Algebra & Trigonometry with Analytic Geometry 11th ed. Swokowski and Cole

Chapter Nine Systems of Equations and Inequalities

Section 9.1

Systems of Equations

Goals

Solve systems of equations, including

the method of substitution;

graphing solutions.

Also give examples.

Solution of a System

We say that the point (a, b) is a solution of the system of equations

y = f ( x) y = g ( x) if b = f(a) and b = g(a) .

Thus the point (a, b) is on both the graphs of y = f(x) and y = g(x) :

Example

If (x, y) is a solution of the system 2

y = x , y = 2x + 3 then we must have both y = 2x + 3 and y = x2 ; therefore x2 = 2x + 3 , or x2 – 2x – 3 = 0 .

Factoring then gives x = –1, 3 .

Example (cont’d)

Then we can use either equation in the system to find the solutions of the system: (– 1, 1) and (3, 9) .

Here are the equations and the solutions graphed together:

Method of Substitution

The preceding example illustrates the method of substitution:

We solved one equation for y in terms of x and then…

substituted into the other equation, obtaining an equation in x alone, which we then were able to solve.

Finally, the corresponding values of y were found from either of the original equations.

Guidelines for Substitution

These guidelines are…

for two equations in two variables;

stated for any two variables u and v , which could be x and y :

Example

We want to solve the system 2

x + y = 6 x + 2 y = 3 and sketch the graph of each equation, showing the points of intersection.

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Example 2

2

x + y = 25 : Next we solve the system 2 x + y = 19

Example (contâ€™d)

Example (contâ€™d)

A System of Three Equations

We can apply the method of substitution also to systems having three equations in three unknowns.

For example, we next solve the system

x − y + z = 2 xyz = 0 2 y + z = 1 for the ordered triple (x, y, z) :

Example (contâ€™d)

Example (contâ€™d)

Section 9.2

Systems of Linear Equations in Two Variables

Goals

Solve systems of linear equations in two variables, including

equivalent systems;

the method of elimination;

consistent/inconsistent systems.

Systems of Linear Equations

A linear equation in two variables x and y is of the form ax + by = c .

We will solve a system of two such equations by finding a simpler system that is equivalent to the given system.

This means that the new system has exactly the same solutions (if any) as the original system.

Theorem on Equivalent Systems ď Ž

The following theorem gives some manipulations (or transformations) that lead to an equivalent system:

Theorem (cont’d)

When applying part (3) of the theorem, we often use the phrase “add to one equation k times any other equation.”

To add two equations means to add corresponding sides of the equations.

Often the theorem is used to eliminate a variable from the system:

Example

We apply the theorem to the system

x + 3 y = −1 : 2 x − y = 5

Example (contâ€™d)

Example (contâ€™d)

Example ď Ž

Here is a linear system with an infinite number of solutions:

Example (contâ€™d)

Example (contâ€™d)

A Linear System With No Solutions

No Solutions (contâ€™d)

Three Possibilities

These examples illustrate the three typical outcomes of solving a linear system:

Exactly one solution

An infinite number of solutions

…so the system is consistent; …so the system is dependent and consistent;

No solution

…so the system is inconsistent.

Possibilities (contâ€™d) ď Ž

In general, exactly one of these possibilities holds for any system of two linear equations in two unknowns:

Example

A motorboat, operating at full throttle, made a trip 4 miles upstream (against a constant current) in 15 minutes.

The return trip (with the same current and at full throttle) took 12 minutes.

Find the speed of the current and the equivalent speed of the boat in still water.

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Section 9.3

Systems of Inequalities

Goals

Discuss inequalities in two variables, including

solutions;

graphing, using test points;

the special case of linear inequalities.

Learn about systems of inequalities.

Inequalities

Some inequalities in the variables x and y :

y2 < x + 4 ;

3x – 4y > 12 ;

x2 + y2 ≤ 16 .

A solution of an inequality in x and y is an ordered pair (a, b) that yields a true statement if a and b are substituted for x and y , respectively.

