Algebra

TABLE OF CONTENTS 

Module One - Algebraic Expressionso The Real Numbers o Exponents o Radicals o Polynomials o Factoring o Complex Numbers

Module Two - Equations and Inequalitieso Linear Equations o Applications of Linear Equations o Quadratic Equations o Applications of Quadratic Equations o Variation o Inequalities o Absolute Value Equations and Inequalities

Module Three – Relations and Functionso Relations o Functions o Linear Functions o Equations of a Line o Graphing Relations and Functionso General Graphing Techniqueso Algebra of Functions

Module Four – Polynomial and Rational Functionso Quadratic Functions o Synthetic Division o Zeros of Polynomial Functions o Graphs of Polynomial Functionso Rational Functions

Module Five – Exponential and Logarithmic Functionso Exponential Functions o Logarithmic Functions o Evaluating Logarithms; Change of Base o Exponential and Logarithmic Equations o Exponential Growth or Decay

Module Six – Systems of Equations and Inequalitieso Linear Systems of Equations o Nonlinear Systems of Equations o Matrix Solution of Linear Equations o Determinants o Cramer’s Rule o Systems of Inequalities; Linear Programming

The Real NumbersSets of Numbers

Natural Numbers {1, 2, 3, 4, . . .}

Whole Numbers {0, 1, 2, 3, 4, . . .}

Integers {. . . , -3, -2, -1, 0, 1, 2, 3, . . .}

Rational Numbers { | p and q are integers and q  0 }

The set of rational numbers contains all numbers that can be written as fractions, or quotients of integers. Integers are also rational numbers since they can be represented as fractions. All decimals that repeat or terminate belong to the set of rational numbers. The following are all rational numbers:

, - , 1 , -5 = , 0 = , 0.125 = , 0.6666 . . . = 

Irrational Numbers {x | x is real but not rational }

The irrational numbers are nonrepeating, nonterminating decimals. They cannot be represented as the quotient of two integers. The following are all irrational numbers:

 ,  , - 

Real Numbers {x | x corresponds to a point on the number line }

The set of real numbers consists of all the rational numbers together with all the irrational numbers.

Example

Given set A = { , - , 0, 2.9, -5, 4, - ,  , -7 ,  }, list all the elements of A that belong to the set of : a) natural numbers, b) whole numbers, c) integers, d) rational numbers, e) irrational numbers, and f) real numbers.

a) 4

b) 0, 4

c) 0, -5, 4

d) , 0, 2.9, -5, 4, - , -7

e) - ,  , 

f) all elements of A are real numbers

Order of Operations

1. Perform operations in grouping symbols (parentheses, brackets, braces, or fraction bars). Start with the innermost and work outward.

2. Calculate powers and roots, working from left to right.

3. Perform multiplication and division in order from left to right.

4. Perform addition and subtraction in order from left to right.

Example

Use order of operations to evaluate:

a) 6(-5) – (-3)(2)

b) 

c) -9 – {6 – 2[12 – (8 – 15)] – 4}

Solution:

a) 6(-5) – (-3)(2)  = 6(-5) – (-3)(16) No grouping symbols; power calculated first

= -30 – (-48) Multiplication performed

= -30 + 48 Subtraction changed to addition

= 18 Addition performed

b) Begin by simplifying the numerator and denominator of fraction.

 =   Calculate powers first

=   Perform multiplications

=   Perform additions and subtractions

=   Simplify

c) -9 – {6 – 2[12 – (8 – 15)] – 4} = -9 – {6 – 2[12 – (-7)] – 4} Start with innermost grouping symbol, parentheses, and subtract

= -9 – {6 – 2[19] – 4} Working outward, perform subtraction in brackets

= -9 – {6 – 38 – 4} Within braces, multiply

= -9 – {-36} Within braces, subtract

= -9 + 36 Change subtraction to addition

= 27 Add

Properties of the Real Numbers

For all real numbers a, b, and c:

1. Commutative Property for Addition: a + b = b + a

2. Commutative Property for Multiplication: ab = ba

The commutative properties state that two numbers may be added or multiplied in any order.

3. Associative Property for Addition: a + (b + c) = (a + b) + c

4. Association Property for Multiplication: a(bc) = (ab)c

For the associative properties, the order of the terms or factors remains the same; only the grouping is changed.

5. Identity Property for Addition: There is a unique real number, 0, such that a + 0 = a and 0 + a = a

The identity property for addition tells us that adding 0 to any number will not change the number.

6. Identity Property for Multiplication: There is a unique real number, 1, such that a·1 = a and 1·a = a

The identity property for multiplication tells us that multiplying any number by 1 will not change the number.

7. Inverse Property for Addition: Each nonzero real number a has a unique additive inverse, represented by –a, such that

a + (-a) = 0 and –a + a = 0

Additive inverses are called opposites.

8. Inverse Property for Multiplication: Each nonzero real number a has unique multiplicative inverse, represented

by  , such that  and 

Multiplicative inverses are called reciprocals.

9. Distributive Property: a(b + c) = ab + ac

Example

Identify the property illustrated in each statement:

a) (x + 7) + 8 = x + (7 + 8)

b) 4x + 0 = 4x

c) 10 · ( x) = (10 · )x

d) (x + 1) ·  = 1

e) 4(x + 5) = 4x + 20

f) 3 · (5 · a) = 3 · (a · 5)

g) -6x + 6x = 0

h) (2 + y) + 5 = 5 + (2 + y)

i) (y + 5)(y – 3) = (y – 3)(y + 5)

j) 5 · 1 = 5

Solution:

a) Associative Property for Addition. Order of terms remains the same. Only the grouping changes.

b) Identity Property for Addition. Adding zero to something does not change it.

c) Associative Property for Multiplication. Order of factors is the same. Only the grouping changes.

d) Inverse Property for Multiplication. The product of reciprocals is 1.

e) Distributive Property.

f) Commutative Property for Multiplication. Order of the factors is changed.

g) Inverse Property for Addition. The sum of opposites is 0.

h) Commutative Property for Addition. The order of the terms is changed.

i) Commutative Property for Multiplication. The order of the factors is changed.

j) Identity Property for Multiplication. Multiplying a number by 1 does not change it.

Exponents

Definition of a : If n is any positive integer and a is any real number, a = a · a · a · a · · · · a

where the factor a occurs n times.

Definition of a : For any nonzero real number a, a  = 1.

Definition of a : If a is a nonzero real number and n is any integer, then a = 

Definition of a : If n is an even positive integer, and if a>0, then a  is the positive real

number whose nth power is a. That is, (a )  = a.

If n is an odd positive integer, and if a is any real number, then a  is

the positive or negative real number whose nth power is a. That is,

(a )  = a.

