Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities

A system of 2 linear equations in 2 variables, x & y consists of 2 equations of the following form: A x + B y = C D x + E y = F A, B, C, D, E and F all represent constant values. A solution of a system of linear equations in 2 variables is an ordered pair (x,y) that satisfies both equations. Example 1: Checking solutions of a linear system. Are ( 2, 2 ) and ( 0, − 1 ) solutions of the following system: 3 x – 2 y = 2 x + 2 y = 6 3 x – 2 y = 2 x + 2 y = 6 3 ( 2 ) – 2 ( 2 ) = 2 ( 2 ) + 2 ( 2 ) = 6 6 – 4 = = 6 2 = 2√ 6 = 6 √ 3 x – 2 y = 2 x + 2 y = 6 3 ( 0 ) – 2 (− 1 ) = 2 ( 0 ) + 2 (− 1 ) = = 2 0 − 2 = 6 2 = 2√ − 2 ≠ 6 Solution works for both Solution does NOT work for both Solving Linear Systems by Graphing 3.1

x2 x – 3 y = 1yx x + y = 3y 02 (0) – 3 y = 1 – 3 y = 1 − y = x – 3 (0) = 103x + 0 = 30 2 x – 3 y = 1 x + y = 3 ( 2, 1 ) 2 x – 3 y = 1 2 ( 2 ) – 3 ( 1 ) = 1 4 – 3 = 1 1 = 1 √ x + y = = 3 3 = 3 √ Solving Linear Systems by Graphing 3.1

Graphical Interpretation Algebraic Interpretations The graph of the system is a pair of lines that intersect in 1 point The system has exactly 1 solution The graph of the system is a single lineThe system has infinitely many solutions The graph of the system is a pair of parallel lines so that there is no point of intersection The system has no solution Exactly 1 solutionInfinitely many solutionsNo solution Number of Solutions of a Linear System 3.1

Substitution Method: 1.Solve for one equation 2.Substitute the expression from step 1 into the other equation, then solve for the other variable 3.Substitute the value from step 2 into the revised equation from step 1, then solve Substitution Method: 1.Solve for one equation 2.Substitute the expression from step 1 into the other equation, then solve for the other variable 3.Substitute the value from step 2 into the revised equation from step 1, then solve Example 1: 3 x + 4 y = − 4[1 st equation ] x + 2 y = 2[2 nd equation ] 1)Solve for x in equation 2 x + 2 y = 2 − 2 y − 2 y x = − 2 y + 2 2)Substitute 3 x + 4 y = − 4 3 (− 2 y + 2 ) + 4 y = − 4 − 6 y y = − 4 − 2 y + 6 = − 4 − 2 y = − 10 y = 5 3). Use value for y to get x: x = − 2 y + 2 x = − 2 ( 5 ) + 2 x = − = − 8 Check: 3 x + 4 y = − 4 3 (− 8 ) + 4 (5 ) = − 4 − = − 4 √ − 4 = − 4 √ Solving Linear System Algebraically 3.2

7 x − 12 y = − 22 − 5 x + 8 y = 14 2 [ 7 x − 12 y = − 22 ] 3 [− 5 x + 8 y = 14 ] 14 x − 24 y = − 44 − 15 x + 24 y = 42 − 1 x = − 2 x = 2 − 5 x + 8 y = 14 − 5 (2) + 8 y = 14 − y = 14 8 y = 24 y = 3 ( x, y ) ( 2, 3 ) Check: 7 x − 12 y = − 22 7 ( 2 ) − 12 ( 3 ) = − − 36 = − 22 √ − 22 = − 22 √ Solving Linear System Algebraically Linear Combination Method: 3.2

2 x + 3 y = 5 x − 5 y = 9 x = 5 y ( 5 y + 9 ) + 3 y = 5 10 y y = 5 13 y + 18 = 5 13 y = − 13 y = − 1 Check: 2 x + 3 y = 5 2 (4) + 3 ( − 1 ) = 5 8 − 3 = 5 √ 5 = 5 √ x − 5 y = 9 x − 5 (− 1) = 9 x + 5 = 9 x = 9 − 5 x = 4 3 x + 5 y = − 16 3 x − 2 y = − 9 − 1 [ ] 3 x + 5 y = − 16 − 3 x + 2 y = y = − 7 y = − 1 3 x + 5 (− 1) = − 16 3 x − 5 = − 16 3 x = − 11 x = − 11 3 Check: 3 x + 5 y = − 16 3 (− 11 ) + 5 (− 1) = − 16 3 − 11 − 5 = − 16 √ − 16 = − 16 √ ( x, y ) (− 11, − 1 ) 3 Solving Linear System Algebraically Linear Combination Method: Substitution Method: 3.2

