1. Copyright © 1999 by the McGraw-Hill Companies, Inc. Barnett/Ziegler/Byleen College Algebra, 6 th Edition Chapter Six Systems of Equations & Inequalities

2. x y 5–5 5 (4, 2) x y 5–5 5 x y 5 5 (A) 2x – 3y = 2 x + 2y = 8 (B) 4x + 6y = 12 2x + 3y = –6 Lines intersect at one point only. Exactly one solution: x = 4, y = 2. Lines are parallel (each has slope –  ). No solution. Lines coincide. Infinitely many solutions. Nature of Solutions to Systems of Equations 6-1-58

3. In general, associated with each linear system of the form a 11 x 1 + a 12 x 2 = k 1 a 21 x 1 + a 22 x 2 = k 2 where x 1 and x 2 are variables, is the augmented matrix of the system: Augmented Matrix 6-1-59

4. Elementary Row Operations Producing Row-Equivalent Matrices 6-1-60

5. A matrix is in reduced form if: 1. Each row consisting entirely of 0’s is below any row having at least one nonzero element. 2. The leftmost nonzero element in each row is 1. 3. The column containing the leftmost 1 of a given row has 0’s above and below the 1. 4. The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row. Reduced Matrix 6-2-61

6. Step 1.Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top. Step 2.Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3.Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4.Repeat step 2 with the entire matrix, including the mentally deleted rows. Continue this process until it is impossible to go further. [Note: If at any point in the above process we obtain a row with all 0’s to the left of the vertical line and a nonzero number n to the right, we stop, since we will have a contradiction: 0 = n, n  0. We can then conclude that the system has no solution.] Gauss-Jordan Elimination 6-2-62

7. x y 1 –1 1 x y 5–5 5 x y 5 5 1. x 2 + y 2 = 5 3x + y= 1 Two real solutions. 2. x 2 – 2y 2 = 2 xy = 2 Two real solutions and two imaginary solutions. (Imaginary solutions cannot be shown on the graph.) 3. x 2 + 3xy + y 2 = 20 xy – y 2 = 0 Four real solutions. Solutions of Nonlinear Systems of Equations 6-3-63

8. (a) y  2x – 3 (b) y > 2x – 3 (c) y  2x – 3 (d) y < 2x – 3 Graph of a Linear Inequality 6-4-64

9. Procedure for Graphing Linear Inequalities 6-4-65

10. Solution of Linear Programming Problems 6-5-66