Definition of a : For all integers m, all positive integers n, and all real numbers a for

which a is a real number, a = (a )

 Example

Evaluate: a) 3  b) -2  c) (-5)   d) (7x)  e) -9  f) (-5)  g) 2  h) 81

i) (-32)

Solution:

a) 3 = 3·3·3·3 = 81

b) -2  = -2·2·2·2·2·2 = -64

c) (-5)  = (-5)(-5)(-5) = -125

d) (7x)  = 1

e) -9  = -1

f) (-5)  =  =  = 

g) 2  =  =  = 

h) 81  = (81 )  = 3  = 27  since 

i) (-32)  =  =  =  =  (-32)  = -2 since (-2) = -32

Rules of Exponents

For all rational numbers r and s, and for all positive numbers a and b:

Example

Use rules of exponents to simplify each expression. Write answers without negative exponents. Assume that all variables represent nonzero real numbers.

a) 

b) 

c) 

d) 

e) 

f) 

Solution:

a) 

b) 

c) 

d) 

e) 

f) 

Radicals

Radical Notation

If a is a real number, n is a positive integer, and   is a real number, then 

If a is a real number, n is a positive integer, and   is a real number, then 

Example

Evaluate each root: a)   b)   c) -  d)   e) -

f) 

Solution:

a)   = 3 since 3 = 81

b)   = -2 since (-2) = -8

c) -  since 

d)   is not a real number. There is no real number that when raised to the 4th

power is equal to a negative number.

e)   since 2 = 64

f)   since 

Rules for Radicals

For all real numbers a and b, and positive integers m and n for which the indicated roots are real numbers,

A radical expression is simplified when the following conditions are satisfied.

1. All possible factors have been removed from the radicand.

2. There is no fraction in the radicand.

3. There are no radicals in the denominator.

4. The index of the radical is reduced.

Rationalizing the denominator

1. If the denominator is a monomial, multiply both the numerator and denominator by the same radical so that the resulting denominator is rational (contains no radical).

For example, 

2. If the denominator is a binomial, multiply both the numerator and denominator by the conjugate of the denominator. For example:

Example

Simplify each radical expression. Assume that all variables represent positive real numbers.

a)   b)   c)   d)   e) 

f)   g)   h)   i) 

Solution:

a) 

b) 

c) 

d) 

e) 

f) 

g)   

h) 

i) 

Example

Rationalize the denominator of the radical expression. Assume that all variables represent nonnegative numbers and that no denominators are zero.

a)   b) 

Solution:

a) 

b) 

Polynomials

A term is an algebraic expression that is either a constant or a product of a constant and one or more variables raised to whole-number powers.

Examples of terms include: 9x , 2xy , 5

A polynomial is a finite sum of one or more terms. Examples of polynomials are:

The degree of a term is the sum of the exponents of its variables. The degree of a nonzero constant is zero. For

example, 2x  has degree 6;   has degree 7;   has degree 6 (s=s )

The degree of a polynomial is the highest degree of any of its terms. For example,

 has degree 5;   has degree 14 (7+5+2=14)

A trinomial is a polynomial containing exactly three terms.

A binomial is a polynomial containing exactly two terms.

A monomial is a polynomial containing exactly one term.

Polynomials are added by adding coefficients of like terms.

Polynomials are subtracted by subtracting coefficients of like terms.

Example.

a) Add: 

b) Subtract: 

Solution:

a)   

b)   

To multiply polynomials, multiply each term of one polynomial by each term of the other polynomial and then combine like terms.

Special Products

(x + y)(x – y) = x  - y

(x + y)  = x  + 2xy + y

(x – y)  = x  - 2xy + y

The FOIL Method

When multiplying two binomials, the letters of FOIL can help you remember which terms are multiplied to complete the product.

F – firsts Multiply the first terms of the binomials

O – outers Multiply the outer terms of the binomials

I – inners Multiply the inner terms of the binomials

L – lasts Multiply the last terms of the binomials

For example, F O I L

(2x – 1)(3x + 5) = (2x)(3x) + (2x)(5) + (-1)(3x) + (-1)(5)

= 6x  + 10x – 3x – 5

= 6x + 7x – 5

Example.

Multiply: a) (2x – 3)(x  + 3x – 5)

b) (3x + 5)(3x – 5)

c) (5x – 6)(4x + 3)

d) (x + 3)

e) (7x – 4)

Solution:

a) (2x – 3)(x + 3x – 5) = 2x(x + 3x – 5) – 3(x  + 3x – 5)

= 2x(x ) + 2x(3x) + 2x(-5) – 3(x ) – 3(3x) –3(-5)

= 2x  + 6x  - 10x – 3x  - 9x + 15

= 2x  + 3x  - 19x + 15

b) (3x + 5)(3x – 5) = (3x)  - 5  Special Product: (x+y)(x-y) = x  - y

= 9x  - 25

F O I L

c) (5x – 6)(4x + 3) = (5x)(4x) + (5x)(3) + (-6)(4x) + (-6)(3)

= 20x  + 15x – 24x - 18

= 20x  - 9x – 18

d) (x + 3)  = x  + 2(x)(3) + 3  Special Product: (x+y) = x +2xy+y

= x  + 6x + 9

e) (7x – 4)  = (7x)  - 2(7x)(4) + 4  Special Product: (x-y)  = x - 2xy + y

= 49x  - 56x + 16

Factoring

Factoring is the process of writing a polynomial as the product of two or more polynomials.

General Strategy for Factoring

o Factor out the greatest common factoro If the polynomial has two terms, determine whether it matches the pattern of one of the special

products:

o Difference of two squares: a  - b  = (a – b)(a + b)

o Difference of two cubes: a  - b  = (a – b)(a  + ab + b )

o Sum of two cubes: a  + b  = (a + b)(a  - ab + b )o If the polynomial has three terms, determine whether it is a perfect square trinomial:

o a  + 2ab + b  = (a + b)

o a  - 2ab + b  = (a - b)

o If the trinomial is not a perfect square, factor as the product of two binomials using the guess-and-check method or the grouping method.

o If the polynomial has more than three terms, try factoring by grouping.

Example. Factor completely: a) 

b) 

c) 

d) 2y + 10 – 3xy – 15x

e) m  - 10m + 16

f) 64 – 27a

g) 12x  + 8x – 15

Solution.

a)   = 3ac( ) Factor out greatest

common factor

= 3ac(2a + 3c)  Perfect square trinomial

b)   = 5rs( ) Factor out greatest common factor

= 5rs(r – 3s)(r + 3s) Difference of two squares

c)   = (3x – 5y)  Perfect square trinomial

d) 2y + 10 – 3xy – 15x = 2(y + 5) – 3x(y + 5) Factor by grouping

= (y + 5)(2 – 3x) Factor out common factor, y + 5

e) m  - 10m + 16 = (m – 2)(m – 8) Trinomial: guess and check

f) 64 – 27a  = ( ) Difference of two cubes

= (4 – 3a)( ) 4 and 3a substituted into formula

= (4 – 3a)(16 + 12a + 9a ) Simplify

g) This example will be done using the grouping method for trinomials. To find

the product number, multiply 12(-15) = -180; the sum number is 8. Two

numbers whose product is –180 and whose sum is 8 are 18 and –10.

12x  + 8x – 15 = 12x  + 18x – 10x – 15 The 8x term is replaced with

the sum, 18x – 10x, using the numbers found above as coefficients

= 6x(2x + 3) – 5(2x + 3) Factor by grouping

= (2x + 3)(6x – 5) Common factor

Complex Numbers 

The imaginary unit, i, is defined by i  = -1.

A complex number is a number that can be written in the form a + bi , where a and b are real numbers, and i is the imaginary unit. a is called the real part of a + bi and b is called the imaginary part.