Solving Inequalities ď Ž

Here are some guidelines for solving an inequality using test points:

Example ď Ž

We find the solutions and sketch the graph of the inequality y2 < x + 4 :

Example (contâ€™d)

Example (contâ€™d)

Linear Inequalities

A linear inequality is an inequality that can be written in one of the following forms, where a , b , and c are real numbers:

ax + by < c ;

ax + by > c ;

ax + by ≤ c ;

ax + by ≥ c .

Linear Inequalities (cont’d)

The line ax + by = c divides the xy-plane into two halfplanes, as shown:

The solutions of the inequality consist of all points in one half-plane.

So for a linear inequality, only one test point is needed.

Example

Here we sketch the graph of the linear inequality 3x – 4y > 12 :

Example (contâ€™d)

Systems of Inequalities

As with equations, we sometimes work with several inequalities simultaneously in two variables…

known as a system of inequalities.

The solutions of such a system are the solutions common to all inequalities in the system.

Example ď Ž

Here we solve a system of linear inequalities and sketch the solution:

Example (contâ€™d)

Example ď Ž

Here we add two more inequalities to the system we just solved (called Example 3):

Example (contâ€™d)

Example ď Ž

The inequalities in the following example are not both linear:

Example (contâ€™d)

Example

The manager of a baseball team wants to buy bats and balls costing $12 and $3 each, respectively.

At least five bats and ten balls are needed, and the total cost is not to exceed $180.

Find a system of inequalities that describes all possibilities, and sketch.

Example (cont’d)

We let…

x denote the number of bats, and

y denote the number of balls purchased.

This means that…

12x = cost of x bats;

3y = cost of y balls.

Therefore 12x + 3y ≤ 180 , or y ≤ – 4x + 60 .

Example (cont’d)

We also need x ≥ 5 and y ≥ 10 .

Here is the graph of the overall system, with the solution shown in purple:

Section 9.4

Linear Programming

Goals

Study linear programming problems, including

objective functions;

constraints;

feasible solutions;

solutions.

Linear Programming Problems

Ingredients of an LPP:

Constraints of the form ax + by ≤ c or ax + by ≥ c that define a region R in the xy-plane;

The points (x, y) in R are the feasible solutions for the problem.

An objective function of the form C = Ax + By + K ,

which we want to maximize or minimize.

Example

We seek the maximum and minimum values of the objective function C = 7x + 3y subject to the following constraints:

x – 2y ≥ –10

2x + y ≤ 10

x≥0

y≥0

Example (cont’d)

The constraints define the region R shown:

According to theory, the maximum and minimum values of C on R must occur at vertices of R .

Example (cont’d)

Therefore we compare the values of C at the vertices of R :

The table shows that the…

minimum value of C occurs at (0, 0) ;

maximum value of C occurs at (5, 0) .

General Guidelines ď Ž

To solve linear programming problems, we can use the following guidelines:

Example

Example (contâ€™d)

Example

A firm manufactures two products, X and Y.

For each product, it is necessary to use three different machines, A, B, and C.

To manufacture one unit of…

product X requires

3 hours on machine A, and

1 hour each on machines B and C.

Example (cont’d)

product Y requires

2 hours on A,

2 hours on B, and

1 hour on C.

The profit on

product X is $500 per unit, and

product Y is $350 per unit.

Example (cont’d)

Machine A is available for a total of 24 hours per day; however

B can be used for only 16 hours, and

C for 9 hours.

Determine the number of units of each product that should be manufactured each day in order to maximize the profit.

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Section 9.5

Systems of Linear Equations in More Than Two Variables

Goals

Extend the method of elimination to systems of linear equations having more than two variables, including homogeneous systems.

Study the general nature of the solutions of such systems.

Apply matrix methods to such systems.

