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Barnett/Ziegler/Byleen
College Algebra, 7th Edition
Chapter Six
Systems of Equations & Inequalities
Copyright © 2001 by the McGraw-Hill Companies, Inc.
y
Nature of Solutions to
Systems of Equations
5
(4, 2)
x
–5
5
y (B) 4x + 6y = 12
2x + 3y = –6
(A) 2x – 3y = 2
x + 2y = 8
5
–5
Lines intersect at one point only.
Exactly one solution: x = 4, y = 2.
–5
y
5
5
(C) 2x – 3y = –6
3
–x + 2 y = 3
Lines coincide.
Infinitely many
solutions.
x
–5
x
–5
5
Lines are parallel
(each has slope – ).
No solution.
–5
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Augmented Matrix
In general, associated with each linear system of the form
a11x1 + a12x2 = k1
a21x1 + a22x2 = k2
where x1 and x2 are variables, is the augmented matrix of the system:
Column 1 ( C1)
Column 2 ( C2)
Column 3 ( C3)
a11

a
 21
a 12
a 22
k1

k 2 
Row 1 ( R 1)
Row 2 ( R 2)
6-1-59
Elementary Row Operations Producing
Row-Equivalent Matrices
An augmented matrix is transformed into a row-equivalent matrix if any of
the following row operations is performed:
1. Two rows are interchanged (Ri  Rj ).
2. A row is multiplied by a nonzero constant (kRi  Ri ).
3. A constant multiple of another row is added to a given
row (kRj + Ri  Ri ).
[Note: The arrow means "replaces."]
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Reduced Matrix
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.
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Gauss-Jordan Elimination
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.]
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Solutions of Nonlinear
Systems of Equations
1. x2 + y2 = 5
3x + y = 1
y
2. x2 – 2y2 = 2
xy = 2
1
–1
y
5
x
–1
1
x
–5
5
y
Two real solutions.
–5
5
3. x2 + 3xy + y2= 20
xy – y2 = 0
x
–5
5
Two real solutions and two
imaginary solutions.
(Imaginary solutions cannot
be shown on the graph.)
–5
Four real solutions.
6-3-63
y
x
0
Graph of a Linear Inequality
(a) y  2x – 3
y
y
y =2x -3
8
–5
x
0
(4, y ) y > 2(4) - 3 = 5;
point in upper half-plane
(4, y ) y = 2(4) - 3 = 5;
point on line
(4, y ) y < 2(4) - 3 = 5;
point in lower half-plane
5
10
(b) y > 2x – 3
y
x
0
x
(c) y  2x – 3
y
0
–8
x
(d) y < 2x – 3
6-4-64
Procedure for Graphing Linear Inequalities
Step 1. Graph Ax + By = C as a dashed line if equality is not
included in the original statement or as a solid line if
equality is included.
Step 2. Choose a test point anywhere in the plane not on the
line. The origin (0, 0) often requires the least
computation. Substitute the coordinates into the
inequality.
Step 3. The graph of the original inequality includes the halfplane containing the test point if the inequality is
satisfied by that point, or the half-plane not containing
that point if the inequality is not satisfied by the point.
6-4-65
Solution of Linear Programming Problems
Step 1. Form a mathematical model for the problem:
(A) Introduce decision variables and write a linear
objective function.
(B) Write problem constraints in the form of linear
inequalities.
(C) Write nonnegative constraints.
Step 2. Graph the feasible region and find the corner points.
Step 3. Evaluate the objective function at each corner point to
determine the optimal solution.
6-5-66