Powers of i

i  = i

i  = -1

i  = i · i = -1 · i = -i

i  = (i ) = (-1)  = 1

i  = i · i = 1 · i = i

i  = i · i  = 1 · -1 = -1

i  = i · i  = 1 · -i = -i

i  = = (i )  = 1  = 1

The powers if i continue to cycle through the numbers i, -1, -i, l. If k is a multiple of 4, i  = 1.

Example. Evaluate: a) i  b) i

Solution. a) i  = i  · i  = 1 · -i = -i

b) i  = i · i = 1 · i = i

Definition of Addition, Subtraction, and Multiplication

For complex numbers a + bi and c + di:

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) – (c + di) = (a – c) + (b – d)i

(a + bi)(c + di) = (ac – bd) + (ad + bc)i

Example. a) (3 – 2i) + (-4 + 7i)

b) (7 + 3i) – (8 – 7i)

c) (1 + i)(-3 + 5i)

Solution. a) (3 – 2i) + (-4 + 7i) = (3 + (-4)) + (-2 + 7)i

= -1 + 5i

b) (7 + 3i) – (8 – 7i) = (7 – 8) + (3 – (-7))i

= -1 + 10i

c) Instead of using the definition, just use FOIL as you would in

multiplying two binomials

(1 + i)(-3 + 5i) = 1(-3) + 1(5i) + i(-3) + i(5i) Multiply using FOIL

= -3 + 5i – 3i + 5i

= -3 + 5i – 3i + 5(-1) Substitute –1 for i

= -3 + 5i – 3i – 5

= -8 + 2i Combine like terms

The conjugate of the complex number a + bi is a – bi.

For any two complex conjugates a + bi and a – bi: (a + bi)(a – bi) = a  + b

To divide two complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

Example. Divide: 

Solution:  ·   =   Multiply, using FOIL in the numerator;

use conjugate product rule in denominator

=   Substitute –1 for i

= 

= 

= 

Linear Equations

General Steps in Solving a Linear Equation

1. Simplify each side of the equation by removing grouping symbols and combining like terms.

2. Isolate variable terms on one side of the equation and constant terms on the other side. To accomplish this, you may add (or subtract) the same real number or variable expression on both sides of the equation.

3. Isolate the variable. To accomplish this, you may multiply (or divide) on both sides of the equation by the same nonzero quantity.

4. Check.

Example. Solve: a) 4(2x + 1) – 29 = 3(2x – 5)

b) 5(x – 9) = 5x – 45

c) 3x + 6 – (3x – 2) = 5 + 4x + 4(3 – x)

d) 

Solution: a) 4(2x + 1) – 29 = 3(2x – 5)

8x + 4 – 29 = 6x – 15 Remove parentheses, using distributive property

8x – 25 = 6x – 15 Combine like terms

8x – 25 – 6x = 6x – 15 – 6x Subtract 6x from both sides; this will

isolate variable terms

2x – 25 = -15 Combine like terms

2x – 25 + 25 = -15 + 25 Add 25 to both sides; this will isolate

constant terms

2x = 10

 Divide both sides by 2 to isolate the

variable

x = 5

Check: 4(2x + 1) – 29 = 3(2x – 5) Original Equation

4(2·5 + 1) – 29 = 3(2·5 – 5) The solution, 5, is substituted for x

4(10 + 1) – 29 = 3(10 – 5) Each side of the equation is simplified

4(11) – 29 = 3(5)

44 – 29 = 15

15 = 15 This true statement indicates that 5 is the correct solution.

b) 5(x – 9) = 5x – 45

5x – 45 = 5x – 45

Since the left side of the equation is identical to the right side of the equation, the given equation is true for every value of x. This equation is an identity. The solution set consists of all real numbers.

c) 3x + 6 – (3x – 2) = 5 + 4x + 4(3 – x)

3x + 6 – 3x + 2 = 5 + 4x + 12 – 4x

8 = 17

This is a false statement. So this equation is a contradiction. There is no solution.

d)   When the equation contains fractions, multiply

both sides by the LCD, 12m

6m  - 12 = m(6m + 5)

6m  - 12 = 6m  + 5m

6m  - 12 - 6m  = 6m  + 5m - 6m

-12 = 5m

Applications of Linear Equations 

Solving a formula for a specified variable

A formula is solved for a variable when that variable is totally isolated on one side of the equation.

Example. Solve for C: V = C - 

Solution:

V = C - 

L · V = L (C -  )

Multiply both sides by LCD, L

LV = LC – L · 

Remove parentheses

LV = LC – (C – S)N Simplify

LV = LC –(CN – SN) Distributive Property

LV = LC – CN + SN Distributive Property

LV – SN = LC – CN + SN – SN Subtract SN from both sides to isolate terms containing C on one side

LV – SN = LC – CN Simplify

LV – SN = C(L – N) Factor out C to isolate it

Divide both sides by L – N

Simplify

Strategy for Solving Verbal Problems

1. Read the problem carefully.

2. Use diagrams or charts if you think it will make the information clearer.

3. Find the relationship or formula relevant to the problem.

4. Identify the unknown quantity (or quantities), and label them, using one variable.

5. Write an equation involving the unknown quantity, using the relationship or formula from step 3.

6. Solve the equation.

7. Answer the question.

8. Check the answer in the original words of the problem.

Example. A total of $20,000 was invested in two bond mutual funds, a junk bond fund and a government bond fund. The junk bond fund is risky and yields 11% interest. The safer government bond fund yields only 5%. The total income for the year from the two investments was $1300. How much was invested in each fund?

Solution. Let x = the amount of money invested in the junk bond fund

20,000 – x = the amount of money invested in the government bond fund

To calculate interest, we use the simple interest formula: interest = principle · rate · time

Since the time is one year, in this problem, interest = principle · rate

Amount invested

(Principal)

Interest

Rate

Interest Earned

Junk Bonds x 0.11 0.11x

Government Bonds 20,000 – x 0.05 0.05(20,000 – x)

Total 20,000 1300

The total interest earned is the sum of the interest earned from each fund. We obtain the following equation from the "Interest Earned" column of the chart:

0.11x + 0.05(20,000 – x) = 1300

0.11x + 1000 – 0.05x = 1300 Distributive Property

0.06x + 1000 = 1300 Combine like terms

0.06x = 300 Subtract 1000 from each side

x = 5000

Divide both sides by 0.06

The amount invested in the junk bond fund, x, is $5000.

The amount invested in the government bond fund, 20000 – x, is $15,000.

Example. Two travelers left a restaurant in Oklahoma City and traveled in opposite directions on Interstate 40. If one driver averaged 65 mph and the other averaged 60 mph, how long was it before they were 400 miles apart?

Solution. This problem involves distance, rate, and time; these variables are related by the distance formula: distance = rate · time

Let t = the time the two travelers drive until they are 400 miles apart

Rate Time Distance

One driver 65 t 65t

Other driver 60 t 60t

Total 400

The sum of the distances of each driver is the total, 400. From the "Distance" column in the chart, we obtain the equation:

65t + 60t = 400

125t = 400 Combine like terms

Divide both sides by 125

t = 3.2 Simplify

The travelers will be 400 miles apart in 3.2 hours.

Example. Braums Dairy mixed two grades of milk, one containing 3% butterfat and the other containing 4.5% butterfat, to obtain 150 gallons of milk that contained 4% butterfat. How many gallons of each were used in the mixture?

