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Chapter 1 Section 8-7 Systems of Equations and Applications 8-7-1 © 2008 Pearson Addison-Wesley. All rights reserved Systems of Equations and Applications • • • • Linear Systems in Two Variables Elimination Method Substitution Method Applications of Linear Systems 8-7-2 © 2008 Pearson Addison-Wesley. All rights reserved Linear System in Two Variables When multiple equations are considered together the set of equations is called a system of equations. For example y = 3x + 4 y = –2x + 2 The point where the graphs intersect is a solution of each of the individual equations. It is also the solution of the system of equations. 8-7-3 © 2008 Pearson Addison-Wesley. All rights reserved Graphs a Linear System (Three Possibilities) 1. The two graphs intersect in a single point. The system is consistent and the equations are independent. y One solution x 8-7-4 © 2008 Pearson Addison-Wesley. All rights reserved Graphs a Linear System (Three Possibilities) 2. The graphs are parallel lines. In this case, the system is inconsistent and the equations are independent. There is no solution common to both equations. y No solution x 8-7-5 © 2008 Pearson Addison-Wesley. All rights reserved Graphs a Linear System (Three Possibilities) 3. The graphs are the same line. The system is consistent and the equations are dependent, because any solution of one equation is also a solution of the other. The solution set is an infinite set of ordered y pairs. Infinite number of x solutions 8-7-6 © 2008 Pearson Addison-Wesley. All rights reserved Elimination Method We can use algebraic methods to solve systems. One method is called the elimination method. The elimination method involves combining the two equations of the system so that one variable is eliminated. We use the fact that If a = b and c = d then a + c = b + d. 8-7-7 © 2008 Pearson Addison-Wesley. All rights reserved Solving Linear Systems by Elimination Step 1 Write both equations in standard form Ax + By = C. Step 2 Make the coefficients of one pair of variable terms opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either x or y is zero. Step 3 Add the new equations to eliminate a variable. The sum should be an equation with just one variable. © 2008 Pearson Addison-Wesley. All rights reserved 8-7-8 Solving Linear Systems by Elimination Step 4 Solve the equation from Step 3. Step 5 Find the other value. Substitute the result of Step 4 into either of the given equations and solve for the other variable. Step 6 Find the solution set. Check the solution in both of the given equations. Then write the solution set. 8-7-9 © 2008 Pearson Addison-Wesley. All rights reserved Example: Elimination Solve the system. 3 x 2 y 4 (1) 2 x y 5 (2) Solution Step 1 Both equations are in standard form. Step 2 Multiply equation (2) by 2 to get opposite coefficients on y. 3x 2 y 4 4 x 2 y 10 8-7-10 © 2008 Pearson Addison-Wesley. All rights reserved Example: Elimination Solution (continued) Step 3 Add the equations to eliminate y. 7 x 14 Step 4 Solve to find x = 2. Step 5 Substitute in one of the original equations to find y. 2(2) y 5 y 1 Step 6 The solution (2, 1) checks in both original equations. © 2008 Pearson Addison-Wesley. All rights reserved 8-7-11 Substitution Method Linear systems can also be solved by the substitution method. This method is useful for solving linear systems in which one variable has coefficient 1 or –1. It is also the best choice for solving many nonlinear systems in advanced algebra courses. 8-7-12 © 2008 Pearson Addison-Wesley. All rights reserved Solving Linear Systems by Substitution Step 1 Solve for one variable in terms of the other. Solve one of the equations for either variable. Step 2 Substitute for that variable in the other equation. The result should be an equation with just one variable. Step 3 Solve the equation from Step 2. 8-7-13 © 2008 Pearson Addison-Wesley. All rights reserved Solving Linear Systems by Substitution Step 4 Find the other value. Substitute the result of Step 3 into the equation from Step 1 to find the value of the other variable. Step 5 Find the solution set. Check the solution in both of the given equations. Then write the solution set. 8-7-14 © 2008 Pearson Addison-Wesley. All rights reserved Example: Substitution Solve the system. 3 x 2 y 4 (1) 2 x y 5 (2) Solution Step 1 Solve equation (2) for y. y = –2x + 5 Step 2 Substitute –2x + 5 for y in equation (1). 3x 2(2 x 5) 4 8-7-15 © 2008 Pearson Addison-Wesley. All rights reserved Example: Substitution Solution (continued) Step 3 Solve the equation 3x 4x 10 4 7 x 14 x2 Step 4 Substitute to find y. y 2(2) 5 1 Step 5 The solution (2, 1) checks in both original equations. © 2008 Pearson Addison-Wesley. All rights reserved 8-7-16 Special Cases – Solving a System Solving the system 3x 2 y 4 6 x 4 y 7 leads to a false statement such as 0 = 15. This indicates that the two equations have no solutions in common. The system is inconsistent with the empty set as the solution set. 8-7-17 © 2008 Pearson Addison-Wesley. All rights reserved Special Cases – Solving a System Solving the system 3x 2 y 4 6 x 4 y 8 leads to a true statement such as 0 = 0. This indicates that the two equations are dependent. Choose one equation, solve for x and the solution set can be written as 2 y 4 , y . 3 8-7-18 © 2008 Pearson Addison-Wesley. All rights reserved Applications of Linear Systems Many problems involve more than one unknown quantity. Although some problems with two unknowns can be solved using just one variable, many times it is easier to use two variables. 8-7-19 © 2008 Pearson Addison-Wesley. All rights reserved Solving an Applied Problem by Writing a System of Equations Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3 Write a system of equations that relates the unknowns. 8-7-20 © 2008 Pearson Addison-Wesley. All rights reserved Solving an Applied Problem by Writing a System of Equations Step 4 Solve the system of equations. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. 8-7-21 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solving a Perimeter Problem A particular field has a perimeter of 320 yards. Its length measures 40 yards more than its width. What are the dimensions of the field? Solution Step 1 We need to find the dimensions of the field. Step 2 Assign variables. Let L = length and W = width. 8-7-22 © 2008 Pearson Addison-Wesley. All rights reserved Example: Solving a Perimeter Problem Solution (continued) Step 3 Perimeter: 2L + 2W = 320 and L = W + 40. Step 4 Solve using substitution to get W = 60 and L = 100. Step 5 The length is 100 yards and the width is 60 yards. Step 6 Check: 2(100) + 2(60) = 320 and 100 – 40 = 60. 8-7-23 © 2008 Pearson Addison-Wesley. All rights reserved