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

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

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

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

Elementary Row Operations Producing Row-Equivalent Matrices

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 The column containing the leftmost 1 of a given row has 0’s above and below the The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row. Reduced Matrix

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

x y 1 –1 1 x y 5–5 5 x y 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

(a) y  2x – 3 (b) y > 2x – 3 (c) y  2x – 3 (d) y < 2x – 3 Graph of a Linear Inequality

Procedure for Graphing Linear Inequalities

Solution of Linear Programming Problems