Solution. Let x = number of gallons of the 3% butterfat milk

150 – x = number of gallons of the 4.5% butterfat milk

We multiply the number of gallons of solution by the percent of butterfat to obtain the number of gallons of butterfat in each solution.

Gallons of

Solution

Percent of

Butterfat

Gallons of

Butterfat

3% Milk x 0.03 0.03x

4.5% Milk 150 – x 0.045 0.045(150 – x)

Mixture 150 0.04 0.04(150)

The total amount of butterfat in the mixture is the sum of the amounts of butterfat in each of the two original milks. Thus, from the "Gallons of Butterfat" column in the chart, we obtain the equation:

0.03x + 0.045(150 – x) = 0.04(150)

0.03x + 6.75 – 0.045x = 6 Distributive Property

-0.015x + 6.75 = 6 Combine like terms

-0.015x = -0.75 Subtract 6.75 from both sides

x = 50

Divide both sides by –0.015

The number of gallons of 3% butterfat milk in the mixture, x, is 50 gallons. The number of gallons of 4.5% butterfat milk in the mixture, 150 – x, is 100 gallons.

Example. A winery has a vat to hold Chardonnay. An inlet pipe can fill the vat in 9 hours, while an outlet pipe can empty the vat in 12 hours. How long will it take to fill the vat if both the outlet and the inlet pipes are open?

Solution. In solving a work problem, begin by using the following fact to express each rate of work: If a job can be

done in t units of time, then the rate of work is   job per unit of time.

Let x = number of hours it will take to fill the vat with both pipes open

 = rate of the inlet pipe

-  = rate of the outlet pipe (This rate is negative because the outlet pipe is working

against the inlet pipe rather than with it)

Rate Time Part of Job Done

Inlet Pipe x

Outlet Pipe x

Part of the job done Part of the job done

by one pipe + by other pipe = 1 (Whole job done)

Multiply both sides by the LCD, 36

Distributive Property

4x – 3x = 36 Simplify

x = 36 Combine like terms

It would take 36 hours to fill the vat with both pipes open.

Quadratic Equations

A quadratic equation is an equation that can be written in the standard form

ax + bx + c = 0, where a, b, and c are real numbers and a 0.

Solving quadratic equations by factoring

When a quadratic equation is in standard form and the nonzero side, ax + bx + c , can be factored, the equation can be solved by factoring. To solve by factoring, the following property is used:

Zero-Factor Property: If a and b are complex numbers, with a  b = 0,

then a = 0 or b = 0 or both equal zero.

Example. Solve: 2x  + 7x = 4

Solution.

2x  + 7x = 4

2x  + 7x – 4 = 0To get the equation in standard form, subtract 4 from each side; one side must be 0 when solving by factoring

(2x – 1)(x + 4) = 0 Factor left side of equation

2x – 1 = 0 or x + 4 = 0 Set each factor equal to 0, using the Zero Factor Property

2x = 1 x = -4

x = 

Solve each of the resulting two equations

The solution set is 

Solving Quadratic Equations Using the Square Root Property

A quadratic equation of the form x  = k can be solved using the following property:

Square Root Property: The solution set of x = k is 

Example. Solve: a) 4x  = 20

b) z  = -49

Solution. a)

4x  = 20

Divide both sides by 4 to isolate x

Simplify

x =  Use the square root property to obtain two solutions

The solution set is 

b)

z = -49

z = Use the square root property

z =  7i Simplify the radical

The solution set is 

Solving Quadratic Equations by Completing the Square

To solve a quadratic equation by completing the square, the equation must be written in the form (x + n)  = k. The following steps are used to solve the equation

ax + bx + c = 0 , a  0 , by completing the square:

1. If a  1, divide both sides of the equation by a.

2. Rewrite the equation so that the constant term is isolated on one side of the equation.

3. Take half the coefficient of x and square this result; add this square to both sides of the

equation.

4. Factor the resulting trinomial as a perfect square; combine like terms on the other

side.

5. Use the square root property to complete the solution.

Example. Solve by completing the square: 2x  + 5x – 4 = 0.

Solution.

2x  + 5x – 4 = 0

Divide both sides by 2

Add 2 to both sides to isolate the constant

Take half the coefficient of x and square it: 

Add   to both sides of the equation

Factor the trinomial; add the constants

Use the square root property

Add -  to both sides and simplify the radical

The solution set is 

Solving Quadratic Equations Using the Quadratic Formula

The solutions of the quadratic equation ax + bx + c = 0 , a  0, are

The following steps are used to solve a quadratic equation using the quadratic formula:

1. Write the equation in standard form, ax + bx + c = 0.

2. Determine the values of a, b, and c; a is the coefficient of x , b is the coefficient of x,

and c is the constant.

3. Substitute these values of a, b, and c into the quadratic formula.

4. Simplify.

Example. Solve using the quadratic formula: 3x  = 2x – 4

Solution.

3x  = 2x – 4

3x  - 2x + 4 = 0Write the equation in standard form.

a = 3, b = -2, c = 4 Determine the values of a, b, and c

The quadratic formula

Substitute the values of a, b, and c into the

formula

Simplify

The solution set is 

Applications of Quadratic Equations

General Strategy for Solving Verbal Problems

1. Read the problem carefully.

2. Use diagrams if you think it will make the information clearer.

3. Find the relationship or formula relevant to the problem.

4. Identify the unknown quantity (or quantities), and label them, using one variable.

5. Write an equation involving the unknown quantity, using the relationship or formula

from step 3.

6. Solve the equation.

7. Answer the question.

8. Check the answer in the original words of the problem.

Example. Meg rowed her boat upstream a distance of 9 miles and then rowed back to the starting point. The total time of the trip was 10 hours. If the rate of the current was 4 mph, how fast would the boat have traveled in still water?

Solution. Let x = the rate of the boat in still water

x-4 = the rate of the boat traveling upstream (against the current)

x+4= the rate of the boat traveling downstream (with the current)

Use the distance formula, d = r · t , solved for t: t = 

distance, d rate, r

time, t = 

Upstream 9 x – 4

Downstream 9 x + 4

Total 10

The sum of the time spent traveling upstream and the time traveling downstream is 10.

Multiply both sides by LCD, (x-4)(x+4)

Use distributive property

Simplify

9x + 36 + 9x – 36 = 10x -160

18x = 10x  - 160

0 = 10x  - 18x – 160Subtract 18x from both sides to get 0 on one side

0 = 5x  - 9x – 80Divide both sides by 2

0 = (5x + 16)(x – 5) Factor

5x + 16 = 0 or x – 5 = 0 Use Zero Factor property

5x = -16 x = 5 Solve each equation

x =   or x = 5

Looking back, we see that x represents the rate of the boat in still water. The negative solution makes no sense in this case. Therefore, the only valid solution is 5, which means that the rate of the boat in still water is 5 mph.

Example. Working together, two roommates can paint their apartment in 10 hours. Working alone, one of the roommates can complete the job in 2 hours less time than the other roommate working alone. How long would each person take working alone?

Solution. Let x = time for one roommate to do the job alone

x – 2 = time for the other roommate to do the job alone

 = rate of one roommate

 = rate of the other roommate

Multiply the time it takes them working together, 10 hours, by each rate to get the fractional part of the job done by each person.

