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CHAPTER 5
College Algebra
5.1 SYSTEMS OF EQUATIONS IN TWO
VARIABLES
VOCABULARY
 A system of equations is composed of two or more equations
considered simultaneously.
 For example:
 x-y = 5,
2x + y =1
 This is an example of a system of two linear equations in two variables
 The solution set of this system consists of all ordered pairs that make
BOTH equations true.
 For our example this would be the point (2,-3)
VOCABULARY
 Graphically
 If a system of equations has at least one solution, it is consistent.
 If the system has no solutions, it is inconsistent.
 In addition, if a system has an infinite number of solutions, the equations are
dependent. Otherwise, they are independent.
METHODS TO SOLVING
 Graphically
 When we graph a system of linear equations, each point at which the graphs
intersect is a solution to BOTH equations and therefore is a solution of the system
of equations.
 Let’s use the equations that we started with:
 x–y=5
 2x + y = 1
METHODS TO SOLVING
Substitution
We begin by using one of the equations to
express one variable in terms of the other; then
we substitute that expression in the other
equation of the system.
It is used most often when a variable is alone on
one side of an equation or when it is easy to
solve for a variable.
METHODS TO SOLVING: EXAMPLE
 Our equations are:
 (a) x – y = 5
 (b) 2x + y = 1
 (a) y = x – 5
 b(a):
2x + (x – 5) = 1
2x + x – 5 = 1
3x = 6
x=2
 (a) 2 – y = 5 then y = -3
 (b) 2(2) + y = 1 then y = -3
 (2, -3)
 We should chose the one that
is easiest to solve for one
variable (a)
 Now we can plug in our
equation (a) into (b)
 Solve for variable
 Plug in your answer to find y
(you can use a or b)
 Write answer as a coordinate
METHODS TO SOLVING
Elimination
Another technique for solving systems of
equations is the elimination method. With this
method, we eliminate a variable by adding two
equations. If the coefficients of a particular
variable are opposites, we can eliminate that
variable simply by adding the original equations.
METHODS TO SOLVING EXAMPLE
 Since the y coefficients in our
example are already opposites
(-1 and 1), we can eliminate y
with ease.
 2x + y = 1
x
3x
-y=5
=6
so x = 2
 Then we can plug x into either
equation to solve for y.
5.2 SYSTEMS OF EQUATIONS IN THREE
VARIABLES
 A linear equation in three variables is an equation equivalent to one of
the form Ax +By + Cz = D, where A, B, C, and D are real numbers and
A, B, and C are not equal to zero.
 A solution of a system of three equations in three variables is an
ordered triple that makes all three equations true. (x, y, z)
 We will solve systems of equations in three variables using an algebraic
method called Gaussian elimination, named for the German
mathematician Karl Gauss.
 Interchange any two equations
 Multiply both sides of one of the equations by a nonzero constant
 Add a nonzero multiple of one equation to another equation
EXAMPLE
SOLUTIONS MEANINGS
 One Solution
 Planes intersect at exactly one point
 No Solution
 Three planes, each intersect each other but at no point do all intersect
 Parallel planes
 Infinitely Many Solutions
 Planes intersect in a line
 Planes are identical
HOMEWORK TIME
5.3 MATRICES AND SYSTEMS OF
EQUATIONS
 Row Equivalent Operations
 Interchange any two rows
 Multiply each entry in a row by the same nonzero constant
 Add a nonzero multiple of one row to another row
 Row Echelon Form
 If a row does not consist of entirely zero’s then the first nonzero element in the row is a
one.
 For any two successive nonzero rows, the leading one in the lower row is farther to the
right than the leading one in the higher row.
 All the rows consisting entirely of zeros are at the bottom of the matrix

Reduced Row Echelon Form: each column that contain a leading one has zero’s everywhere else.

