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

Linear Combination Method: 1.Multiply one or both equations by a constant to obtain coefficients that differ only in sign for one of the variables. 2.Add the revised equations from step 1, combining like terms will eliminate one of the variables. Now solve for the remaining variable. 3.Substitute the values obtained in step 2 into either one of the original equations and solve for the other variable. Linear Combination Method: 1.Multiply one or both equations by a constant to obtain coefficients that differ only in sign for one of the variables. 2.Add the revised equations from step 1, combining like terms will eliminate one of the variables. Now solve for the remaining variable. 3.Substitute the values obtained in step 2 into either one of the original equations and solve for the other variable. Example 2: 2 x − 4 y = 13 4 x − 5 y = 8 − 2 [ 2 x − 4 y = 13 ] 4 x − 5 y = 8 − 4 x + 8 y = − 26 4 x − 5 y = 8 3 y = − 18 y = − 6 2 x − 4 y = 13 2 x − 4 (− 6 ) = 13 2 x + 24 = 13 − 24 − 24 2 x = − 11 x = − 11 2 ( x, y ) (− 11, − 6 ) 2 Solving Linear System Algebraically 3.3

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 (0) = 30 (2,3) C = 5 (8) + 6 (0) = 28 (6,0) C = 5 (0) + 6 (8) = 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