Method of Elimination

We apply this method to solve the system

x − 2 y + 3z = 4 2x + y − 4z = 3 . −3 x + 4 y − z = −2

As before, we use the theorem on equivalent systems to eliminate variables:

Elimination (cont’d)

Adding – 2 times the first equation to the second equation causes the following change:

x − 2 y + 3z = 4 x − 2 y + 3z = 4 5 y − 10 z = −5 2x + y − 4z = 3 → −3 x + 4 y − z = −2 −3 x + 4 y − z = −2

Elimination (cont’d)

Adding 3 times the first equation to the third equation gives

x − 2 y + 3z = 4 x − 2 y + 3z = 4 5 y − 10 z = −5 → 5 y − 10 z = −5 −3 x + 4 y − z = −2 −2 y + 8 z = 10

Elimination (cont’d)

Multiplying the second equation by ⅕ and the third equation by – ½ gives

x − 2 y + 3z = 4 x − 2 y + 3z = 4 y − 2 z = −1 5 y − 10 z = −5 → −2 y + 8 z = 10 y − 4 z = −5

Elimination (cont’d)

Adding – 1 times the second equation to the third equation gives

x − 2 y + 3z = 4 x − 2 y + 3z = 4 y − 2 z = −1 → y − 2 z = −1 y − 4 z = − 5 − 2 z = −4

Elimination (cont’d)

Finally, multiplying the third equation by – ½ gives

x − 2 y + 3z = 4 x − 2 y + 3z = 4 y − 2 z = −1 → y − 2 z = −1 − 2 z = − 4 z=2

Elimination (cont’d)

This last system is easy to solve using back substitution:

The last equation gives z = 2 .

Substituting this value into the second equation gives y = 3 .

Substituting both these values into the first equation gives x = 4 .

So there is one solution, namely (4, 3, 2) .

Nature of Solutions

As with systems of two equations in two unknowns, systems of linear equations in three (or more) variables have either…

a unique solution

an infinite number of solutions

…so the system is consistent; …so the system is dependent; or

no solution

…so the system is inconsistent.

Matrices

Idea: In the method of elimination, the actual symbols for the variables don’t matter…

…only the coefficients do!

x − 2 y + 3z = 4 2x + y − 4z = 3 −3 x + 4 y − z = −2

3 4 1 −2 So instead of the system 2 1 −4 3 shown above, we can 4 −1 −2 study the matrix shown −3 below:

Terminology ď Ž

The matrix shown above is called the augmented matrix of the system.

ď Ž

If we delete the last column, we get the coefficient matrix of the system:

Definition of Matrix

We use double subscript notation to locate the numbers in a matrix.

For example, the number that appears in the 3rd row and 2nd column of a matrix will be denoted by a32 .

We can use this notation to give a definition of matrices in general:

Definition (cont’d)

We say the matrix is of size m × n . If m = n , then we say the matrix is square with main diagonal elements a11 , a22 , … , ann .

Examples ď Ž

Here are some matrices together with their sizes:

Matrix Row Transformations ď Ž

Idea: Since a system can be represented by its augmented matrix, we can apply operations (called row transformations) to the matrix rather than to the system itself.

ď Ž

Theorem on Matrix Row Transformations:

Symbolism

We use the arrow → to indicate “replaces”. For example, the symbol kR i → R i says that the constant multiple kRi replaces Ri :

Example ď Ž

We now rework the preceding example using matrices.

ď Ž

Compare the two solutions to see that we are doing the same thing in both cases:

Example (contâ€™d)

Example (contâ€™d)

Echelon Form ď Ž

The final matrix in the preceding solution is in echelon form.

ď Ž

In general, a matrix is in this form if it satisfies the following conditions:

Examples ď Ž

Here are some matrices in echelon form. ď Ž

The symbols aij represent real numbers.

Guidelines for Finding the Echelon Form of a Matrix

Using Echelon Form

There are three steps in the use of echelon form in solving a system of equations:

Use elementary row operations to transform the matrix of the system to echelon form.

Use the echelon form to produce a system of equations equivalent to the original system.

Find the solution of the given system by back substitution.

Example

We apply the preceding steps in the solution of the system

−2 x + 3 y + 4 z = −1 x − 2 z + 2w = 1 , y + z − w = 0 3 x + y − 2 z − w = 3 which has four equations in four unknowns.