y ≥ ─ 3 x ─ 1 xy 0−1 3 0 Test ( 0, 0 ) > y > ─ 3 x ─ 1 > 0 > ─ 3 ( 0 ) ─ 1 > √ 0 > ─ 1 √ y < x + 2 xy 02 − 20 Test ( 0, 0 ) y < x < ( 0 ) + 2 √ 0 < 2 √ y = ─ 3 x ─ 1 y = x + 2 BLUE ONLY BOTH RED & BLUE NEITHER RED nor BLUE A solution of a system of linear inequalities is an ordered pair that is a solution of each inequality in the system or a coordinate in BOTH solutions of each inequality y < x + 2 y ≥ ─ 3 x ─ 1 RED ONLY Graphing and Solving of Linear Inequalities 3.3

x ≥ 0 y ≥ 0 4 x + 3 y ≤ 24 x ≥ 0 y ≥ 0 4 x + 3 y ≤ 24 Test Point (2, 2) 2 ≥ 0 √ 4 (2) + 3 (2) ≤ ≤ ≤ ≤ 24 √ 14 ≤ 24 √ Graphing a System of 3 Inequalities 3.3

Real life problems involve a process called OPTIMIZATION = finding the maximum or minimum value of some quantity. Linear Programming is the process of optimizing a linear objective function subject to a system of linear inequalities called constraints. The graph of the system of constraints is called the feasible region. Optimal solution of a Linear Programming problem If an objective function has a maximum or a minimum value, then it must occur at a vertex of the feasible region. If the objective function is bounded,, then it has both a maximum and a minimum value. Optimal solution of a Linear Programming problem If an objective function has a maximum or a minimum value, then it must occur at a vertex of the feasible region. If the objective function is bounded,, then it has both a maximum and a minimum value. Bounded Region Unbounded Region Linear Programming (a type of optimization) 3.4

Optimum value, maximum or minimum, occur at vertices of a feasible region. Maximum = (3.6) or C = 2 x + 4 y “R” C = 2 (3) + 4 (6) “R” C = 2 (3) + 4 (6) C = = 30 C = = 30 Minimum = (0.0) or C = 2 x + 4 y “O” C = 2 (0) + 4 (0) “O” C = 2 (0) + 4 (0) C = = 0 C = = 0 C = 2 x + 4 y (0,5)(0,5)(0,5)(0,5) (0,0)(0,0)(0,0)(0,0) (6,0)(6,0)(6,0)(6,0) (3,6)(3,6)(3,6)(3,6) O P R V (0,5)(0,5)(0,5)(0,5) “P” C = 2 (0) + 4 (5) C = = 20 C = = 20 (6,0) “V” C = 2 (6) + 4 (0) C = = 12 C = = 12 Optimum Value of a Feasible Region feasible region 3.4

(8,0)(8,0)(8,0)(8,0) (0,8)(0,8)(0,8)(0,8) (0,0)(0,0)(0,0)(0,0) Objective Function: C = 3 x + 4 y x ≥ 0 Contraints: y ≥ 0 x + y ≤ 8 { Vertices: C = 3 x + 4 y (0,0) C = 3 (0) + 4 (0) = 0 (8,0) C = 3 (8) + 4 (0) = 24 (0,8) C = 3 (0) + 4 (8) = 32 minimum value maximum value Find Maximum and Minimum Values Ex 1 3.4

(6,0)(6,0)(6,0)(6,0) (0,5)(0,5)(0,5)(0,5) (2,3)(2,3)(2,3)(2,3) Objective Function: C = 5 x + 6 y x ≥ 0 y ≥ 0 Contraints: y ≥ 0 x + y ≥ 5 x + y ≥ 5 3x + 4 y ≥ 18 { Vertices: C = 5 x + 6 y (0,5) C = 5 (0) + 6 (5) = 30 (2,3) C = 5 (2) + 6 (3) = 28 (6,0) C = 5 (6) + 6 (0) = 30 minimum value No maximum value x + y = 5 3x + 4y = 18 x = 0 Unbounded Region y = 0 y = 0 Find Maximum and Minimum Values Ex 2 3.4

(5, 3, ─ 4 ) + y + x + z ─ z─ z─ z─ z ─ x─ x─ x─ x ─ y─ y─ y─ y ( x, y, z ) 5 ─ 4 3 ( x, y, z ) is an ordered triple, where 3 axes, taken 2 at a time, determine 3 coordinate planes that divide into eight octants. Graphing Linear Equations in 3 variables 3.5

(3, ─ 4, ─ 2 ) + y + x + z ─ z─ z─ z─ z ─ x─ x─ x─ x ─ y─ y─ y─ y ( x, y, z ) 3 ─ 4 ─ 2 Linear equation in three variables, ( x, y, z ), is an ordered equation of the form: A x + B y + C z = D [Where A, B, C and D are constants] [Where A, B, C and D are constants] Linear equation in three variables, ( x, y, z ), is an ordered equation of the form: A x + B y + C z = D [Where A, B, C and D are constants] [Where A, B, C and D are constants] The graph of a an ordered triple is a plane. Graphing Linear Equations in 3 variables 3.5

