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Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab

Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab. Chapter 3: Linear Systems and Matrices. BIG IDEAS: Solving systems of equations using a variety of methods Graphing systems of equations and inequalities Using Matrices. Lesson 1: Solve Linear Systems by Graphing.

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Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab

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  1. Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab

  2. Chapter 3:Linear Systems and Matrices BIG IDEAS: Solving systems of equations using a variety of methods Graphing systems of equations and inequalities Using Matrices

  3. Lesson 1: Solve Linear Systems by Graphing

  4. Essential question How do you solve a system of linear equations graphically?

  5. VOCABULARY • Consistent: A system of equations that has at least one solution • Inconsistent: A system of equations that has no solutions • Dependent: A consistent system of equations that has infinitely many solutions • Independent: A consistent system that has exactly one solution

  6. Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, –4). You can check this algebraically as follows. EXAMPLE 1 Solve a system graphically Graph the linear system and estimate the solution. Then check the solution algebraically. 4x + y = 8 Equation 1 2x – 3y = 18 Equation 2 SOLUTION

  7. 4(3) +(–4) 8 8 = 8 2(3) – 3(– 4) 18 6 + 12 18 12 –4 8 18 = 18 ? ? ? ? = = = = EXAMPLE 1 Solve a system graphically Equation 2 Equation 1 4x+ y= 8 2x– 3y= 18 The solution is (3, –4).

  8. The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent. EXAMPLE 2 Solve a system with many solutions Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. 4x – 3y = 8 Equation 1 8x – 6y = 16 Equation 2 SOLUTION

  9. The graphs of the equations are two parallel lines. Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent. EXAMPLE 3 Solve a system with no solution Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. Equation 1 2x + y = 4 Equation 2 2x + y = 1 SOLUTION

  10. Infinitely many solutions; consistent and dependent ANSWER for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent. 4. 2x + 5y = 6 4x + 10y = 12

  11. 5. 3x – 2y = 10 3x – 2y = 2 no solution; inconsistent ANSWER for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

  12. ANSWER (–1, 3); consistent and independent –2x + y = 5 6. y = –x + 2 for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

  13. Essential question How do you solve a system of linear equations graphically? Graph the equations. The point at which the graphs meet is the solution. CHECK your solution!

  14. Split a sheet of graph paper with a partner. Then solve the system by graphing:x+y = 22x + y = 3

  15. Lesson 2: Solve Linear Systems Algebraically

  16. Essential question How do you solve a system of linear equations algebraically?

  17. VOCABULARY • Substitution Method: A method of solving a system of equations by solving one of the equations for one of the variables and then substituting the resulting expression in the other equation(s) • Elimination Method: A method of solving a system of equations by multiplying equations by constants, then adding the revised equations to eliminate a variable

  18. Solve Equation 2 for x. STEP 1 EXAMPLE 1 Use the substitution method Solve the system using the substitution method. 2x + 5y = –5 Equation 1 x + 3y = 3 Equation 2 SOLUTION x = –3y + 3 Revised Equation 2

  19. EXAMPLE 1 Use the substitution method STEP 2 Substitute the expression for xinto Equation 1 and solve for y. 2x+5y = –5 Write Equation 1. 2(–3y + 3) + 5y = –5 Substitute –3y + 3 for x. y = 11 Solve for y. STEP 3 Substitute the value of yinto revised Equation 2 and solve for x. x = –3y+ 3 Write revised Equation 2. x = –3(11) + 3 Substitute 11 for y. x = –30 Simplify.

  20. ANSWER The solution is (– 30, 11). 2(–30) + 5(11) –5 –5 = –5 3 = 3 –30+ 3(11) 3 ? ? = = EXAMPLE 1 Use the substitution method CHECK Check the solution by substituting into the original equations. Substitute for xand y. Solution checks.

  21. 6x – 8y = 8 EXAMPLE 2 Use the elimination method Solve the system using the elimination method. 3x – 7y = 10 Equation 1 6x – 8y = 8 Equation 2 SOLUTION STEP 1 Multiply Equation 1 by – 2 so that the coefficients of xdiffer only in sign. –6x + 14y = 220 3x – 7y = 10 6x – 8y = 8

  22. 4 – x= 3 EXAMPLE 2 Use the elimination method STEP 2 6y = –12 Add the revised equations and solve for y. y= –2 STEP 3 Substitute the value of yinto one of the original equations. Solve for x. 3x – 7y= 10 Write Equation 1. 3x – 7(–2) = 10 Substitute –2 for y. 3x + 14 = 10 Simplify. Solve for x.

  23. ANSWER 1. 4x + 3y = –2 The solution is (1,–2). x + 5y = –9 ANSWER 3x + 3y = –15 2. 5x – 9y = 3 – The solution is ( , –2) 3 for Examples 1 and 2 GUIDED PRACTICE Solve the system using the substitution or the elimination method.

