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Section 8.1 Solving Systems of Linear Equations by Graphing OBJECTIVES A Solve a system of two equations in two variables by graphing. OBJECTIVES B Determine whether a system of equations is consistent, inconsistent, or dependent. OBJECTIVES C Solve an application. SOLVING SYSTEMS OF SIMULTANEOUS EQUATIONS There are three possible solutions. POSSIBLE SOLUTION 1 Consistent and independent equations: The graphs of the equations intersect at one point, whose coordinates give the solution of the system. POSSIBLE SOLUTION 2 Inconsistent equations: The graphs of the equations are parallel lines; there is no solution for the system POSSIBLE SOLUTION 3 Dependent equations: The graphs of the equations coincide (are the same). There are infinitely many solutions for the system. COMPARISONS Intersecting Lines Have different slopes COMPARISONS Intersecting Lines Have one solution COMPARISONS Intersecting Lines Form a consistent system COMPARISONS Parallel Lines Have the same slopes COMPARISONS Parallel Lines Have different y-intercepts COMPARISONS Parallel Lines Have no solution COMPARISONS Parallel Lines Form inconsistent systems COMPARISONS Coinciding Lines Have the same slope COMPARISONS Coinciding Lines Have the same y-intercept COMPARISONS Coinciding Lines Have infinite solutions COMPARISONS Coinciding Lines Form a dependent system A HELPFUL HINT For y = m1x + b1 y = m2 x + b2 When m1 m2, the lines intersect. When m1 = m2 and b1 = b2, there is only one line. A HELPFUL HINT For y = m1x + b1 y = m2 x + b2 When m1 = m2 and b1 b2, the lines are parallel. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.1 Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.1 Exercise #1 Use the graphical method to solve the system. x + 2y = 4 2y – x = 0 y 5 –4 6 –5 x Use the graphical method to solve the system. x y x + 2y = 4 0 2 2y – x = 0 4 0 y 5 –4 6 –5 x Use the graphical method to solve the system. x y x + 2y = 4 0 0 2y – x = 0 4 2 y 5 (2, 1) –4 6 –5 x Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.1 Exercise #2 Use the graphical method to solve the system. y – 2x = 2 2y – 4x = 8 y 6 –5 5 –4 x Use the graphical method to solve the system. x y y – 2x = 2 0 2 2y – 4x = 8 –1 0 y 6 –5 5 –4 x Use the graphical method to solve the system. x y y – 2x = 2 0 4 2y – 4x = 8 –2 0 y 6 Inconsistent: No Solution. –5 5 –4 x Section 8.2 Solving Systems of Equations by Substitution OBJECTIVES A Solve a system of equations in two variables. OBJECTIVES B Determine whether a system of equations is consistent, inconsistent, or dependent. OBJECTIVES C Solve an application. PROCEDURE: Solving a system of equations by the Substitution Method. 1. Solve one of the equations for x or y. PROCEDURE: Solving a system of equations by the Substitution Method. 2. Substitute the resulting expression into the other equation. PROCEDURE: Solving a system of equations by the Substitution Method. 3. Solve the new equation for the variable. PROCEDURE: Solving a system of equations by the Substitution Method. 4. Substitute the value of the variable and solve to get the value for the second variable. PROCEDURE: Solving a system of equations by the Substitution Method. 5. Check the solution by substituting the numerical values of the variables in both equations. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.2 Exercise #3 Use the method of substitution to solve the system (if possible). x = 6 – 3y x + 3y = 6 2x + 6y = 8 2 6 – 3 y + 6y = 8 12 – 6y + 6y = 8 12 = 8 12 8 No Solution (inconsistent) Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.2 Exercise #4 Use the method of substitution to solve the system (if possible). x = 6 – 3y x + 3y = 6 3x + 9y = 18 3 6 – 3 y + 9y = 18 18 – 9y + 9y = 18 18 = 18 identity Dependent (infinitely many solutions). Section 8.3 Solving Systems of Equations by Elimination OBJECTIVES A Solve a system of equations in two variables. OBJECTIVES B Determine whether a system of equations is consistent, inconsistent, or dependent. OBJECTIVES C Solve an application. ELIMINATION METHOD One or both equations in a system of simultaneous equations can be multiplied (or divided) by any nonzero number to obtain an equivalent system. ELIMINATION METHOD In the equivalent system, the coefficients of x (or y) are opposites, thus eliminating x or y when the equations are added. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.3 Exercise #5 Solve the system (if possible). 2x + 3y = – 8 Multiply by –1 3x + y = – 5 Multiply by 3 9x + 3y = – 15 + –2x – 3y = 8 7x = – 7 x = –1 Solve the system (if possible). 2x + 3y = – 8 3x + y = – 5 3 –1 + y = – 5 –3 + y = – 5 y = –2 (–1, –2) x = –1 Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.3 Exercise #6 Solve the system (if possible). 