Rate x Time = Part of job accomplished 

Rate Time Part of job accomplished

One Roommate 10

Other Roommate 10

The sum of the parts of the job done by each roommate must equal 1:

 +   = 1

x(x – 2)  = 1· x (x – 2)

Multiply both sides by the LCD, x(x – 2)

Use distributive property

10(x – 2) + 10x = x  - 2x

10x – 20 + 10x = x  - 2x

20x – 20 = x  - 2x

Simplify

0 = x  - 22x + 20Get 0 one one side

x = 

Use quadratic formula

x = 

Simplify

x = 

x represents the time it would take one roommate to do the job alone, and x – 2 represents the time it would take the other. If x = 0.95, then x – 2 = -1.05. This negative time does not make sense. So the only valid solution for x is 21.05. Thus, it would take one roommate 21.05 hours to do the job alone, and it would take the other roommate 19.05 hours to do the job alone.

Example. A rectangular pool 20 feet wide and 60 feet long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 square feet, how wide is the walkway?

Let x = the width of the walkway

The length of the outer rectangle is 60 + x + x = 60 + 2x

The width of the outer rectangle is 20 + x + x = 20 + 2x

The area of the outer rectangle is: A = lw = (60 + 2x)(20 + 2x)

The area of the inner rectangle is: A = lw = 60(20) = 1200

Area of the outer rectangle = Area of inner rectangle + Area of Walkway

(60 + 2x)(20 + 2x) = 1200 + 516

1200 + 120x + 40x + 4x  = 1716Simplify

4x  + 160x – 516 = 0Get 0 on one side of the equation

x  + 40x – 129 = 0Divide both sides by 4

(x – 3)(x + 43) = 0 Factor

x – 3 = 0 or x + 43 = 0 Use Zero Factor property

x = 3 x = -43 Solve

x represents the width of the walkway. The negative solution, -43, makes no sense. So, the solution is 3, which means that the width of the walkway is 3 feet.

Pythagorean Theorem

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

Example. Find the length of the sides of the right triangle shown in the figure below:

Pythagorean Theorem

64 + x  - 4x + 4 = x  + 8x + 16Square the binomials

x  - 4x + 68 = x  + 8x + 16Simplify

52 = 12xSubtract x  from both sides; add 4x to both sides; subtract 16 from both sides

 = x

Divide both sides by 12

Looking back at the figure, the unknown sides of the triangle are represented by x – 2 and x + 4. So, the lengths of these sides are:

x – 2 =   and x + 4 = 

Variation

Direct Variation

"y varies directly as x" means that there is a nonzero constant k such that y = kx.

"y is directly proportional to x" means the same as "y varies directly as x"

k is called the constant of proportionality

Inverse Variation

"y varies inversely as x" means that there is a nonzero constant k such that y = 

"y is inversely proportional to x" means the same as "y varies inversely as x"

Joint Variation

"z varies jointly as x and y" means that there is a nonzero constant k such that z = k x y

Solving Variation Problems

1. Using one of the definitions above, write the general relationship among the variables as an equation.

2. Substitute the given values of the variables; solve for k.

3. Substitute this value of k into the equation from Step 1.

4. Solve for the required unknown.

Example:

Given the formula for the area of a circle with radius r:

A =  r . Fill in the blanks with the appropriate response:

The area of a circle varies ____________ as the ___________ of its ______________. The constant of variation is ___________.

Solution: The area of a circle varies directly as the square of its radius, r. The constant of variation is  .

Example: If a varies jointly as c and d and a = 20 when c = 2 and d = 4, find d when a = 25 and c = 8.

Solution:

a = kcd Joint variation formula

20 = k  2  4 Substitute initial values for a, c, and d.

20 = 8k Solve for k.

2.5 = k

a = 2.5 cd Substitute 2.5 for k in joint variation formula.

25 = 2.5  8d Substitute 25 for a and 8 for c.

25 = 20d Solve for d

1.25 = d

Example: If r is inversely proportional to s and r = 5 when s = 4, find s when r = 7

Solution:

r = 

Formula for inverse variation

5 = 

Substitute initial values for r and s

4  5 = 4  

Multiply both sides by 4 to isolate

20 = k

r = 

Substitute 20 for 1c in formula for inverse variation

7 = 

Substitute 7 for r.

s  7 = s  

Solve for s

7s = 20

s = 

Example.

The volume, V, of a gas varies directly as its temperature, T, and inversely as its pressure, P. If 40 m  of a gas

yields a pressure of 20 kg/m  at a temperature of 200 k, what will be the volume of the same gas if the pressure is

decreased 10 kg/m  and the temp. is increased to 250 k?

Solution:

V = 

Variation formula

40 = 

Substitute initial values for V, T and P

40 = 10 k Solve for k.

4 = k

V = 

Substitute 4 for k in variation formula

V = 

Substitute 10 for P and 250 for T.

V = 100 m

Inequalities

Properties of Inequalities

1. Addition and Subtraction Properties

If a < b, then a + c < b + c

If a < b, then a – c < b – c

The same quantity may be added to or subtracted from both sides of an inequality without changing the solution set

2. Positive Multiplication and Division Properties

If a < b and c is positive, then a · c < b · c

If a < b and c is positive, then   < 

Both sides of an inequality may be multiplied or divided by the same positive number without changing the solution set.

3. Negative Multiplication and Division Properties

If a < b and c is negative, then a · c > b · c

If a < b and c is negative, then   > 

Both sides of an inequality may be multiplied or divided by the same negative number without changing the solution set as long as the direction of the inequality symbol is reversed.

REMEMBER: Always reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

Example. Solve: 7(x + 4) – 13 > 12 + 13(3 + x)

Solution:

7(x + 4) – 13 > 12 + 13(3 + x)

7x + 28 – 13 > 12 + 39 + 13x Distributive Property

7x + 15 > 13x + 51 Combine like terms

7x + 15 – 13x > 13x + 51 – 13x Subtract 13x from both sides

-6x + 15 > 51 Combine like terms

-6x + 15 – 15 > 51 – 15 Subtract 15 from both sides

-6x > 36 Combine like terms

 < 

Divide both sides by –6 and reverse the direction of the inequality symbol

x < -6 Simplify

The solution set consists of all real numbers that are greater than –6. Interval notation for the solution set is (- ,-6].

Example. Solve: 1 <   < 9

Solution.

To solve this three-part inequality, apply the properties of inequalities, working with three parts of the inequality at the same time. The goal is to isolate the variable in the middle part.

1 <   < 9

-2 · 1 > -2 ·   > -2 · 9

Multiply all three parts of the inequality by –2; reverse the direction of both inequality symbols

-2 > 4m – 5 > -18 Simplify

-2 + 5 > 4m – 5 + 5 > -18 + 5 Add 5 to all three parts of the inequality

3 > 4m > -13 Simplify

 >   > 

Divide all three parts by 4. Since we are dividing by a positive number, the inequality symbols are not reversed.

 > m > 

Simplify

The solution set consists of all real numbers between   and  . Interval notation for this solution set

is  .

Quadratic Inequalities

A quadratic inequality is an inequality that can be written in the form ax  + bx + c < 0, where a, b, and c are real numbers, and c  0. (The symbol < can be replaced with >, <, or >.)

