Solving Systems of Equations: Applications and Examples, Schemes and Mind Maps of Linear Algebra

Download Solving Systems of Equations: Applications and Examples and more Schemes and Mind Maps Linear Algebra in PDF only on Docsity! 16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations 1 The two methods we have used to solve systems of equations are substitution and elimination. Either method is acceptable for solving the systems of equations that we will be working with in this lesson. Method of Substitution: 1. solve one equation for one variable (it doesn’t matter which equation you choose or which variable you choose). 2. substitute the solution from step 1 into the other equation. 3. solve the new equation from step 2. 4. back substitute to solve the equation from step 1. Method of Elimination: 1. multiply at least one equation by a nonzero constant so the coefficients for one variable will be opposites (same absolute value) 2. add the equations so the variable with the opposite coefficients will be eliminated. 3. take the result from step 2 and solve for the remaining variable. 4. take the solution from step 3 and back substitute to any of the equations to solve for the remaining variable. Steps for solving applications: 1. assign variables to represent the unknown quantities 2. set-up equations using the variables from step 1 3. solve using substitution or elimination; it makes no difference which method you use Just as in the previous lesson, all of these application problems should result in a system of equations with two equations and two variables: { 𝑎𝑥 + 𝑏𝑦 = 𝑐 𝑑𝑥 + 𝑒𝑦 = 𝑓 The equations will usually be linear, but not always. 16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations 2 Example 1: Set-up a system of equations and solve using any method. A salesperson purchased an automobile that was advertised as averaging 25 miles gallon of gasoline in the city and 40 miles gallon on the highway. A recent sales trip that covered 1800 miles required 51 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city and how many miles were driven on the highway? How many gallons of gasoline were used in the city? How many gallons of gasoline were used on the highway? Write an equation to represent the total number of gallons of gasoline used on the trip. How many miles were driven in the city, assuming the advertised mileage of 25 miles gallon is correct? 25 miles gallon ∙ 𝑥 gallons = How many miles were driven on the highway, assuming the advertised mileage of 40 miles gallon is correct? 40 miles gallon ∙ 𝑦 gallons = Write an equation to represent the total number of miles driven on the trip. 16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations 5 Example 3: Set-up a system of equations and solve using any method. For a particular quadratic function 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, 𝑓(−3) = −13 and 𝑓(2) = 7. Find the values of 𝑏 and 𝑐, given that 𝑎 = 5. 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑓(−3) = 𝑎(−3)2 + 𝑏(−3) + 𝑐 𝑓(2) = 𝑎(2)2 + 𝑏(2) + 𝑐 −13 = 9𝑎 − 3𝑏 + 𝑐 7 = 4𝑎 + 2𝑏 + 𝑐 At this point I have two equations, but both equations have three unknowns (𝑎, 𝑏, and 𝑐). Since we’re told in the direction that 𝑎 = 5, I can replace 𝑎 with 5 in both equations, and this will leave me with two equations with only two unknowns. −13 = 9(5) − 3𝑏 + 𝑐 7 = 4(5) + 2𝑏 + 𝑐 −13 = 45 − 3𝑏 + 𝑐 7 = 20 + 2𝑏 + 𝑐 −58 = −3𝑏 + 𝑐 −13 = 2𝑏 + 𝑐 { −58 = −3𝑏 + 𝑐 −13 = 2𝑏 + 𝑐 −58 = −3𝑏 + 𝑐 + 13 = −2𝑏 − 𝑐 −45 = −5𝑏 + 0 −45 = −5𝑏 𝟗 = 𝒃 −13 = 2𝑏 + 𝑐 −13 = 2(9) + 𝑐 −13 = 18 + 𝑐 −𝟑𝟏 = 𝒄 16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations 6 The next two problems involve objects traveling a certain distance, at a certain rate, for certain time. Anytime I encounter an application problem involving distance, rate, and time, I will set-up a table and use the formula 𝑑 = 𝑟 ∙ 𝑡, which represents distance equals rate times time. Remembering the formula 𝑑 = 𝑟 ∙ 𝑡 might be easier if you think about it in a different form. When you’re driving a car, the speedometer tells you the rate as miles per hour, which means 𝑟 = 𝑑 𝑡 . Example 4: Set-up a system of equations and solve using any method. Two people can canoe 18 miles in an hour and a half when going with the current. Against the same current, the same two people can only canoe 3 miles in 1 hour. Using this information, determine how fast the people can move the canoe on their own, and how fast the current is moving. miles = miles hour ∙ hours 𝒅 = 𝒓 ∙ 𝒕 With Current Against Current 16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations 7 { 18 = (𝑥 + 𝑦) ∙ 1.5 3 = (𝑥 − 𝑦) ∙ 1 18 = (𝑥 + 𝑦) ∙ 1.5 12 = 𝑥 + 𝑦 12 = 𝑥 + 𝑦 + 3 = 𝑥 − 𝑦 15 = 2𝑥 + 0 15 = 2𝑥 15 2 = 𝑥 𝑥 = 7.5 Since 𝑥 represents how fast the people can move the boat, that means the people can row the boat 7.5 miles per hour. To find the value of 𝑦 (the speed of the current), I’ll back substitute to one of the prior equations by replacing 𝑥 with 7.5. 3 = 𝑥 − 𝑦 3 = 7.5 − 𝑦 𝑦 = 7.5 − 3 𝑦 = 4.5 Since 𝑦 represents the speed of the current, that means the current is moving at 4.5 miles per hour.