PRECALCULUS I

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PRECALCULUS I
SOLVING SYSTEMS OF EQUATIONS
Dr. Claude S. Moore Cape Fear Community College
Chapter 8
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PRECALCULUS I
TWO-VARIABLE LINEAR SYSTEMS
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GRAPHICAL METHOD
1. Graph each equation on the same
coordinate (x-y) plane. 2. Find the point(s) of
intersection, if any exist. 3. Check the
solution(s) in each of the original equations.
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INTERPRETING GRAPHS of two linear equations in
two variables
Number of solutions 1. Exactly one consistent
independent 2. Infinitely many consistent
dependent 3. No solution inconsistent
Graph interpretations 1. Intersect in one
point 2. Lines are identical 3. Lines are
parallel
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GRAPHICAL EXAMPLE 1
Graph and solve (1) 8x 9y 42(2) 6x -
y 16 (1) (0,42/9) (42/8,0) (2) (0,-16)
(16/6,0)
2
1
The solution is (3,2) which checks in both
equations.
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ELIMINATION METHOD
1. Get coefficients of x (or y) to be
opposites of each other. 2. Add equations to
eliminate variable. 3. Back-substitute into
either equation. 4. Check your solution in both
of the original equations.
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ELIMINATION EXAMPLE 2
Solve by elimination (1) 5u 6v 32 (2) 3u
5v 22 -3(1) -15u - 18v - 96 5(2) 15u
25v 110 (3) 7v 14 v
2
In equation (2), substitute v 2 3u 5(2)
22 3u 10 22 3u 12
u 4Solution is (4,2).
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PRECALCULUS I
MULTIVARIABLE LINEAR SYSTEMS
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ROW-ECHELON FORM
The Gausian elimination process was named for
Carl Friedrich Gauss (1777-1855) a German
mathematician who developed the row-echelon form .
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ROW-ECHELON FORM
x - 2y 3z 9 -x 3y - 4 2x -
5y 5z 17 This system is in its original form.
x - 2y 3z 9 y 3z 5
z 2 This equivalent system is in Row-Echelon
form.
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ROW OPERATIONS EQUIVALENT SYSTEMS
1. Interchange two equations. 2. Multiply one
equation by non-zero constant. 3. Add a
non-zero multiple of one equation to a
non-zero multiple of another equation.
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INCONSISTENT SYSTEM
Solve the system of equations
E1 3x - 2y - 6z -4 E2 -3x 2y 6z
1 E3 x - y - 5z -3
e1 x - y - 5z -3 -3e1e2 y
9z 5 3e1e3 - 1y - 9z -8
e1 x - y - 5z -3 e2 3x - 2y - 6z
-4 e3 -3x 2y 6z 1
e1 x - y - 5z -3 e2 y 9z
5 e2e3 0z -3
Since 0 -3 is never true, there is no solution.
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INFINITE SOLUTION
Solve the system of equations
E1 x 2y - 7z -4 E2 2x y z
13 E3 3x 9y -36z -33
e1 x 2y - 7z -4 e2-2e1 -3y 15z
21 e3-3e1 3y - 15z -21
e1 x 2y - 7z -4 (-1/3)e2 y -
5z -7 e3e1 0z 0
Since 0z 0 is always true, the solution
(-3a10, 5a-7, a) is infinite.
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NON-SQUARE SYSTEM
Solve the system of equations
E1 x - 3y 2z 18 E2 5x - 13y
12z 80
e1 x - 3y 2z 18 (1/2)e2 y
z -5
So y -z - 5. Let z a. Then y -a - 5.
e1 x - 3y 2z 18 -5e1e2 2y
2z -10
Substitute z a and y -a -5 into equation 1
and get the solution (-5a 3, -a - 5, a).
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PRECALCULUS I
TWO-VARIABLE NON-LINEAR SYSTEMS
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SUBSTITUTION METHOD
1. Solve one equation for one variable. 2.
Substitute into other equation. 3. Solve equation
from Step 2. 4. Back-substitute into Step 1. 5.
Check the solution in each equation.
Section 9-4
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SUBSTITUTION EXAMPLE 1
Solve x y 0 and x3 - 5x - y 0. Substitute
y -x to get x3 - 5x - (-x) 0 x3 - 5x x 0
or x3 - 4x 0 x(x2 - 4) 0 or x(x 2)(x - 2)
0 Thus x 0, x -2, and x 2 giving the
solutions (-2,2), (0,0), and (2,-2).
Section 9-4
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GRAPHICAL METHOD
1. Graph each equation on the same
coordinate (x-y) plane. 2. Find the point(s) of
intersection, if any exist. 3. Check the
solution(s) in each of the original equations.
Section 9-4
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GRAPHICAL EXAMPLE 2

Graph and solve y ex x - y -1 x - y -1
yields y x 1.
Solution is (0,1) which checks in both equations
1 e0 and 0 - 1 -1.
Section 9-4
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GRAPHICAL EXAMPLE 3
Graph and solve (1) 2x - y 1(2) x2
y 2 (1) y 2x - 1(2) y -x2 2.

Solutions are (-3,-7) (1,1) which check in both
equations.
Section 9-4