Procedure for Solving Quadratic Inequalities

1. Write the inequality in standard form: ax  + bx + c < 0. ( The symbol might also be >, <, or >.)

2. Replace the inequality symbol with =, and solve the equation ax  + bx + c = 0.

3. Locate the solutions obtained in step 2 on a number line; the solution(s) divide the number line into test intervals.

4. Choose one number from each test interval, and substitute it into the original inequality. If the substitution produces a true statement, then all real numbers in the test interval are in the solution set. If the substitution produces a false statement, then the real numbers in the test interval are not in the solution set.

Example. Solve: 2x  + x > 15

Solution.

2x  + x > 15

2x  + x - 15 > 0Subtract 15 from both sides to obtain standard form

2x  + x - 15 = 0Replace > with =

(2x – 5)(x + 3) = 0

2x – 5 = 0 or x + 3 = 0

2x = 5 or x = -3

x = 

Solve by factoring

The solutions obtained above divide the number line into three test intervals:

(- , -3) , (-3,  ) , ( ,  ).

We will take one number from each test interval and substitute that number into the original inequality:

(- , -3) Test number: -4 (Any number less than –3 could be chosen)

2x  + x > 15 Original inequality

2(-4) +(-4) > 15 Substitution of –4

2(16) – 4 > 15

32 – 4 > 15

28 > 15 True

Conclusion: (- , -3) belongs to the solution set.

(-3,  ) Test number: 0

2x  + x > 15

2(0) + (0) > 15

0 + 0 > 15

0 > 15 False

Conclusion: (-3,  ) does not belong to the solution set.

( ,  ) Test number: 3

2x  + x > 15

2(3)  + 3 > 15

2(9) + 3 > 15

18 + 3 > 15

21 > 15 True

Conclusion: ( ,  ) belongs to the solution set.

Thus, the solution set is (- , -3)  ( ,  )

Rational Inequalities

A rational inequality involves quotients of algebraic expressions.

Procedure for Solving Rational Inequalities

1. Write the inequality so that one side is 0.

2. If necessary, combine terms on the other side into a single quotient.

3. Find the number(s) which make the numerator and denominator equal to zero.

4. Locate the numbers obtained in step 3 on the number line; these numbers divide the number line into test intervals.

5. Choose one number from each test interval and substitute it into the original inequality. If the substitution produces a true statement, then all real numbers in the test interval are in the solution set. If the substitution produces a false statement, then the real numbers in the test interval are not in the solution set.

Example. Solve: 

Solution.

Subtract 3 from both sides to get 0 on one side

Get a common denominator; subtract fractions to obtain a single quotient

7 – 5x = 0 or 2x – 1 = 0

7 = 5x or 2x = 1

 or x = 

Find numbers which make the numerator and the denominator equal to 0; solve each equation

The solutions obtained above divide the number line into three test intervals:

(- , ) , ( ,  ) , ( ,  )

We will take one number from each test interval and substitute it into the original inequality:

(- , ) Test number: 0

 Original inequality

-4 > 3 False

Conclusion: (- , ) does not belong to the solution set.

( ,  ) Test number: 1

 Original inequality

5 > 3 True

Conclusion: ( ,  ) belongs to the solution set.

( ,  ) Test number: 2

 Original inequality

2 > 3 False

Conclusion: ( ,  ) does not belong to the solution set.

The interval : ( ,  ) is in the solution set. Because the inequality symbol is >, we must include values which

make the expression   equal to 0. This would be the value which makes the numerator, 7 – 5x, equal to 0;

that number is  . Thus, the solution set is the interval ( ,  ].

Relations

Definition of a Relation

A relation is a set of ordered pairs. The set of all first coordinates is the domain of the relation, and the set of all second coordinates is the range of the relation.

For example, given the relation { (3,-2), (7,4), (7,1), (6,3), (9,3), (9,2)}, the domain is

{ 3, 7, 6, 9} and the range is { -2, 4, 1, 3, 2}.

Example. For the relation defined by the equation y =  , find three ordered pairs that belong to the relation, and state the domain and range of the relation.

Solution. Choose three values of x, and use the given equation to find the corresponding y to make an ordered pair.

Let x = 7. y =   (7,3)

Let x = 2. y =   (2,2)

Let x = -2. y =   (-2,0)

To find the domain, we must determine the set of all x values. Since y must be a real number, and only nonnegative numbers have real solutions, the radicand x + 2 must be nonnegative. Therefore, x + 2 must be greater than or equal to 0. We solve the inequality: x + 2 > 0

x > -2

Thus, the domain is [-2, )

Since y equals the square root of a nonnegative number, y will equal a nonnegative number. Therefore, the range is [0,  ).

Circles

A circle is the set of all points whose distance from a fixed point is constant. The fixed point is the center, and the distance from the center to the circle is the radius.

Since a circle is a set of ordered pairs, it is a relation.

Center-Radius Form of the Equation of a Circle

The circle with center (h,k) and radius r has equation (x – h) + (y – k) = r

Example. Find the equation of the circle with center (-2,4) and radius 3.

Solution. Since the center is (-2,4), we have h = -2 and k = 4. The radius is 3; so r = 3. Substituting the values into the center radius form:

(x – h) + (y – k) = r

(x – (-2)) + (y – 4) = 3

(x + 2) + (y – 4) = 9

Example. Find the center and radius of the circle with equation

 - 42 = 0.

Solution. The equation must be written in center-radius form. To accomplish this, we will complete the square for both variables.

 - 42 = 0

 = 42Add 42 to both sides

(x  - 14x ) + (y  - 6y ) = 42Separate the terms containing x from the terms containing y

(x - 14x + 49) + (y - 6y + 9)= 42 + 49 + 9Complete each square

(x – 7)  + (y – 3)  = 100Factor each square

h = 7 , k = 3; r  = 100, so r = 10.

The center is (7,3) and the radius is 10.

Functions

Definition of a Function

A function is a relation is which for each element in the domain thee corresponds exactly one element in the range.

All functions are relations, but not all relations are functions. A function is a special relation in which no two ordered pairs have the same first coordinate.

For example, given the relation  (3,5), (4,2), (3,6), (5,7) , there are two ordered pairs, (3,5) and (3,6), with the same first coordinate. So, this relation is not a function.

Example. Determine whether or not the following relation is a function:

 (x,y)  y = x + 5

Solution. The ordered pairs of this relation are found by choosing values for x, and substituting them into the equation, y = x + 5, to obtain the corresponding y. For each x value, there is only one possible "answer" for y. So there could be no two ordered pairs with the same x-coordinate. Thus, the given relation is a function.

The graph of a relation is the set of points in the plane that correspond to the ordered pairs of the relation. You can determine whether a given graph is the graph of a function by using the Vertical Line Test.

Vertical Line Test

If each vertical line intersects a graph at no more than one point, the graph is the graph of a function.

In other words, if a vertical line intersects a graph at more than one point, the graph is not the graph of a function.

Function Notation

Consider the function defined by the equation y = x + 5. We use function notation to name the function in the following way: f(x) = x + 5.

The notation f(x) is read "f of x". f is the name of the function, and x is an element of the domain of f. The value of the function corresponding to x is f(x).

Also, f(x) is just another notation for y. You can always replace f(x) with y or y with f(x).