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Reduced Echelon Form ď Ž

After obtaining an echelon form, it is often convenient to apply additional row operations so that 0 also appears above the first 1 in each row.

ď Ž

The resulting matrix is said to be in reduced echelon form.

Examples ď Ž

Here are some examples of reduced echelon form, with the zeros highlighted in blue:

Example ď Ž

Beginning with the echelon form obtained in the preceding example, we can apply additional row operations as follows:

Example (cont’d)

The system of equations corresponding to the reduced echelon form now gives the solution x=3,y=–1,z=2,w=1 immediately, without using back substitution.

Example

In the system

2 x + 3 y + 4 z = 1 , 3x + 4 y + 5z = 3 the number of equations is greater than the number of variables.

However, the same matrix techniques apply:

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Homogeneous Systems

A system of linear equations is homogeneous if all the constant terms are zero.

The constant terms are those that do not contain variables.

Such a system always has the trivial solution where each variable is zero.

Sometimes such a system also has nontrivial solutions.

Example

We solve the homogeneous system

x − y + 4z = 0 2x + y − z = 0 : − x − y + 2 z = 0

Example (contâ€™d)

Example (contâ€™d)

Example

The homogeneous system

x − y + 4z = 0 2x + y − z = 0 − x − y + 2 z = 0 turns out to have only the trivial solution:

Example (contâ€™d)

Example (contâ€™d)

Example ď Ž

A merchant wants to mix two grades of peanuts costing $3 and $4 per pound, respectively, with cashews costing $8 per pound, to obtain 140 pounds of a mixture costing $6 per pound.

ď Ž

If the merchant also wants the amount of lower-grade peanuts to be twice that of the higher-grade peanuts, how many pounds of each variety should be mixed?

Example

Section 9.6

The Algebra of Matrices

Goals

Learn arithmetic operations with matrices, including

addition;

product of a real number and a matrix;

matrix multiplication.

Also discuss row and column matrices.

Matrix Equality and Addition

Below we define…

equality of matrices, and

matrix addition.

(The notation (aij) denotes an m × n matrix A .)

Remarks

Note that two matrices are equal if and only if…

they have the same size, and

corresponding elements are equal.

We can use the parentheses notation for matrices to write the definition of addition as (aij) + (bij) = (aij + bij) .

Examples

Further Definitions ď Ž

The m Ă— n zero matrix, denoted by O , is the matrix with m rows and n columns in which every element is 0 :

Definitions (cont’d)

The additive inverse –A of the matrix A = (aij) is the matrix (–aij) obtained by changing the sign of each nonzero element of A .

Here is an example:

Theorem on Matrix Properties

We also define subtraction of two m × n matrices by A – B = A + (–B) .

Subtraction (cont’d)

We can express this using parentheses notation: (aij) – (bij) = (aij) + (–bij) = (aij – bij) .

Here is an example:

Product of a Real Number and a Matrix

We can multiply a real number by a matrix as follows:

Here is an example with c = 3 :

Theorem on Matrix Properties ď Ž

The following theorem gives properties of multiplication of a real number and a matrix:

Product of Two Matrices

We next describe the product AB of two matrices A and B .

The definition may seem unusual, but is actually very useful throughout mathematics and its applications.

To form AB it is not necessary that A and B have the same size; however…

Product (cont’d)

…there is a restriction on the sizes of A and B, namely: The number of columns of A must equal the number of rows of B .

Thus, if A is m × n , then B must be n × p for some p .

We will see that AB must then be m × p .

Product (cont’d)

Let C = AB . The guidelines on the next slide will show how to find the element cij in row i and column j of C .

As above, we assume that

A is of size m × n , and

B is of size n × p .

Guidelines

Definition of AB ď Ž

This leads to the following definition of the product AB , where A and B are as above:

Diagram of Matrix Sizes ď Ž

The following diagram shows the relationships among the sizes of A , B , and AB :

Illustration of Sizes ď Ž

Here are some samples of the size of AB , given the sizes of A and of B . ď Ž

Note the case where AB cannot be defined.

Example

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d) ď Ž

Here is the final result:

Row and Column Matrices

A row matrix has only one row;

Similarly a column matrix has only one column.