Gauss-Jordan Elimination
MATRICES TO SOLVE SYSTEM OF
EQUATIONS
 2x – y + 4z = -3
 x – 2y – 10z = -6
 3x + 4z = 7
5.4 MATRIX OPERATIONS
 Addition and Subtraction of Matrices
 Given two m x n matrices A and B their sum is A + B and their difference is A – B
 Example:
ADDITIVE INVERSE
 The opposite, or additive inverse, of a matrix is obtained by replacing
each entry with its opposite or additive inverse.
 Find – A such that A + (-A) = 0 given A =
é2 ù
ê ú
ë-8û
 A matrix having zeros for all its entries is called the zero matrix.
 When a zero matrix is added to a second matrix of the same order, the second
matrix is unchanged
 Therefore the zero matrix is the additive identity
MATRIX MULTIPLICATION
 Scalar Product
 The scalar product of a number k and a matrix A is the matrix denoted kA, obtained
by multiplying each entry of A by the number k. The number k is called a scalar.
 Properties
 For any m x n matrices A, B, and C and any scalars k and l:
 A + B = B + A Commutative Property of Addition
 A + (B + C) = (A + B) + C Associative Property of Addition
 (kl)A = k(lA)
Associative Property of Scalar Multiplication
 k(A+B) = kA + kB
Distributive Property
 (k+l)A = kA + lA
Distributive Property
 There exists a unique matrix 0 such that: A + 0 = 0 + A = A
 There exists a unique matrix -A such that: A + (-A) = -A + A = 0
PRODUCT OF MATRICES
 Note that we can multiply two matrices only when the number of
columns in the first matrix is equal to the number of rows in the second
matrix.
 Therefore matrices are not generally commutative.
 ORDER MATTERS!
 Properties of Matrix Multiplication
 For matrices A, B, and C assuming the indicated operations are possible –
 A(BC) = (AB)C
Associative Property of Multiplication
 A(B + C) = AB + AC
 (B + C)A = BA + CA
Distributive Property
MATRIX MULTIPLICATION
MATRIX EQUATIONS
Write the matrix equation equivalent to 2x – y = 0 and –x + 2y = 3
HOMEWORK
5.5 INVERSES OF MATRICES
 For an n x n matrix A, if there is a matrix A-1 for which A-1A = I = AA-1
 Then A-1 is the inverse of A
IDENTITY MATRIX
SOLVING SYSTEM OF EQUATIONS
5.6 DETERMINANTS AND CRAMER’S RULE
The determinate of a 2x2 matrix is ad - bc
MINORS AND COFACTORS
 For the matrix to the right find the following
 M11
 A23
-8
0
6
4
-6
7
-1
-3
5
DETERMINANT OF ANY SQUARE MATRIX
CRAMERS RULE FOR 2X2
EXAMPLE CRAMERS 2X2
 Solve using Cramer’s Rule: 2x + 5y = 7 and 5x – 2y = -3
HOMEWORK TIME
5.7 SYSTEMS OF INEQUALITIES AND
LINEAR PROGRAMMING
 Steps to graphing a linear inequality with two variables:
 Replace the inequality symbol with an equals sign and graph this related equation.
 If the inequality symbol does not have an “or equal to” then the line should be dashed
 If the inequality symbol does have an “or equal to” then the line should be solid
 The graph consists of a half plane on one side of the line and if the line is solid, it
includes the line as well.
 To determine which half plane to shade, test a point not on the line or in the original
inequality.
 If that point is a solution, shade the plane that contains that point
 If that point is not a solution, shade the plane that does not contain that point.
 You can always double check by using a point in the other half plane.
GRAPHING AN INEQUALITY
GRAPHING A SYSTEM OF INEQUALITIES
LINEAR PROGRAMMING
 To find the maximum and minimum value of a linear objective function
subject to a set of constraints
 Graph the region of feasible solutions
 The constraints
 Determine the coordinates of the vertices of the region
 Corners
 Evaluate the objective function at each vertex. The largest and the smallest of those
values are the maximum and the minimum values of the function, respctively.
LINEAR PROGRAMMING
 The function we are trying to maximize is P = 40x + 75y
 Given the constraints shown in the graph below
X
Y
P
0
2
150
2
1
3
0
120
0
0
0
5.8 PARTIAL FRACTIONS
 Procedure:
 Consider any rational expression P(x)/Q(x) such that P(x) and Q(x)
have no common factor other than 1 or -1
 If the degree of P(x) is greater than or equal to the degree of Q(x), divide to express
P(x)/Q(x) as a quotient plus remainder/Q(x) and follow steps 2-5 to decompose the
resulting rational expression
 If the degree of P(x) is less than the degree of Q(x) into linear factors of the form
(px + q)n and/or quadratic factors of the form (ax2 + bx + c )m. Any quadratic factor
must be irreducible (cannot be factored into linear factors with rational coefficients)
 Assign to each linear factor the sum of the n partial fractions:
A1
A2
An
+
+... +
2
n
px + q (px + q)
(px + q)
PARTIAL FRACTIONS PROCEDURE
 Assign to each quadratic factor (ax2+bx+c)m the sum of m partial
fractions:
B1 x + C1
B2 x + C2
Bm x + Cm
+
+... +
2
2
2
ax + bx + c (ax + bx + c)
(ax 2 + bx + c)m
 Apply algebraic methods to find the constants in the numerators of the
partial fractions
PARTIAL FRACTION EXAMPLE
Decompose into partial fractions ---
7x - 29x + 24
A
B
C
=
+
+
2
2
(2x -1)(x - 2) 2x -1 x - 2 (x - 2)
2
Find a common denominator on the right side to obtain:
7x 2 - 29x + 24 A(x - 2)2 + B(2x -1)(x - 2) + C(2x -1)
=
2
(2x -1)(x - 2)
(2x -1)(x - 2)2
PARTIAL FRACTIONS EXAMPLE
 Equate the numerators to get:
 7x2 - 29x + 24 = A(x-2)2 + B(2x-1)(x-2) +C(2x-1)

= A(x2 – 4x + 4) + B(2x2 – 5x + 2) + C(2x-1)

= Ax2 – 4Ax + 4A + 2Bx2 – 5Bx + 2B + 2Cx –C
 Group like terms together

= (A + 2B)x2 + (-4A – 5B +2C)x + (4A + 2B – C)
 Equate corresponding coefficients
 7 = A + 2B
-29 = (-4A -5B + 2C)
24 = 4A + 2B – C
 Solve the system of equations to get
 A = 5, B = 1 and C = -2 therefore the decomposition is:
7x - 29x + 24
5
1
2
=
+
2
(2x -1)(x - 2) 2x -1 x - 2 (x - 2)2
2
HOMEWORK TIME