Most of the time, the letters f, g, h, and k are used to name functions. Examples of functions are:

k(x) =   , g(t) = t - 4t + 7, and h(a) = 

Note: The parentheses in the notation f(x) do not indicate multiplication. f(x) is "f of x", not "f times x".

To evaluate a function, we determine the value of f(x) for a specific value of x. f(a) is the value of f(x) when a is substituted for x in f(x).

Example. a) Given f(x) =  , find f(-4).

b) Given h(x) =  3 – 2x , find h(5).

Solution. a) f(x) = 

f(-4) = 

= 

= -18

b) h(x) =  3 – 2x

h(5) =  3 – 2(5)

=  3 – 10

=  –7 

= 7

Linear Functions 

Definition of Linear Function

A function f is a linear function if f(x) = ax + b , for real numbers a and b.

When a linear function is written in the form Ax + By = C, it is said to be in standard form.

The graph of a linear function is a straight line. To graph a linear function, find at least two of its ordered pairs, plot them, and draw a line through them.

Example. Graph 5x – 2y = 10

Solution.

Let x = 0. 5(0) – 2y = 10 Letting x = 0 will give us the y-intercept

- 2y = 10 Solve for y.

y = -5

(0,-5)

Let y = 0. 5x – 2(0) = 10 Letting y = 0 will give us the x-intercept.

5x = 10

x = 2

(2,0)

Let x = 4. 5(4) – 2y = 10 Third point is a check

20 – 2y = 10

- 2y = -10

y = 5

(4,5)

These points would be plotted and a line drawn through them to complete the graph.

Slope of a Line

The slope is a numerical measure of the steepness of a line. Slope compares the vertical change (the rise) to the horizontal change (the run) when moving from one point to another along the line. To calculate the slope, we use a

ratio comparing the change in y,  y, to the change in x,  x. The slope m of the line through the points   

and   is 

Example. Find the slope of the line through the points (-3, -1) and (-2,4).

Solution.   = (-3, -1) and   = (-2, 4).

The slope of a vertical line is undefined.

The slope of a horizontal line is 0.

The Difference Quotient

If h is any nonzero number, then the quotient   is called the difference quotient. The difference quotient is used in calculus to find the steepness of a curve at a point.

Example. For the function f(x) = x  - 4x + 7, find  .

Solution. f(x + h) is found by replacing x with x + h in the formula for f(x).

f(x + h) = (x + h)  - 4(x + h) + 7 = x  + 2xh + h  - 4x – 4h + 7

We will substitute this result into the Difference Quotient formula.

= 

Difference Quotient; substitution of

f(x + h) and f(x)

Simplify

= 

= 

= 2x + h – 4

Equations of a Line

If a particular point on a line and the slope of a line are known, we can write an equation for that line, using Point-Slope Form:

An equation of the line with slope m passing through the point   is

Example. Write an equation of the line through (-5,6) with slope -2.

Solution.  , and m = -2.

Point-Slope Form

Substitute values of  , and m

y – 6 = -2(x + 5)

y – 6 = -2x - 10

Simplify

y = -2x – 4 Solve for y

The equation of the line is y = -2x – 4.

The Point-Slope Form can also be used to find an equation for a line when two points on the line are known.

Example. Find an equation of the line through (1,-2) and (-3,-5).

Solution. We must first find the slope of the line, using the formula for slope:

Slope formula

Substitute: 

Simplify

m = 

m = 

Now we have the slope, m =  , and a point, (1,-2). (We can use either of the given points, (1, -2) or (-3, -5), for the point.)

Point-Slope Form

Substitute: 

Simplify

Solve for y

Another special form of a linear equation is the Slope-Intercept Form:

An equation of the line with slope m and y-intercept b is y = mx + b.

Example. Find an equation of the line with slope 4 and y-intercept -3.

Solution.

y = mx + b Slope-Intercept Form

y = 4x – 3 Substitute: m = 4 and b = -3

Example. Find the slope and y-intercept of the line with equation 2x + 3y = 15.

Solution. We will start by solving the given equation for y to obtain Slope-Intercept Form.

2x + 3y = 15

3y = -2x + 15 Add –2x to both sides

Divide both sides by 3

Simplify

Comparing this result to Slope-Intercept Form, y = mx + b, we can determine that m =   and b = 5.

Thus, the slope is   and the y-intercept is 5.

Two other special forms of the linear equation are as follows:

Equation of a Vertical Line

An equation of the vertical line through the point (a, b) is x = a.

Equation of a Horizontal Line

An equation of the horizontal line through the point (a, b) is y = b.

Example. a) Find the equation of the vertical line through (3,-1).

b) Find the equation of the horizontal line through (-5, -3).

Solution. a) a = 3, so the equation is x = 3.

b) b = -3, so the equation is y = -3.

Slope and Parallel Lines

1. If two nonvertical lines are parallel, then they have the same slope.

2. If two distinct nonvertical lines have the same slope, then they are parallel.

3. Two vertical lines are parallel.

Example. Write an equation of the line passing through (-3,2) and parallel to the line with equation y = 2x + 1.

Solution. The equation y = 2x + 1 is in Slope-Intercept Form, with m = 2. So, the slope of this line is 2. Since our line is parallel to this given line, it must have the same slope (part 1 of Slope and Parallel Lines). Therefore, we are to find the equation of a line with slope 2 passing through the given point (-3,2).

Point-Slope Form

Substitute: m = 2, 

y – 2 = 2(x + 3) Simplify

y – 2 = 2x + 6

y = 2x + 8 Solve for y

The equation of the line is y = 2x + 8.

Slope and Perpendicular Lines

1. If two nonvertical lines are perpendicular, then the product of their slopes is –1.

2. If the product of the slopes of two lines is –1, then the lines are perpendicular.

3. A horizontal line is perpendicular to a vertical line.

Example. Write an equation of the line passing through (1,-3) and perpendicular to the line with equation -x + 2y = 1.

Solution. We can find the slope of the line -x + 2y = 1 using Slope-Intercept Form;

-x + 2y = 1 Solve for y

2y = x + 1 Add x to both sides

Divide both sides by 2

Simplify

From the Slope-Intercept Form, we determine that m =  . The slope of the given line is  . Because our line is perpendicular to this given line, the slope of our line is –2.

[ (-2) = -1]. We now use Point-Slope Form with m = -2,  , and  .

Point-Slope Form

Substitute: m = -2,  , and  .

y + 3 = -2x + 2 Simplify

y = -2x – 1 Solve for y

The equation of the line is y = -2x – 1.

Algebra of Functions; Composite Functions

Let f and g be two functions. The sum, f + g, the difference, f – g, the product, fg, and the quotient,   are functions defined as follows:

1. Sum: (f + g)(x) = f(x) + g(x)

2. Difference: (f – g)(x) = f(x) – g(x)

3. Product: (fg)(x) = f(x)  g(x)

4. Quotient: 

The domains of the sum, difference, and product functions are the set of all real numbers common to the domains of

both f and g. The domain of   includes all real numbers common to the domains of f and g for which g(x) 0.

Example. Let f(x) = 4x  - 1 and g(x) = (2x + 1) . Find: a) (f + g)(5), b) (f – g)(2),

c) (fg)(-1), and d)  .

