Download 3x3 equations - hrsbstaff.ednet.ns.ca

SYSTEMS OF THREE EQUATIONS
Solving systems of three equations and three variables can be done
in two ways through the ADDITION/SUBTRACTION method, and through
the use of MATRICES and ROW REDUCTION.
Systems of three equations are used in such fields as mathematics,
chemistry, biology, physics and statistics. Therefore it is useful to know
how to use them. Both methods will work for most systems, however some
are easier to use in certain situations. Therefore it is important to practice
both methods.
How to solve a system with three equations and three variables by the
Addition/Subtraction method.
1) Pick any two pairs of equations from the system.
2) Eliminate the same variable from each pair using the Addition/
Subtraction method.
3) Solve the system of the two new equations using the addition/Subtraction
method.
4) Substitute the solution back into one of the original equations and solve
for the third value.
5) Check to see if the question is right by plugging the solution into one of
the other original questions.
Example:
Question: 4x - 3y + z = -10
2x + y + 3z = 0
-x + 2y - 5z = 17
1) Pick any two pairs:
4x - 3y + z = -10
2x + y + 3z = 0
and
2x + y + 3z = 0
-x + 2y -5z =17
2) Eliminate the same variable from each system:
4x - 3y + z = -10
2x + y + 3z = 0
=
4x - 3y + z = -10
(-2) -4x - 2y - 6z = 0
=
-5y - 5z = -10
2x + y + 3z = 0
-x + 2y - 5z = 17
=
2x + y + 3z = 0
(2)
-2x + 4y - 10z = 34
=
5y - 7z =34
3) Solve the system of the two new equations:
-5y - 5z = -10
5y - 7z = 34
-12z = 24
z = 24/-12 thus z = -2
-5y - 5(-2) = -10
-5y + 10 = -10
-5y = -10 - 10
-5y = -20
y = -20/-5 thus y = 4
4) Substitute into one of the original equations:
-x + 2y - 5z = 17
-x + 2(4) - 5 (-2) = 17
-x + 8 +10 = 17
-x + 18 = 17
-x = 17 - 18
-x = -1
x=1
Therefore (x, y, z) equals (1,4, -2)
5) Check using one of the three original questions:
does 2(1) + 4 + 3(-2) = 0
2 + 4 -6 = 0
6-6=0
0=0
Solving using Matrices and Row Reduction
Systems with three equations and with three variables can also be
solved using matrices and row reduction. First, you have to arrange the
system in the following form:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Where a1, 2, 3, b1, 2, 3, and c1, 2, 3 are the x, y, and z coefficients are, and d1, 2, 3
are constants.
Next, you have to create a 3×4 matrix. Placing the x coefficients
in the 1st column, the y coefficients in the 2nd column, and the z
coefficients in the 3rd column. The constants (which is d), in the 4th
column, with a line solid or broken separating the 3rd column and the 4th
column:
a1 b1 c1
d1
a2 b2 c2
d2
a3 b3 c3
d3
This is correspondent to
a1 b1 c1
x
a2 b2 c2
y
a3 b3 c3
z
d1
=
d2
d3
Which is correspondent to the original equations.
Finally, you can row reduce the 3×4 matrix using the elementary row
operations. The result should be the identity matrix on the left side of the
line and a column of constants on the right side of the line. These constants
are the solution to the system of equations:
1 0 0
x
0 1 0
y
0 0 1
z
If the system row reduces to
1 0 0
x
0 1 0
y
0 0 0
e1
Then the system is inconsistent.
If the system row reduced to
1 0 0
x
0 1 0
y
0 0 0
0
Then the system has multiple solutions. Leaving you with the above
answer.
Test
Solve the following using the adding and subtracting method:
1)
2x + y + 3z = 10
-3x - 2y + 7z = 5
3x + 3y - 4z = 7
Answer (x, y, z) = (-1, 6, 2)
2)
x + y - 2z = 5
2x + 3y + 4z = 18
-x - 2y - 6z = -13
Answer (x, y, z) = (2, 4, [1/2])
3)
3x + 2y = 7
-4x + 3z = 5
-2y - 6z = -10
Answer (x, y, z) = (3, -1, 2)
4)
4x + 2y - 5z = -21
2x - 2y + z = 7
4x + 3y - z = -1
Answer (x, y, z) = (1, 0, 5)
5)
x + y + z = 10
2x + 2y + 3z = 14
2x - 3y + 2z = -5
Answer (x, y, z) = (-2, 3, 4)
Solve the following system using matrices and row reduction:
1)
2x - y + 3z = -15
x + 2y - z = 15
2x - 3y - 3z = 7
Answer (x, y, z) = (2, 4, -5)
2)
2x + 2y - 5z = -9
5x - 2y + z = 4
7x + 4y - 4z = 3
Answer (x, y, z) = (1, 2, 3)
3)
2x + y + z = -11
-3x - y + 2z = -1
x+ 2y - z = 2
Answer (x, y, z) = (-3, 0, -5)
4)
5x + y + z = 5
2x + 2y + 2z = -6
3x - y + 4z = -9
Answer (x, y, z) = (2, -1, -4)
5)
x - 2y - z = 0
2x + y - 3z = 7
-3x - y + 4z = -11
Answer (x, y, z) = (7, 2, 3)
Bibliography
1) http://www.sparknotes.com/math/algebra2/systemsofthreeequations/s
ummary.html
2) http://www.sci.sdsu.edu/sciences/summer_bridge/coursemats/math/le
cture2/LECTURE2.HTM
3) http://www.sosmath.com/index.html
4) http://mathforum.org/library/results.html?textsearch=systems+of+thre
e+equation&textsearch_bool_type=and&textsearch_whole_words=no
&topics=&ed_topics=&resource_types=&levels=