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N O SS LE 9 Solving Systems of Equations Solving Linear Systems of Equations UNDERSTAND Recall that a system of equations is a set of two or more equations with the same variables. The solution to a system of linear equations is the set of values of the variables that make all equations in the system true. For a system with 2 variables, the solutions are also the points on the coordinate plane where the graphs of the two equations intersect. y The graph on the right shows y 5 22x 2 1 and y 5 4x 1 5. The graphs intersect at the point (21, 1). The point gives the x- and y-values of the solution. Test to see that the point makes both equations true. y 5 22x 2 1 y 5 4x 1 5 1 5 22(21) 2 1 1 5 4(21) 1 5 15221 1 5 24 1 5 151 151 6 4 2 –6 –4 –2 0 2 4 6 x –2 y 4x 5 –4 y 2x 1 –6 UNDERSTAND A linear system can have no solution, one solution, or infinitely many solutions. If the graphs do not intersect at all, the system has no real solution and is said to be inconsistent. If an equation is a constant multiple of another, the system has infinitely many solutions. Use elimination to solve the system above. Eliminate y by subtracting the second equation from the first. y 5 22x 2 1 2(y 5 4x 1 5) Remember to distribute the negative sign. 0 5 26x 2 6 6x 5 26 x 5 21 The value of x is the same as the one found on the graph. To find the corresponding value of y, substitute the value of x into one of the original equations. y 5 22(21) 2 1 5 1 78 Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC UNDERSTAND A system can also be solved algebraically using substitution or elimination. To use elimination, add or subtract the equations to write an equation in one variable. Unit 1: Polynomial, Rational, and Radical Relationships M_569NASE_ALGII_PDF.indd 78 21/07/15 12:54 pm Connect Solve the system using elimination and by graphing. 3x 1 2y 5 4 4x 2 3y 5 11 1 Solve the system using elimination. Choose a variable to eliminate. Eliminate y. 3(3x 1 2y 5 4) 9x 1 6y 5 12 2(4x 2 3y 5 11) 8x 2 6y 5 22 Add the two equations to eliminate the y term. 9x 1 6y 5 12 1 (8x 2 6y 5 22) 17x 5 34 2 Solve the system by graphing. x 5 2 Draw both graphs on the same coordinate plane. Identify the point of intersection. Use one of the original equations to solve for y. y 3x 1 2y 5 4 3(2) 1 2y 5 4 6 2y 5 22 4x 3y 11 y 5 21 3x 2y 4 4 2 –6 –4 –2 0 –2 2 4 (2, 1) 6 x Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC –4 –6 3 Identify and test the solution. Using elimination, the solution is (2, 21). The point of intersection of the graphs is (2, 21). Substitute the point into each equation to test. 3x 1 2y 5 4: 3(2) 1 2(21) 5 4 4x 2 3y 5 11: 4(2) 2 3(21) 5 11 ▸ The solution to the system is (3, 22). TRY Solve the system using elimination and by graphing. 3y 2 6x 5 12 y 2 2x 5 4 Lesson 9: Solving Systems of Equations M_569NASE_ALGII_PDF.indd 79 79 21/07/15 12:54 pm Solving a Linear-Quadratic System of Equations UNDERSTAND In a system of equations, the equations may not all be linear. Another type of system is a linear-quadratic system. In this system one equation is linear and the other is quadratic. Recall that the graph of a quadratic equation is a special curve called a parabola. A linear-quadratic system can be solved by graphing or using substitution. The graph on the right shows y 5 x2 and y 5 2x. The graphs have 2 points of intersection: (0, 0) and (2, 4). Test to see that they make both equations true. (0, 0) x 5 0 and y 5 0 2 y 10 y x2 8 y 2x (2, 4) 6 x 5 2 and y 5 4 4 2 0 50 2 54 0 5 2(0) 4 5 2(2) (2, 4) 2 –6 –4 –2 0 2 4 6 x –2 UNDERSTAND A line can intersect a parabola at two points, one point, or no points. So, a system of linear quadratic equations can have two solutions, one solution, or no solutions. UNDERSTAND A linear-quadratic system can also be solved algebraically using substitution. Use substitution to solve the system above. Substitute 2x for y in the quadratic equation to get an equation in one variable. 2x 5 x2 0 5 x2 2 2x Solve the equation by factoring. x 5 0 and x 5 2 The values of x are the same as the ones found on the graph. To find the corresponding values of y, substitute the values of x into one of the original equations. y 5 x2 y 5 x2 y 5 02 y 5 22 y 5 0 y54 (0, 0) and (2, 4) are solutions to the system. These solutions are the same as the ones found on the graph. 