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Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities

Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities. 3.1. Solving Linear Systems by Graphing. 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.

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Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities

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

  2. 3.1 Solving Linear Systems by Graphing 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 = 2 2 + 4 = 6 2 = 2√ 6 = 6 √ Solution works for both 3 x – 2 y = 2 x + 2 y = 6 3 ( 0 ) – 2 (− 1 ) = 2 ( 0 ) + 2 (− 1 ) = 6 0 + 2 = 2 0 − 2 = 6 2 = 2√ − 2 ≠ 6 Solution does NOT work for both

  3. 3.1 Solving Linear Systems by Graphing 2 x – 3 y = 1 2 ( 2 ) – 3 ( 1 ) = 1 4 – 3 = 1 1 = 1 √ x + y = 3 • ( 2 , 1 ) • • • x + y = 3 2 + 1 = 3 3 = 3 √ • 2 x – 3 y = 1

  4. 3.1 Number of Solutions of a Linear System Exactly 1 solution Infinitely many solutions No solution

  5. 3.2 Solving Linear System Algebraically Substitution Method: Solve for one equation Substitute the expression from step 1 into the other equation, then solve for the other variable Substitute the value from step 2 into the revised equation from step 1, then solve Example 1: 3 x + 4 y = − 4 [1st equation ] x + 2 y = 2 [2nd equation ] Solve for x in equation 2 x + 2 y = 2 − 2y− 2y x=− 2 y + 2 Substitute 3 x + 4 y = − 4 3 (− 2 y + 2 ) + 4 y = − 4 − 6 y + 6 + 4 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 = − 10 + 2 = − 8 Check: 3 x + 4 y = − 4 3 (− 8 ) + 4 (5 ) = − 4 − 24 + 20 = − 4 − 4 = − 4 √

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

  7. 3.2 Solving Linear System Algebraically Linear Combination Method: 7 x − 12 y = − 22 − 5 x + 8 y = 14 14 x − 24 y = − 44 − 15 x + 24 y = 42 − 1 x =− 2 x = 2 2 [ 7 x − 12 y = − 22 ] 3 [− 5 x + 8 y = 14 ] − 5 x + 8 y = 14 − 5 (2) + 8 y = 14 − 10 + 8 y = 14 8 y = 24 y= 3 ( x , y ) ( 2, 3 ) Check: 7 x − 12 y = − 22 7 ( 2 ) − 12 ( 3 ) = − 22 14 − 36 = − 22 − 22 = − 22 √

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

  9. Graphing and Solving of Linear Inequalities y < x + 2 RED ONLY y = x + 2 y = ─ 3 x ─ 1 Test ( 0 , 0 ) y < x+ 2 0 < ( 0 ) + 2 0 < 2√ y ≥ ─ 3 x ─ 1 NEITHER RED norBLUE • • • BOTHRED &BLUE • y ≥ ─ 3 x ─ 1 BLUE ONLY y < x + 2 Test ( 0 , 0 ) y > ─ 3x─ 1 0 > ─ 3 ( 0 ) ─ 1 0 > ─ 1√ 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 3.3

  10. 3.3 Graphing a System of 3 Inequalities • x ≥ 0 y ≥ 0 4 x + 3 y ≤ 24 4 x + 3 y ≤ 24 x ≥ 0 Test Point (2, 2) 2 ≥ 0 √ 2 ≥ 0 √ 4 (2) + 3 (2) ≤ 24 8 + 6 ≤ 24 14 ≤ 24 √ • • y ≥ 0

  11. 3.4 Linear Programming (a type of optimization) 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. • • • • • • • Bounded Region Unbounded Region

  12. 3.4 Optimum Value of a Feasible Region 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) C = 6 + 24 = 30 Minimum = (0.0) or C = 2 x + 4 y “O” C = 2 (0) + 4 (0) C = 0 + 0 = 0 C = 2 x + 4 y (3,6) • R • (0,5) P (0,5) “P” C = 2 (0) + 4 (5) C = 0 + 20 = 20 (6,0) “V” C = 2 (6) + 4 (0) C = 12 + 0 = 12 feasible region V O • • (0,0) (6,0)

  13. 3.4 Find Maximum and Minimum Values • (0,8) Ex 1 Objective Function: C = 3 x + 4 y x ≥ 0 Contraints: y ≥ 0 x + y ≤ 8 { • • (0,0) (8,0) 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

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

  15. 3.5 Graphing Linear Equations in 3 variables ( x , y , z ) + y ─ z (5 , 3 , ─ 4 ) • ─ 4 3 ─ x + x 5 ( x , y , z ) is an ordered triple, where 3 axes, taken 2 at a time, determine 3 coordinate planes that divide into eight octants. + z ─ y

  16. 3.5 Graphing Linear Equations in 3 variables + y The graph of a an ordered triple is a plane. ─ z 3 ─ x + x • ( x , y , z ) ─ 4 (3 , ─ 4 , ─ 2 ) ─ 2 + z 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] ─ y

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

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

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

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

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

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

  23. 3.5 Evaluating a function of 2 variables as a function of x & y 3 x + 2 y + 4 z = 12 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 4 4 z = 12 – 3 x – 2 y 4 f (x,y) = 12 – 3 x – 2 y 4 Replace z with f (x,y) Evaluate function when x = 1 and y = 3z = f (1 , 3 ) = 12 – 3 x – 2 y 4 z = f (1 , 3 ) = 12 – 3 (1) – 2 ( 3) 4 z = f (1 , 3 ) = 12 – 3 – 6 = 3 4 4 Graph has a solution = (1 , 3 , 3) 4

  24. 3.5 Graphing Linear Equations in 3 variables If 3 planes intersect at a single point, the system has 1 solution If 3 planes intersect in a line, the system has infinitely many solutions If 3 planes have no point of intersection, the system has no solution Linear Combination Method: Rewrite the linear system from 3 variables to 2 variables Solve the new linear system for both of its variables 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

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

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