possible for a point plotted in the Cartesian 2. Describe the process for finding the x-intercept and coordinate system to not lie in one of the four the y-intercept of a graph algebraically. quadrants? Explain. 3. Describe in your own words what the y-intercept of 4. When using the distance formula —— 2 2 a graph is. d = √(x − x ) + (y − y ) , explain the correct 2 1 2 1 order of operations that are to be performed to obtain the correct answer. Algeb RAIC For each of the following exercises, find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept. 5. y = −3x + 6 6. 4y = 2x − 1 7. 3x − 2y = 6 8. 4x − 3 = 2y 2 3 _ _ 9. 3x + 8y = 9 10. 2x − = y + 3 3 4 For each of the following exercises, solve the equation for y in terms of x. 11. 4x + 2y = 8 12. 3x − 2y = 6 13. 2x = 5 − 3y 14. x − 2y = 7 15. 5y + 4 = 10x 16. 5x + 2y = 0 For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers. 17. (−4, 1) and (3, −4) 18. (2, −5) and (7, 4) 19. (5, 0) and (5, 6) 20. (−4, 3) and (10, 3) 21. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth. (19, 12) and (41, 71) For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 22. (−5, −6) and (4, 2) 23. (−1, 1) and (7, −4) 24. (−5, −3) and (−2, −8) 25. (0, 7) and (4, −9) 26. (−43, 17) and (23, −34) gRAPhICAl For each of the following exercises, identify the information requested. 27. What are the coordinates of the origin? 28. If a point is located on the y-axis, what is the x-coordinate? 29. If a point is located on the x-axis, what is the y-coordinate? For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line). 30. (4, 1)(−2, −3)(5, 0) 31. (−1, 2)(0, 4)(2, 1) y y 5 5 4 4 3 3 2 2 1 1 x x –5 –4 –3–2 –1 14 2 3 5 –5 –4 –3–2 –1 14 2 3 5 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5SECTION 2.1 s ectio N e xercises 85 32. (−3, 0)(−3, 4)(−3, −3) 33. Name the coordinates of the points graphed. y y 5 5 4 4 B 3 3 A 2 2 1 1 C x x –5 –4 –3–2 –1 14 2 3 5 –5 –4 –3–2 –1 14 2 3 5 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 34. Name the quadrant in which the following points would be located. If the point is on an axis, name the axis. a. (−3, −4) b. (−5, 0) c. (1, −4) d. (−2, 7) e. (0, −3) For each of the following exercises, construct a table and graph the equation by plotting at least three points. 1 _ 35. y = x + 2 36. y = −3x + 1 37. 2y = x + 3 3 nUmeRIC For each of the following exercises, find and plot the x- and y-intercepts, and graph the straight line based on those two points. x − 3 _____ 38. 4x − 3y = 12 39. x − 2y = 8 40. y − 5 = 5x 41. 3y = −2x + 6 42. y = 2 For each of the following exercises, use the graph in the figure below. y 43. Find the distance between the two endpoints using the distance formula. Round to three decimal places. 5 4 3 44. Find the coordinates of the midpoint of the line 2 segment connecting the two points. 1 x –5 –4 –3–2 –1 14 2 3 5 45. Find the distance that (−3, 4) is from the origin. –1 –2 –3 46. Find the distance that (5, 2) is from the origin. –4 Round to three decimal places. –5 47. Which point is closer to the origin? TeChn Ol Ogy For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. nd After graphing it, use the 2 CALC button and 1:value button, hit ENTER. At the lower part of the screen you will see “x=” and a blinking cursor. You may enter any number for x and it will display the y value for any x value you input. Use this and plug in x = 0, thus finding the y-intercept, for each of the following graphs. 3x − 8 x + 5 ______ _____ 48. Y = −2x + 5 49. Y = 50. Y = 1 1 1 4 2 For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. nd After graphing it, use the 2 CALC button and 2:zero button, hit ENTER. At the lower part of the screen you will see “le b ft ound? ” and a blinking cursor on the graph of the line. Move this cursor to the left of the x-intercept, hit ENTER. Now it says “right bound?” Move the cursor to the right of the x-intercept, hit ENTER. Now it says “guess?” Move your cursor to the left somewhere in between the left and right bound near the x-intercept. Hit ENTER. At the bottom of your screen it will display the coordinates of the x-intercept or the “zero” to the y-value. Use this to find the x-intercept.86 CHAPTER 2 e aqu tio Ns a Nd iN qeu alitise Note: With linear/straight line functions the zero is not really a “guess,” but it is necessary to enter a “guess” so it will search and find the exact x-intercept between your right and left boundaries. With other types of functions (more than one x-intercept), they may be irrational numbers so “guess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries. 3x + 5 ______ 51. Y = −8x + 6 52. Y = 4x − 7 53. Y = R ound your answer to the nearest thousandth. 