Transcript Systems of Equations Sections 1-3

PRECALCULUS I
SOLVING SYSTEMS
OF EQUATIONS
Dr. Claude S. Moore
Cape Fear Community College
Chapter 8
PRECALCULUS I
TWO-VARIABLE
LINEAR SYSTEMS
672
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 Graph interpretations
1. Exactly one
1. Intersect in one
consistent; independent
point
2. Infinitely many
2. Lines are identical
consistent; dependent
3. No solution
3. Lines are parallel
inconsistent
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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)
1
2
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=4
Solution is (4,2).
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PRECALCULUS I
MULTIVARIABLE
LINEAR SYSTEMS
688
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 - 2y + 3z = 9
-x + 3y
=-4
y + 3z = 5
2x - 5y + 5z = 17
z=2
This system is in This equivalent
its original form. 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
-3e1+e2
3e1+e3
x -
y - 5z = -3
y + 9z = 5
- 1y - 9z = -8
e1
x - y - 5z = -3
e2 3x - 2y - 6z = -4
e3 -3x + 2y + 6z = 1
e1
e2
e2+e3
x -
y - 5z = -3
y + 9z = 5
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
e2-2e1
e3-3e1
x + 2y - 7z = -4
-3y + 15z = 21
3y - 15z = -21
e1
(-1/3)e2
e3+e1
x + 2y - 7z = -4
y - 5z = -7
0z = 0
Since 0z = 0 is always true,
the solution (-3a+10, 5a-7, a) is infinite.
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NON-SQUARE SYSTEM
Solve the system of equations:
E1
E2
x - 3y + 2z = 18
5x - 13y + 12z = 80
e1
x - 3y + 2z = 18
-5e1+e2
2y + 2z = -10
e1
x - 3y + 2z = 18
(1/2)e2
y + z = -5
So y = -z - 5. Let z =a.
Then y = -a - 5.
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
672
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
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
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
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
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