  24. 3x – 6y = 9 3. –4x + 7y = –16 for Examples 1 and 2 GUIDED PRACTICE Solve the system using the substitution or the elimination method. ANSWER The solution is (11, 4)

  25. Essential question How do you solve a system of linear equations algebraically? By using the Substition Method Elimination Method

  26. On a half sheet of graph paper (share with a partner) Graph the inequality:y≤x+2

  27. Lesson 3: Graph Systems of Linear Inequalities

  28. Essential question How do you find the solution to a system of linear inequalities?

  29. VOCABULARY • No new vocab!!

  30. EXAMPLE 1 Graph a system of two inequalities Graph the system of inequalities. y > –2x – 5 Inequality 1 y <x + 3 Inequality 2

  31. STEP 1 Graph each inequality in the system. Use red for y > –2x – 5and bluefory≤ x + 3. STEP 2 Identify the region that is common to both graphs. It is the region that is shaded purple. EXAMPLE 1 Graph a system of two inequalities SOLUTION

  32. 2 y < – x + 4 3 EXAMPLE 2 Graph a system with no solution Graph the system of inequalities. 2x + 3y < 6 Inequality 1 Inequality 2

  33. STEP 1 Graph each inequality in the system. Use red for2x + 3y <6and bluefory > – x + 4. 2 3 STEP 2 Identify the region that is common to both graphs. There is no region shaded both red and blue. So, the system has no solution. EXAMPLE 2 Graph a system with no solution SOLUTION

  34. Graph the system of inequalities. 1. y < 3x – 2 y > – x + 4 for Examples 1, 2 and 3 GUIDED PRACTICE

  35. 1 2.2x – y > 4 2 4x – y < 5 for Examples 1, 2 and 3 GUIDED PRACTICE

  36. Essential question How do you find the solution to a system of linear inequalities? The solution to a system of linear inequalities is found by graphing each inequality. The solution is the overlapping shaded region.

  37. Solve the system using substitution or elimination:3x + 4y = -253x – 2y = -1

  38. Lesson 4: Solve Systems of Linear Equations in Three Variables

  39. Essential question How do you solve a system of linear equations in three variables?

  40. VOCABULARY • Ordered Pair: A set of two numbers (x,y) that represent a point in space • Ordered Triple: A set of three numbers of the form (x,y,z) that represent a point in space

  41. STEP 1 Rewrite the system as a linear system in two variables. 4x + 2y + 3z = 1 Add 2 times Equation 3 12x – 2y + 8z = –2 to Equation 1. EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z = –14 Equation 2 6x – y + 4z = –1 Equation 3 SOLUTION 16x + 11z = –1 New Equation 1

  42. Add – 3 times Equation 3 to Equation 2. –18x + 3y –12z = 3 STEP 2 Solve the new linear system for both of its variables. Add new Equation 1 –16x – 7z = –11 and new Equation 2. EXAMPLE 1 Use the elimination method 2x – 3y + 5z = –14 –16x – 7z = –11 New Equation 2 16x+ 11z = –1 4z = –12 z = –3 Solve for z. x = 2 Substitute into new Equation 1 or 2 to find x.

  43. Substitute x = 2 and z = – 3 into an original equation and solve for y. STEP 3 EXAMPLE 1 Use the elimination method 6x–y + 4z = –1 Write original Equation 3. 6(2) –y + 4(–3) = –1 Substitute 2 for xand –3 for z. y = 1 Solve for y.

  44. –4x – 4y – 4z = –12 Add –4 times Equation 1 4x + 4y + 4z = 7 to Equation 2. EXAMPLE 2 Solve a three-variable system with no solution Solve the system. x + y + z = 3 Equation 1 4x + 4y + 4z = 7 Equation 2 3x – y + 2z = 5 Equation 3 SOLUTION When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. 0 = –5 New Equation 1

  45. EXAMPLE 2 Solve a three-variable system with no solution Because you obtain a false equation, you can conclude that the original system has no solution.

  46. STEP 1 Rewrite the system as a linear system in two variables. x + y + z = 4 Add Equation 1 x + y – z = 4 to Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y + z = 4 Solve the system. Equation 1 x + y – z = 4 Equation 2 3x + 3y + z = 12 Equation 3 SOLUTION 2x + 2y = 8 New Equation 1

  47. 3x + 3y + z = 12 Solve the new linear system for both of its variables. STEP 2 –4x – 4y = –16 Add –2 times new Equation 1 4x + 4y = 16 to new Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y – z = 4 Add Equation 2 to Equation 3. 4x + 4y = 16 New Equation 2 Because you obtain the identity 0 = 0, the system has infinitely many solutions.

  48. EXAMPLE 3 Solve a three-variable system with many solutions STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = –x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, –x + 4, 0) is a solution of the system.

  49. 1. 3x + y – 2z = 10 6x – 2y + z = –2 x + 4y + 3z = 7 (1, 3, –2) ANSWER for Examples 1, 2 and 3 GUIDED PRACTICE Solve the system.

  50. 2. x + y – z = 2 2x + 2y – 2z = 6 5x + y – 3z = 8 ANSWER no solution for Examples 1, 2 and 3 GUIDED PRACTICE

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