1 1 1 4 exTen SIOnS 54. A man drove 10 mi directly east from his home, 55. If the road was made in the previous exercise, how made a left turn at an intersection, and then traveled much shorter would the man’s one-way trip be 5 mi north to his place of work. If a road was made every day? directly from his home to his place of work, what would its distance be to the nearest tenth of a mile? 56. Given these four points: A(1, 3), B(−3, 5), C(4, 7), 57. Aer fin ft ding the two midpoints in the previous and D(5, −4), find the coordinates of the midpoint exercise, find the distance between the two _ _ of line segments AB an   d CD  . midpoints to the nearest thousandth. 58. Given the graph of the rectangle shown and the 59. In the previous exercise, find the coordinates of the coordinates of its vertices, prove that the diagonals midpoint for each diagonal. of the rectangle are of equal length. y 6 (–6, 5) (10, 5) 5 4 3 2 1 x –7 –5 –4 –3–2 –1 14 2 3 56 78 9 11 –1 (–6, –1) (10, –1) –2 –3 ReAl-W ORld A PPl ICATIOnS 60. e co Th ordinates on a map for San Francisco are 61. If San Jose’s coordinates are (76, −12), where the (53, 17) and those for Sacramento are (123, 78). coordinates represent miles, find the distance Note that coordinates represent miles. Find the between San Jose and San Francisco to the distance between the cities to the nearest mile. nearest mile. 62. A small craft in Lake Ontario sends out a distress 63. A man on the top of a building wants to have a guy signal. The coordinates of the boat in trouble were wire extend to a point on the ground 20 ft from the (49, 64). One rescue boat is at the coordinates building. To the nearest foot, how long will the wire (60, 82) and a second Coast Guard craft is at have to be if the building is 50 ft tall? coordinates (58, 47). Assuming both rescue craft (20, 50) travel at the same rate, which one would get to the distressed boat the fastest? 50 20 (0, 0) 64. If we rent a truck and pay a 75/day fee plus .20 for every mile we travel, write a linear equation that would express the total cost y, using x to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi.SECTION 2.2 l i Near e aqu tio Ns i N oN e v aria ble 87 l eARnIng Obje CTIveS In this section you will: • Solve equations in one variable algebraically. • Solve a rational equation. • Find a linear equation. • Given the equations of two lines, determine whether their graphs are parallel or perpendicular. • Write the equation of a line parallel or perpendicular to a given line. 2.2 lI ne AR eQUATIOnS In One vARIAble Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays 15.00/hr, and she opened a savings account with an initial deposit of 400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately 2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1. y 3000 2500 2000 1500 1000 500 x 0 20 40 60 80 100 120140 160180 200 Hours Worked Figure 1 Solving l inear equations in One variable A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form ax + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation. 