(4, ─ 6, 3 ) and (─ 7, 5, ─ 2) Plot points (4, ─ 6, 3 ) and (─ 7, 5, ─ 2) + y + x + z ─ z─ z─ z─ z ─ x─ x─ x─ x ─ y─ y─ y─ y 4 (4, ─ 6, 3 ) ─ 7 ─ ─ 2 (─ 7, 5, ─ 2)

(0, 7, 0) (0, 0, 4) (─ 5,, 0, 0) (0, 0, ─ 5) (0, ─ 7, 0) (9, 0, 0) + y ─ y─ y─ y─ y + x ─ x─ x─ x─ x ─ z─ z─ z─ z + z

A B C D EF ( 3, 4, 2 ) ( x, y, z ) A = (,, ) B = (,, ) C = (,, ) D = (,, ) E = (,, ) F = (,, ) + y ─ y─ y─ y─ y + x ─ x─ x─ x─ x ─ z─ z─ z─ z + z

A B C D EF ( 3, 4, 2 ) ( x, y, z ) A = (,, ) B = (,, ) C = (,, ) D = (,, ) E = (,, ) F = (,, ) + y ─ y─ y─ y─ y + x ─ x─ x─ x─ x ─ z─ z─ z─ z + z 3.5

(0, 6, 0) (0, 0, 3) (4,, 0, 0) + y ─ y─ y─ y─ y + x ─ x─ x─ x─ x ─ z─ z─ z─ z + z 3.5

(0, 6, 0) (0, 0, 3) (4,, 0, 0) + y ─ y─ y─ y─ y + x ─ x─ x─ x─ x ─ z─ z─ z─ z + z 3 x + 2 y + 4 z = 12 XYZ NOTES: Page 41, Section 3.5 Graphing Linear Equations in three variables ( 3 dimensions ) 3.5

3 x + 2 y + 4 z = 12 – 3 x – 2 y – 3 x – 2 y 4 z = 12 – 3 x – 2 y 4 z = 12 – 3 x – 2 y z = 12 – 3 x – 2 y z = 12 – 3 x – 2 y 4 f (x,y) = 12 – 3 x – 2 y f (x,y) = 12 – 3 x – 2 y 4 3 x + 2 y + 4 z = 12 Replace z with f (x,y) z = f (1, 3 ) = 12 – 3 x – 2 y Evaluate function when x = 1 and y = 3 z = f (1, 3 ) = 12 – 3 x – 2 y 4 4 z = f (1, 3 ) = 12 – 3 (1) – 2 ( 3) 4 4 z = f (1, 3 ) = 12 – 3 – 6 = Graph has a solution = (1, 3, 3 ) 4 Evaluating a function of 2 variables as a function of x & y 3.5

1.If 3 planes intersect at a single point, the system has 1 solution 2.If 3 planes intersect in a line, the system has infinitely many solutions 3.If 3 planes have no point of intersection, the system has no solution Linear Combination Method: 1.Rewrite the linear system from 3 variables to 2 variables 2.Solve the new linear system for both of its variables 3.Substitute values found in step 2 into the original equation and solve for the remaining variable Linear Combination Method: 1.Rewrite the linear system from 3 variables to 2 variables 2.Solve the new linear system for both of its variables 3.Substitute values found in step 2 into the original equation and solve for the remaining variable If you obtain an identity, such as 0 = 0, then the system has infinitely many solutions If you obtain an identity, such as 0 = 1, in any of the steps, then the system has no solution Graphing Linear Equations in 3 variables 3.5

+ z 3 x + 2 y + 4 z = 11 z 2 x − y + 3 z = 4 z 5 x − 3 y + 5 z = − 1 { { + z 3 x + 2 y + 4 z = 11 z 2 (2 x − y + 3 z = 4)→→ z − 3 ( 2 x − y + 3 z = 4 ) z 5 x − 3 y + 5 z = − 1→→ + z 3 x + 2 y + 4 z = 11 z 4 x − 2 y + 6 z = 8 z − 6 x + 3 y − 9 z = − 12 z 5 x − 3 y + 5 z = − 1 z 7 x + 10 z = 19 z − x − 4 z = − 13 z 7 x + 10 z = 19 z 7 ( − x − 4 z = − 13) →→ z 7x + 10 z = 19 z − 7x − 28 z = − 91 z − 18 z = − 72 z = 4 z 7x + 10 z = 19 (4) 7x + 10 (4) = 19 7x + 40 = 19 7x = − 21 x = − 3 z 2 x − y + 3 z = 4 y = 2 ( x, y, z ) (− 3, 2, 4 ) − 3) (4) 2 (− 3) − y + 3 (4) = 4 − 6 − 6 − y + 12 = 4 − y + 6 = 4 − y = − 2 Solving Systems Using Linear Combination ( 1 solution ) 3.6