Check the entries in these examples:

Is AB the same as BA ?

Sometimes the answer is “no” simply because AB is defined but BA is not.

For example, if A is 2 × 3 and B is 3 × 4 , then AB exists but BA does not.

However, the next example shows that even when AB and BA are both defined, they may still be different matrices:

Example

Two Properties

The preceding example shows that matrix multiplication is not commutative.

However, matrix multiplication is…

Associative: If A is m × n , B is n × p , and C is p × q , then A(BC) = (AB)C ;

Distributive: If A1 and A2 are m × n and if B1 and B2 are n × p , then: A1(B1 + B2) = A1B1 + A1B2 (A1 + A2)B1 = A1B1 + A2B1

Section 9.7

The Inverse of a Matrix

Goals

Introduce the identity matrix;

Study inverses of matrices;

Use matrix inverses to solve systems of equations.

The Identity Matrix

We define the identity matrix of order n to be the square matrix of order n that has…

1 in each position on the main diagonal, and

0 elsewhere.

Examples:

Multiplying by In

We can show that for any square matrix of order n , AIn = A = InA .

For example, we can verify this for n = 2 :

Inverse of a Matrix

Every nonzero number b has a multiplicative inverse b–1 such that b · b–1 = 1 .

A similar situation sometimes occurs with matrices:

Remarks

If a square matrix A has an inverse, then we say that A is invertible.

Note that…

only square matrices can be invertible!

the inverse A–1 of an invertible matrix A is never written as 1/A .

Finding A

–1

If A is invertible, then we can find A–1 using elementary row operations.

If A = (aij) is n × n , then we begin with the n × 2n matrix formed by adjoining In to A :

Finding A

–1

(cont’d)

Next we apply a succession of elementary row operations (as we did earlier to find reduced echelon forms) until we arrive at a matrix of the form

Finding A

–1

(cont’d)

It can be shown that if A is invertible, then the n × n matrix (bij) is the inverse of A .

As an example we use this method to find A–1 if

3 5 A= : 1 4

Example (contâ€™d)

Example (contâ€™d)

Example

Next we apply the method to the 3 × 3 matrix

1 −1 3 A = 2 5 0 : 3 1 −2

Example (contâ€™d)

Example (contâ€™d)

Solving Systems

We can use inverses of matrices to solve systems of linear equations.

For example, the system

a11x + a12 y = k1 a21x + a22 y = k2 can be expressed in terms of matrices as

a11x + a12 y k1 a x + a y = k . 21 22 2

Solving (cont’d)

If we let

a11 A= a21

a12 k1 x , X = , and B = , a22 y k2

then a matrix form for the system is AX = B .

If A–1 exists, then multiplying both sides of the preceding equation gives

Solving (cont’d) A–1AX = A–1B .

Since A–1A = I2 and I2X = X , this gives X = A–1B .

This inverse method can also be applied to systems of n linear equations in n unknowns when n is greater than 2 .

Example

Example (contâ€™d)

Section 9.8

Determinants

Goals

Introduce the determinant of a square matrix, including

the method of minors and cofactors;

the special cases of 2 × 2 and 3 × 3 matrices.

Study the connection between determinants of square matrices and their invertibility.

Introduction

We associate with each square matrix A a number called the determinant of A , denoted by either

|A| (that is, using the absolute value symbol), or

det A .

On the next slide we begin explaining how the number det A is calculated:

Special Cases of |A|

If A is of order 1 , then A has only one element a11 .

In this case, we define |A| = a11 .

If A is of order 2 , then

a11 A= a21

a12 , a22

and we define |A| = a11a22 – a21a12 .

Special Cases (cont’d)

We also denote this as follows:

a11 A= a21

a12 = a11a22 − a21a12 . a22

Here is an example:

Minors and Cofactors ď Ž

To find determinants of matrices of higher order it is helpful introduce the following terms:

Illustrations

Example

Example (contâ€™d)

Matrices of Order 3

We can use minors and cofactors to define |A| when A is a square matrix of order 3 :

Actually here we are expanding |A| by the first row, however…

Order 3 (cont’d)

…we can also expand about other rows or columns.