3x = 2x + x The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x will make the equation true. A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5x + 2 = 3x − 6, we have the following: 5x + 2 = 3x − 6 2x = −8 x = −4 The solution set consists of one number: −4. It is the only solution and, therefore, we have solved a conditional equation. An inconsistent equation results in a false statement. For example, if we are to solve 5x − 15 = 5(x − 4), we have the following: 5x − 15 = 5x − 20 5x − 15 − 5x = 5x − 20 − 5x Subtract 5x from both sides. −15 ≠ −20 False statement Savings Account Balance88 CHAPTER 2 e aqu tio N as Nd iN qeu alitise Indeed, −15 ≠ −20. There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. linear equation in one variable A linear equation in one variable can be written in the form ax + b = 0 where a and b are real numbers, a ≠ 0. How To… Given a linear equation in one variable, use algebra to solve it. e f Th ollowing steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = , if x is the unknown. There is no set order, as the steps used depend on what is given: 1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero. 2. Apply the distributive property as needed: a(b + c) = ab + ac. 3. Isolate the variable on one side of the equation. 4. When the variable is multiplied by a coec ffi ient in the final stage, multiply both sides of the equation by the reciprocal of the coec ffi ient. Example 1 Solving an Equation in One Variable Solve the following equation: 2x + 7 = 19. Solution This equation can be written in the form ax + b = 0 by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. 2x + 7 = 19 2x = 12 Subtract 7 from both sides. 1 _ x = 6 Multiply both sides by o r divide by 2. 2 e s Th olution is 6. Try It 1 Solve the linear equation in one variable: 2x + 1 = −9. Example 2 Solving an Equation Algebraically When the Variable Appears on Both Sides Solve the following equation: 4(x − 3) + 12 = 15 − 5(x + 6). Solution Apply standard algebraic properties. 4(x − 3) + 12 = 15 − 5(x + 6) 4x − 12 + 12 = 15 − 5x − 30 Apply the distributive property. 4x = −15 − 5x Combine like terms. 9x = −15 Place x- terms on one side and simplify. 15 1 _ _ x = − Multiply both sides by , t he reciprocal of 9. 9 9 5 _ x = − 3 Analysis This problem requires the distributive property to be applied twice, and then the properties of algebra are used 5 _ to reach the final line, x = − . 3SECTION 2.2 l i Near e aqu tio N is N oN e v aria ble 89 Try It 2 Solve the equation in one variable: −2(3x − 1) + x = 14 − x. Solving a Rational equation In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation. 2 7 _ _ Recall that a rational number is the ratio of two numbers, such as or . A rational expression is the ratio, or quotient, 3 2 of two polynomials. Here are three examples. x + 1 1 4 _ _ _ , , or 2 2 x − 3 x − 4 x + x − 2 Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out. Example 3 Solving a Rational Equation 7 5 22 _ _ _ Solve the rational equation: − = . 2x 3x 3 Solution We have three denominators; 2x, 3x, and 3. The LCD must contain 2 x, 3x, and 3. An LCD of 6x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6x. 7 5 22 _ _ _ (6x) − = (6x)     2x 3x 3 7 5 22 _ _ _ (6x) − (6x) = (6x) Use the distributive property.       2x 3x 3 7 5 22 _ _ _    ( 6x ) − (6 x ) = ( 6 x) Cancel out the common factors.          2 x 3x 3 3(7) − 2(5) = 22(2x) Multiply remaining factors by each numerator. 21 − 10 = 44x 11 = 44x 11 _ = x 44 1 _ = x 4 A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as (x + 1). Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x, x − 1, and 3x − 3. First, factor all denominators. We then have x, (x − 1), and 3(x − 1) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of (x − 1). e Th x in the first denominator is separate from the x in the (x − 1) denominators. An ee ff ctive way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x, one factor of (x − 1), and the 3. Thus, the LCD is the following: x(x − 1)3 = 3x(x − 1) So, both sides of the equation would be multiplied by 3x(x − 1). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.90 CHAPTER 2 e aqu tio Ns a Nd iN qeu alitise 2 Another example is a problem with two denominators, such as x and x + 2x. Once the second denominator is factored 2 as x + 2x = x(x + 2), there is a common factor of x in both denominators and the LCD is x(x + 2). Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation. a c _ _ = b d We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign. a c a c _ _ _ _ If = , then = . b d b d Multiply a(d) and b(c), which results in ad = bc. Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. rational equations A rational equation contains at least one rational expression where the variable appears in at least one of the denominators. How To… Given a rational equation, solve it. 1. Factor all denominators in the equation. 2. Find and exclude values that set each denominator equal to zero. 3. Find the LCD. 4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left. 5. Solve the remaining equation. 6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator. Example 4 Solving a Rational Equation without Factoring Solve the following rational equation: 2 3 7 _ _ _ − = x 2 2x Solution We have three denominators: x, 2, and 2x. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2x. Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2x. 2 3 7 _ _ _ 2x − = 2 x     x 2 2x 2 3 7 _ _ _    2 x − 2x = 2 x Distribute 2x.          x 2 2x 2(2) − 3x = 7 Denominators cancel out. 4 − 3x = 7 −3x = 3 x = −1 or −1 The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1, or −1 written in set notation. Try It 3 2 1 1 _ _ _ Solve the rational equation: = − . 3x 4 6xSECTION 2.2 l i Near e aqu tio Ns i N oN e v aria ble 91 Example 5 Solving a Rational Equation by Factoring the Denominator 1 1 3 _ _ _ Solve the following rational equation: = − . x 10 4x Solution First find the common denominator. The three denominators in factored form are x, 10 = 2 ⋅ 5, and 4x = 2 ⋅ 2 ⋅ x. Th e smallest expression that is divisible by each one of the denominators is 20x. Only x = 0 is an excluded value. Multiply the whole equation by 20x. 3 1 1 ___ _ _ 20x = − 20x     x 10 4x 20 = 2x − 15 35 = 2x 35 ___ = x 2 35 _ e s Th olution is . 2 Try It 4 5 3 7 _ _ _ Solve the rational equation: − + = − . 2x 4x 4 Example 6 Solving Rational Equations with a Binomial in the Denominator Solve the following rational equations and state the excluded values: 3 5 x 5 1 x 5 1 _ _ _ _ _ _ _ _ a. = b. = − c. = − x x − 6 x − 3 x − 3 2 x − 2 x − 2 2 Solution a. e den Th ominators x and x − 6 have nothing in common. Therefore, the LCD is the product x(x − 6). However, for this problem, we can cross-multiply. 3 5 _ _ = x x − 6 3x = 5(x − 6) Distribute. 3x = 5x − 30 −2x = −30 x = 15 e s Th olution is 15. The excluded values are 6 and 0. b. e L Th CD i s 2(x − 3). Multiply both sides of the equation by 2(x − 3). x 5 1 _____ _____ __ 2(x − 3) = − 2(x − 3)     x − 3 x − 3 2    2( x − 3) x 2( x − 3)5 2 (x − 3) ________ ________ _______ = −    x − 3 x − 3 2 2x = 10 − (x − 3) 2x = 10 − x + 3 2x = 13 − x 3x = 13 13 ___ x = 3 13 _ e s Th o lution is . The ex cluded value is 3. 392 CHAPTER 2 e aqu tio N as Nd iN qeu alitise c. e le Th ast common denominator is 2( x − 2). Multiply both sides of the equation by x(x − 2). x 5 1 _ _ _ 2(x − 2) = − 2( x − 2)     x − 2 x − 2 2 2x = 10 − (x − 2) 2x = 12 − x 3x = 12 x = 4 e s Th olution is 4. The excluded value is 2. Try It 5 −3 4 _ _ Solve = . S tate the excluded values. 2x + 1 3x + 1 Example 7 Solving a Rational Equation with Factored Denominators and Stating Excluded Values 2x 2 1 _ _ _ Solve the rational equation after factoring the denominators: − = . State the excluded values. 2 x + 1 x − 1 x − 1 2 Solution We must factor the denominator x − 1. We recognize this as the difference of squares, and factor it as (x − 1)(x + 1). Th us, the LCD that contains each denominator is (x − 1)(x + 1). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation. 2 1 2x _ _ __ (x − 1)(x + 1) − = (x − 1)(x + 1)     x + 1 x − 1 (x − 1)(x + 1) 2(x − 1) − 1(x + 1) = 2x 2x − 2 − x − 1 = 2x Distribute the negative sign. −3 − x = 0 −3 = x e s Th olution is −3. The excluded values are 1 and −1. Try It 6 2 1 1 _ _ _ Solve the rational equation: + = . 2 x − 2 x + 1 x − x − 2 Finding a l inear equation Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = mx + b, where m = slope and b = y-intercept. Let us begin with the slope. The Slope of a Line The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run. y − y 2 1 __ m = x − x 2 1 If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2. The lines indicate the following slopes: m = −3, 1 _ m = 2, and m = . 3

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