# Teaching Slope and the Equation of a Line – Part 2 My last post discussed the mathematics of linear growth/decay from an analytic geometry point of view. If you have not read that post, you might find it helpful to read it before continuing. This post will focus on finding a linear equation that expresses a relationship between two variables by first creating a table of data-pairs. After your students inspect the table, encourage them to write the slope-intercept or the point-slope equation of a line that contains the data. Once the equation is found, it can be manipulated and used to answer a wide variety of questions. The general procedure for finding the linear relationship between independent variable I and dependent variable D is described below:

• Set up a table of data values so that the independent variable I is in ascending order.
• From the table, determine the steady rate of change between I and D which equals slope m.
• From the table, determine equation constant b by finding D when I = 0.
• The equation D = mI + b expresses the relationship between D and I.
• The equation I = (D – b)/m expresses the inverse relationship between I and D.

If it is not possible to determine equation constant b from the table, pick any point (p, q) in the table. The equations D – q = m(I – p)  or D = mI – mp + q express the linear relationship. Of course, if the data pairs in the table don’t show a steady rate of change, then the relationship between the variables is not linear.

Most discussions about finding the equation of a linear relationship don’t start by looking at data-pairs in a table. From my past experience, the table-data-pair approach works because it enables students to immediately see and understand the relationship. I routinely use this approach when writing computer programs. It helps me better understand the problem and helps to clarify my thinking. I have selected four familiar linear relation type math problems to demonstrate the table data-pairs approach.

Problem 1: Brian’s electricity provider charges him \$0.14 per kWh (kilowatt-hour) of electricity, plus a basic connection charge of \$17.50 per month. Let n = the number of kilowatt-hours of electricity Brian used in a month and c = the amount charged for n kilowatt-hours of electricity.
(a) Find an equation that expresses the relationship between the variables n and c.
(b) If Brian used 861 kWh of electricity in July, how much was he billed for July?
(c) If Brian was billed \$155.82 for August, how many kWh of electricity did he use in August?

Problem 1 Solution: Use the information stated in the problem to construct a data-pair-table similar to the table shown in the text box below. The table immediately tells us that the steady rate of change constant m = \$0.14/kWh, and c = 17.50 when n = 0. Using the slope-intercept form of the equation of a line, the equations c = 0.14n + 17.50 and n = (c – 17.50)/0.14 follow. If Brian used 861 kWh of electricity in July, he was billed \$138.04 for July. If Brian was billed \$155.82 for August, he used 988 kWh of electricity in August. (Refer to the text box below.) Problem 2: Bubba’s doctor put Bubba on a strict diet that was designed to lose weight at a steady average rate of 10 pounds every 4 weeks. When Bubba started the diet, he weighed 405 pounds. Let n = the number of weeks that Bubba is on the diet and w = Bubba’s expected weight after n weeks on the diet. (Expected weight approximately equals the actual weight on the diet.)
(a) Find an equation that expresses the relationship between the variables n and w.
(b) Find Bubba’s expected weight after 1 year (52 weeks) on the diet.
(c) At one point in his diet, Bubba weighed 335 pounds. Approximately how many weeks was he on the diet?

Problem 2 Solution: Use the information stated in the problem to construct a table of data-pairs similar to the table shown in the text box below. The table tells us that the steady weight loss rate m = -10/4 = -2.5 pounds per week and w = 405 pounds when n = 0. Using the slope-intercept form, the equations w = -2.5n + 405 and n = -(w – 405)/2.5 = -w/2.5 + 162 follow. Bubba’s expected weight after one year of the diet = 275 pounds. If Bubba weighed 355 pounds at one point in his diet, he had been on the diet about 20 weeks. (Refer to the text box below.) Problem 3: A partial handicap chart for a bowling league is shown in the text box below. Bowlers who average more than 210 pins per game receive no handicap, and bowlers who average 210 or less pins per game receive a handicap as indicated in the chart. The variable A represents the bowler’s current average, and H represents the bowler’s handicap. (a) Based on the data in the chart above, find a formula that expresses H as a function of A.
(b) Find a bowler’s handicap if his average score is 147. Round down to nearest integer.
(c) Find a bowler’s average score if he has a 100 pin handicap. Round down to nearest integer.

Problem 3 Solution: The table tells us that the steady rate of change constant m = 9/(-10) = -0.9. This tells us that the handicap drops 0.9 pins for every point increase in the bowler’s average score. Using the point-slope form of the equation of a line with the point (210, 9), the relationship between A and H can be expressed as H – 9 = -0.9(A – 210) or H = -0.9A + 198. Solving the equation for A, we have the equation A = (198 – H)/0.9. The handicap for a 147 bowler = -0.9*147 + 198 = 65.7 = 65 pins. If a bowler has a 100 pin handicap, his average bowling score = (198 – 100)/0.9 = 108.88889 = 108.

Problem 4: At 320F and 00C water freezes. At 2120F and 1000C water boils. Use the fact that the relationship between F and C is linear to find the following:
(a) Find an equation that expresses F as a function of C.
(b) Find an equation that expresses C as a function of F.
(c) Convert 680F (average room temperature) to degrees Celsius.
(d) Convert 98.60F (average body temperature) to degrees Celsius.

Problem 4 Solution: Use the information stated in the problem to construct a data-pair-table similar to the table shown in the text box below. The table immediately tells us that the steady rate of change m = 180/100 = 9/5 and F = 32 when C = 0. Using the slope-intercept form of the equation of a line, the equations F = 9C/5 + 32 and C = 5(F – 32)/9 follow. When F = 680F, C = 200C. When F = 98.60F, C = 370C. 