For example, here we expand about the second column (which gives the same result):

Example

Example (contâ€™d)

Matrices of Order n ď Ž

In a similar way we can define |A| when A is a square matrix of arbitrary order n :

Order n (cont’d)

As the following theorem says, we can find |A| using any row or column:

This is especially helpful if A has a row or column with many zeros:

Example

Example (contâ€™d)

Invertibility of A ď Ž

The preceding theorem leads immediately to this Theorem on a Row of Zeros:

ď Ž

This gives us a very useful criterion for whether a square matrix A is invertible:

Section 9.9

Properties of Determinants

Goals

Learn to simplify finding |A| by applying row and column transformations to A .

Discuss further properties of determinants;

Study Cramer’s rule for solving systems of equations.

More on Determinants ď Ž

It is very difficult to find |A| by minors and cofactors when A is of high order. ď Ž

ď Ž

For example, finding the determinant of a matrix of order 10 by the method of minors and cofactors requires the evaluation of 1,814,400 determinants of order 2 !

However, we can apply transformations to A that do not change its determinant but make it easier to find.

Row and Column Transformations ď Ž

The following theorem gives the effects of various transformations upon |A| :

Illustrations

Illustrations (contâ€™d)

ď Ž

We also have the following Theorem on Identical Rows:

Example

Example (contâ€™d)

Example

Example

Determinants and Systems

We want to apply determinants to systems of linear equations, for example

a11x + a12 y = k1 . a21x + a22 y = k2

We assume that at least one nonzero coefficient appears in each equation…

…and that a11 ≠ 0 ; else a12 ≠ 0 , and we could view y as the first variable rather than x .

Systems (contâ€™d) ď Ž

We can use elementary row operations to obtain the matrix of an equivalent system with a21 = 0 :

Systems (cont’d)

So the given system is equivalent to

a11x + a12 y = k1 , ( a11a22 − a21a12 ) y = a11k2 − a21k1 which may also be written

a11 x + a12 y = k1 a11 a11 a12 y= a a21 21 a22

k1 . k2

Systems (cont’d)

a11 If a21

a12 ≠ 0 , then we can solve the a22

second equation for y , obtaining

a11 k1 a21 k2 y= . a11 a12 a21 a22

Systems (cont’d)

Then the first equation gives the corresponding value for x :

k1 a12 k2 a22 x= . a11 a12 a21 a22

The preceding two formulas are called Cramer’s rule for two variables.

Cramer’s Rule

There is an easy way to remember Cramer’s rule. Let

a11 D= a21

a12 a22

be the coefficient matrix of the system.

Also let from D

Dx denote the matrix obtained by replacing the coefficients a11 ,

a21 of x by k1 , k2 , respectively.

Cramer’s Rule (cont’d)

Similarly we let Dy denote the matrix obtained from D by replacing the coefficients a12 , a22 of y by k1 , k2 , respectively.

Thus

k1 Dx = k2

a12 a11 , Dy = a22 a21

k1 . k2

Cramer’s Rule (cont’d)

If |D| ≠ 0 , then the solution (x, y) is given by the formulas

Dx x= , D

Here is an example:

y=

Dy D

.

Example

More generally…

Cramer’s rule can be extended to systems of n linear equations in n variables x1 , x2 , … , xn , where the ith equation has the form ai1x1 + ai2x2 + ··· + ainxn = ki .

Again we let D denote the coefficient matrix, and we let Dxj denote the matrix obtained by replacing the coefficients of xj in D by the numbers k1 , k2 , … , kn .

Cramer’s Rule (General Form)

Cramer’s rule says that if |D| ≠ 0 , then

Dx 1 Dx 2 Dxn x1 = , x2 = , , xn = . D D D

Example

Example (contâ€™d)

Section 9.10

Partial Fractions

Goals

Use systems of equations to decompose rational expressions into partial fractions.

Give examples illustrating the various cases that can occur.