3x – 2y = 6 Multiply by 2 – 6x + 4y = – 2 + 6x – 4y = 12 0 = 10 0 ° 10 No solution (inconsistent) Section 8.4 Coin, General, Motion, and Investment Problems OBJECTIVES A Solve word problems involving coins. B Solve word problems of a general nature. OBJECTIVES C Solve word problems using the distance formula: D = RT. OBJECTIVES D Solve word problems involving the interest formula: I = Pr Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.4 Exercise #8 Eva has $2 in nickels and dimes. She has twice as many dimes as nickels. How many nickels and how many dimes does she have? Let n = number of nickels. d = number of dimes. 0.05n + 0.10d = 2.00 d = 2n Substitute 2n for d 0.05 n + 0.10 2n = 2.00 0.05n + 0.20n = 2.00 Multiply by 100 Eva has $2 in nickels and dimes. She has twice as many dimes as nickels. How many nickels and how many dimes does she have? 5n + 20n = 200 25n = 200 Divide by 25 n=8 d = 2n Substitute 8 for n d = 2 8 d = 16 She has 8 nickels and 16 dimes. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.4 Exercise #9 The sum of two numbers is 140. Their difference is 90. What are the numbers? Let x = the larger number. y = the smaller number. x + y = 140 + x – y = 90 2x = 230 Divide by 2 x = 115 115 + y = 140 y = 25 The numbers are 115 and 25. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.4 Exercise #10 A plane flies 600 miles with a tailwind in 2 hours. It takes the same plane 3 hours to fly the 600 miles when flying against the wind. What is the plane’s speed in still air? Let r = plane's speed in still air. w = wind's speed. r + w = speed with tail wind. r – w = speed against wind. 2 r + w = 600 Divide by 2 3 r – w = 600 A plane flies 600 miles with a tailwind in 2 hours. It takes the same plane 3 hours to fly the 600 miles when flying against the wind. What is the plane’s speed in still air? Let r = plane's speed in still air. w = wind's speed. r + w = speed with tail wind. r – w = speed against wind. r + w = 300 3 r – w = 600 Divide by 3 A plane flies 600 miles with a tailwind in 2 hours. It takes the same plane 3 hours to fly the 600 miles when flying against the wind. What is the plane’s speed in still air? Let r = plane's speed in still air. w = wind's speed. r + w = speed with tail wind. r – w = speed against wind. r + w = 300 add + r – w = 200 2r = 500 r = 250 The plane’s speed in still air is 250 mi/hr. Let r = plane's speed in still air. w = wind's speed. r + w = speed with tail wind. r – w = speed against wind. r + w = 300 add + r – w = 200 2r = 500 r = 250 Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.4 Exercise #11 Herbert invests $10,000, part at 5% and part at 6%. How much money is invested at each rate if his annual interest is $568? Let x = amount invested at 5%; 0.05x = interest y = amount invested at 6%; 0.06y = interest x + y = 10,000 Multiply by –5 0.05x + 0.06y = 568 Herbert invests $10,000, part at 5% and part at 6%. How much money is invested at each rate if his annual interest is $568? Let x = amount invested at 5%; 0.05x = interest y = amount invested at 6%; 0.06y = interest – 5x – 5y = – 50,000 0.05x + 0.06y = 568 Multiply by 100 Herbert invests $10,000, part at 5% and part at 6%. How much money is invested at each rate if his annual interest is $568? Let x = amount invested at 5%; 0.05x = interest y = amount invested at 6%; 0.06y = interest – 5x – 5y = – 50,000 + 5x + 6y = 56,800 y= 6,800 add x + 6,800 = 10,000 x = 3,200 $3,200 is invested at 5% and $6,800 is invested at 6%. Let x = amount invested at 5%; 0.05x = interest y = amount invested at 6%; 0.06y = interest – 5x – 5y = – 50,000 + 5x + 6y = 56,800 y= 6,800 add x + 6,800 = 10,000 x = 3,200 Section 8.5 Systems of Linear Inequalities OBJECTIVES A Solve a system of linear inequalities by graphing. PROCEDURE: Solving a System of Inequalities. Graph each inequality on the same set of axes using the following steps. PROCEDURE: Solving a System of Inequalities. 1. Graph the line that is the boundary of the region. If the inequality involves or draw a solid line; if it involves < or > draw a dashed line. PROCEDURE: Solving a System of Inequalities. 2. Use any point (a,b) not on the line as a test point. Substitute the values of a and b for x and y. PROCEDURE: Solving a System of Inequalities. If a true statement results, shade the side of the line containing the test point. PROCEDURE: Solving a System of Inequalities. If a false statement results, shade the other side. PROCEDURE: Solving a System of Inequalities. The solution set is the set of points that satisfies all the inequalities in the system. Chapter 8 Solving Systems of Linear Equations and Inequalities Section 8.5 Exercise #12 Graph the solution set of the system. x – 2y > 4 and 2x – y 6 y 3 –3 7 –7 x Graph the solution set of the system. x – 2y > 4 x – 2y = 4 Let x = 4, y = 0 y Let x = 0, y = – 2 3 – 3 (0, –2) (4, 0) –7 7 x Graph the solution set of the system. 2x – y 6 2x – y = 6 Let x = 3, y = 0 y Let x = 0, y = – 6 3 (3, 0) – 3 (0, –2) (4, 0) (0, –6) –7 7 x Graph the solution set of the system. x – 2y > 4 and 2x – y 6 y 3 (3, 0) – 3 (0, –2) (4, 0) (0, –6) –7 7 x