80 Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC 0 5 x(x 2 2) Unit 1: Polynomial, Rational, and Radical Relationships M_569NASE_ALGII_PDF.indd 80 21/07/15 12:54 pm Connect Solve the system using substitution and by graphing. y 5 x2 2 2x 2 3 y 5 2x 2 1 1 Solve the system using substitution. Substitute 2x 2 1 for y in the quadratic equation. Then solve for x. y 5 x2 2 2x 2 3 2 2x 2 1 5 x 2 2x 2 3 0 5 x2 2 x 2 2 0 5 (x 1 1)(x 2 2) 2 Solve the system by graphing. x 5 21 and x 5 2 Use the linear equation to solve for y. x 5 21: y 5 2(21) 2 1 5 0 Draw both graphs on the same coordinate plane. Identify any points of intersection. y x 5 2: y 5 2(2) 2 1 5 23 6 4 2 (1, 0) –6 –4 0 –2 2 4 6 x –2 y x 2 2x 3 (2, 3) –4 Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC –6 y x 1 3 Identify the solution. Using substitution, the solutions are (21, 0) and (2, 23). The graphs intersect at two points: (21, 0) and (2, 23). ▸ The solutions to the system are (21, 0) and (2, 23). CHECK Substitute both solution points into each equation to check that they are solutions. Lesson 9: Solving Systems of Equations M_569NASE_ALGII_PDF.indd 81 81 21/07/15 12:54 pm EXAMPLE A Solve the system using substitution and by graphing. x 2 2y 5 7 4x 1 y 5 10 1 Solve the system using substitution. Solve the first equation for x: x 5 2y 1 7 Substitute 2y 1 7 for x in the second equation. Then solve for y. 4(2y 1 7) 1 y 5 10 8y 1 28 1 y 5 10 9y 5 218 y 5 22 Use the first equation to solve for x. 2 Solve the system by graphing. x 2 2y 5 7 x 2 2(22) 5 7 x1457 Draw both graphs on the same coordinate plane. Identify the point of intersection. y x53 6 4x y 10 4 2 –4 –2 0 –2 x 2y 7 2 4 6 x (3, 2) –4 –6 3 Identify the solution. Using elimination, the solution is (3, 22). The point of intersection of the graphs is (3, 22). ▸ The solution to the system is (3, 22). SC U S S DI Which algebraic method would you use to solve the system? Explain. 82 Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC –6 Unit 1: Polynomial, Rational, and Radical Relationships M_569NASE_ALGII_PDF.indd 82 21/07/15 12:54 pm EXAMPLE B Solve the system. x2 1 y2 5 10 y 5 2x 1 2 1 Solve by substitution. Substitute 2x 1 2 for y in the first equation. x2 1 y2 5 10 x2 1 (2x 1 2)2 5 10 Simplify and solve for x. x2 1 x2 2 4x 1 4 5 10 2 Graph both equations on your graphing calculator, adjusting the window as needed. 2x2 2 4x 2 6 5 0 2(x2 2 2x 1 3) 5 0 (x 1 1)(x 2 3) 5 0 x 5 21 and x 5 3 Use the linear equation to solve for y. x 5 21: y 5 2(21) 1 2 5 3 x 5 3: y 5 2(3) 1 2 5 21 There appear to be 2 points of intersection. Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC Hit 2nd TRACE to open the CALC menu, and select option 5: intersect. Move the cursor to one of the points of intersection and hit ENTER 3 times. The screen shows that the intersection point is (21, 3). Repeat the process for the other intersection point, (3, 21). 3 Identify the solutions. Using substitution, the solutions are (21, 3) and (3, 21). The points of intersection of the graphs are (21, 3) and (3, 21). ▸ The solutions to the equations are (21, 3) and (3, 21). SC U S S DI How many different solutions can there be for a system composed of a circle and a line? Explain. Lesson 9: Solving Systems of Equations M_569NASE_ALGII_PDF.indd 83 83 21/07/15 12:54 pm Practice Use the graphs to solve the system. REMEMBER Intersections of the graphs represent solutions to the system. 1. 2. y –6 –4 –2 y 6 6 4 4 2 2 0 2 4 6 x –6 –4 –2 0 –2 –2 –4 –4 –6 –6 2 4 6 x Solve the system using algebra. 3.2x 1 5y 5 24 4. HI 5. Look for a way to eliminate one of the variables. y 5 x2 1 x 2 1 y5x 84 6. y 5 x2 1 2x 2 8 y5x26 Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC y 1 x 5 23 NT 3x 2 y 5 11 y 2 3x 5 4 Unit 1: Polynomial, Rational, and Radical Relationships M_569NASE_ALGII_PDF.indd 84 21/07/15 12:54 pm Solve each system. 7. x 1 2y 5 10 8. 3x 2 y 5 2 9. 4x 2 2y 5 9 y 5 x2 y 5 2x 1 3 10. y 5 x2 2 5x 1 6 3x 2 y 5 5 2y 5 x 2 2 Solve. 11. EXTEND On the same coordinate plane, graph the following three equations. y x2 1 y2 5 16 x1y54 6 4 y 5 2x 1 4 2 Use the graph to solve each system. –6 –4 –2 0 2 4 6 x Duplicating this page is prohibited by law. © 2016 Triumph Learning, LLC –2 –4 –6 a. x1y54 b. y 5 2x 1 4 c. 2 2 x 1 y 5 16 y 5 2x 1 4 x2 1 y2 5 16 x1y54 d. 2 2 x 1 y 5 16 x1y54 y 5 2x 1 4 Lesson 9: Solving Systems of Equations M_569NASE_ALGII_PDF.indd 85 85 21/07/15 12:54 pm