Introduction

We want to write rational expressions as sums of simpler ones.

For example, it is easy to check that

2 1 −1 = + , all x ≠ ±1 . 2 x −1 x−1 x +1

We say that we have decomposed the lefthand side into partial fractions.

In General…

If f(x) and g(x) are polynomials with the degree of f less than the degree of g , then in theory we can always write

f ( x) = F1 + F2 + + Fr , g ( x) where each of the Fk has one of the forms

A m or ( px + q )

( ax

Ax + B

2

+ bx + c )

n

.

In General (cont’d)

Here…

A and B are real numbers,

m and n are nonnegative integers, and

the quadratic polynomial ax2 + bx + c is irreducible over …

which means that ax2 + bx + c has no real zero.

Each Fk is called a partial fraction.

On the Degrees of f and g

It is crucial that the degree of f be less than the degree of g , as noted above.

If this is not the case, then we can first apply long division; for example, 3

2

x − 6 x + 5x − 3 6x − 9 = x−6+ 2 . 2 x −1 x −1

6x − 9 . Then we find the decomposition of 2 x −1

Guidelines for Finding Partial Fraction Decompositions

Guidelines (contâ€™d)

ď Ž

We now apply these rules to a series of examples that illustrate the various cases:

Linear Denominators

We use the guidelines to find the partial fraction decomposition of 2

4 x + 13 x − 9 : 3 2 x + 2 x − 3x 1.

The degree of the numerator, 2, is less than 3 , the degree of the denominator; thus long division is not required.

2.

Factoring the denominator gives

x 3 + 2 x 2 − 3 x = x ( x 2 + 2 x − 3 ) = x ( x + 3 ) ( x − 1) .

Example (cont’d) 3.

Each factor of the denominator has the form stated in Rule A with m = 1 . Thus the partial fraction decomposition has the form

4 x 2 + 13 x − 9 A B C = + + . 3 2 x + 2 x − 3x x x + 3 x − 1 4.

To find A , B , and C , we multiply both sides of the above by the least common denominatorx ( x + 3 ) ( x − 1) , giving

Example (cont’d) 4 x 2 + 13x − 9 = A ( x + 3 ) ( x − 1) + Bx ( x − 1) + Cx ( x + 3 ) = A ( x2 + 2 x − 3 ) + B ( x 2 − x ) + C ( x 2 + 3x ) = ( A + B + C ) x 2 + ( 2 A − B + 3C ) x − 3 A . Equating the coefficients of like powers of x on each side of the preceding equation gives

A+ B+C = 4 2 A − B + 3C = 13 . −3 A = − 9

Example (cont’d) Solving this system gives A = 3 , B = –1 , and C = 2 . Hence, the partial fraction decomposition is

4 x 2 + 13 x − 9 3 −1 2 = + + . 3 2 x + 2 x − 3x x x + 3 x − 1

There is an alternative way to find A , B , and C if all factors of the denominator are linear and nonrepeated (as they were here):

Example (cont’d)

Instead of equating coefficients and using a system of equations, we begin with the equation

4 x + 13 x − 9 = A ( x + 3 ) ( x − 1) + Bx ( x − 1) + Cx ( x + 3 ) . 2

We next substitute values for x that make the factors x , x – 1 , and x – 3 , equal to zero:

Letting x = 0 gives –9 = –3A , or A = 3 ;

Letting x = 1 gives 8 = 4C , or C = 2 ;

Letting x = –3 gives –12 = 12B , or B = –1 .

Repeated Linear Factor

Next we find the partial fraction decomposition of 2

x + 10 x − 36 : 2 x ( x − 3)

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Irreducible Quadratic Factor

Now we find the partial fraction decomposition of

4 x 3 − x 2 + 15 x − 29 : 3 2 2x − x + 8x − 4

Example (contâ€™d)

Example (contâ€™d)

Example (contâ€™d)

Repeated Quadratic Factor

Finally we find the partial fraction decomposition of

5x3 − 3x 2 + 7 x − 3

(x

2

+ 1)

2

:

Example (contâ€™d)

Example (contâ€™d)