Solution. a) f(5) = 4(5)  - 1 = 4(125) – 1 = 500 – 1 = 499

g(5) = (2(5) +1) = (10 + 1) = 11 = 121

(f + g)(5) = f(5) + g(5) = 499 + 121 = 620

b) f(2) = 4(2)  - 1 = 4(8) – 1 = 32 – 1 = 31

g(2) = (2(2) +1) = (4 + 1) = 5 = 25

(f – g)(2) = f(2) – g(2) = 31 – 25 = 6

c) ) f(-1) = 4(-1)  - 1 = 4(-1) – 1 = -4 – 1 = -5

g(-1) = (2(-1) +1) = (-2 + 1) = (-1) = 1

(fg)(-1) = f(-1)  g(-1) = -5(1) = -5

d) f(0) = 4(0)  - 1 = 4(0) – 1 = 0 – 1 = -1

g(0) = (2(0) +1) = (0 + 1) = 1 = 1

=

Example. Given f(x) =   and g(x) = x  - 5x , find f + g, f – g, fg, and   and their domains.

Solution. (f + g)(x) = f(x) + g(x) =   + x  - 5x

(f - g) (x) = f(x) – g(x) =   - (x  - 5x) =   - x  + 5x

(fg)(x) = f(x)  g(x) =  (x  - 5x)

 =  x

The domain of f consists of values for which the radicand, x – 1, is nonnegative; that is

x – 1 > 0. Solving this inequality gives us x > 1. So, the domain of f is [1, ).

The domain of g is the set of all real numbers because any real number substituted for x in the expression x  - 5 yields a result that is a real number.

Numbers common to the domains of both f and g are in the interval [1,  ). So the domain of f + g, f – g, and fg is [1,  ).

To find the domain of  , we must find values for which g(x) = 0:

x  - 5x = 0Set g(x) equal to 0

x(x – 5) = 0 Solve by factoring

x = 0 or x – 5 = 0

x = 5

0 and 5 must be excluded from the domain. Numbers in the domain are in the interval [1, ), but not including 0 or 5. We can write this set of numbers in the following way:

[1,5)  (5,  ).

Composition of Functions

Given two functions f and g, the composition of f and g, denoted by  , is defined by ( )(x) = f[g(x)] and

the composition of g and f, denoted by  , is defined by ( )(x) = g[f(x)].

Example. Let f(x) =   and g(x) = 2x – 1 . Find a) ( )(x) , b) ( )(x),

c) ( )(5), and d) ( )(16).

Solution.

a) ( )(x) = f[g(x)] Definition of 

= f[2x – 1] Replace g(x) with 2x – 1

= Use f(x), and replace x with 2x – 1

b) ( )(x) = g[f(x)] Definition of 

= g[ ] Replace f(x) with 

= 2  - 1 Use g(x), and replace x with 

c) ( )(5) = f[g(5)] Definition of 

= f[9] Find g(5) by substitution:

g(5) = 2(5) – 1 = 10 – 1 – 9

= 3 Find f(9) by substitution:

f(9) =   = 3

d) ( )(16) = g[f(16)]. Definition of 

= g[4] Find f(16) by substitution:

f(16) =  = 4

= 7 Find g(4) by substitution:

g(4) = 2(4) – 1 = 8 – 1 - 7

Quadratic Functions

Definition of a Quadratic Function

A function f is a quadratic function if f(x) = ax + bx + c, where a, b, and c are real numbers, with a  0.

Properties of Graphs of Quadratic Functions

1. The graph of a quadratic function f(x) = ax + bx + c is called a parabola.

2. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward.

3. As  a increases, the parabola becomes narrower; as  a decreases, the parabola becomes wider.

4. The lowest point of a parabola (when a > 0) or the highest point (when a < 0) is called the vertex.

5. The domain of a quadratic function is (- ,  ), because the graph extends indefinitely to the right and to the left. If (h, k) is the vertex of the parabola, then the range of the function is [k, ) when a > 0 and (- , k] when a < 0.

6. The graph of a quadratic function is symmetric with respect to a vertical line containing the vertex. This line is called the axis of symmetry. If (h, k) is the vertex of a parabola, then the equation of the axis of symmetry is x = h.

Vertex and Intercepts

1. To determine the vertex of the graph of a quadratic function, f(x) = ax + bx + c, we can either:

a) use the method of completing the square to rewrite the function in the form

f(x) = a(x – h)  + k. The vertex is (h, k). , or

b) use the formula x =   to find the x-coordinate of the vertex; the

y- coordinate of the vertex can be determined by evaluating  .

2. If a > 0, then the y-coordinate of the vertex represents the minimum value of the function; if a < 0, then the y-coordinate of the vertex represents the maximum value of the function.

3. To find the y-intercept of the graph of f(x) = ax + bx + c, find f(0); to find the x-intercepts, solve the quadratic

equation ax + bx + c = 0.

Example. For each of the following functions, find the vertex, axis, domain, range, intercepts, and sketch the graph.

a) f(x) = -2(x – 3)  + 2

b) f(x) = x  - 8x + 10

c) f(x) = 2x + 12x + 17

Solution. a) According to part 1 of Vertex and Intercepts, a function written in the form

f(x) = a(x – h) + k has vertex (h, k). So, f(x) = -2(x – 3)  + 2 has vertex (3,2).

The axis of symmetry is a vertical line through the vertex, so its equation is x = 3.

The domain is (- ,  ). Since a < 0 (a = -2), the range is (- , 2]. (Property 5)

To find the y-intercept, we find f(0).

f(0) = -2(0 – 3) + 2 = -2(-3) + 2 = -2(9) + 2 = -18 + 2 = -16

The y-intercept is (0, -16).

To find the x-intercept, we solve the equation, -2(x – 3)  + 2 = 0

-2(x – 3)  + 2 = 0

-2(x - 6x + 9) + 2 = 0Square the binomial

-2x  + 12x – 18 + 2 = 0Simplify

-2x  + 12x – 16 = 0

x  - 6x + 8 = 0Divide both sides by -2

(x – 4)(x – 2) = 0 Solve by factoring

x – 4 = 0 or x – 2 = 0

x = 4 x = 2

The x-intercepts are (4,0) and (2,0).

The graph is a parabola. Since a < 0, the parabola opens downward. Since  a =  -2 = 2, the parabola is narrower

than the graph of f(x) = x . (See figure 1, page 276 in text).

To draw the graph, plot the vertex and intercepts; then draw a parabola shape opening downward from the vertex.

b) In this example, we will find the vertex by completing the square.

f(x) = x  - 8x + 10

f(x) = (x  - 8x ) + 10Group terms containing x together

f(x) = (x  - 8x + 16 – 16) + 10Complete the square by adding 16

(-8) = -4; (-4)  = 16

You must also subtract 16 so the value of the expression is not changed

f(x) = (x  - 8x + 16) – 16 + 10Regroup

f(x) = (x – 4)  - 6Factor the perfect square

Comparing this result to the form f(x) = (x – h)  + k, we see that h = 4 and k = -6; so, the vertex is (4, -6).

The axis of symmetry is x = -4.

The domain is (- ,  ). The range is [-6, ).

To find the y-intercept, let x = 0:

f(0) = 0 - 8(0) + 10 = 10. So, the y-intercept is (0, 10).

To find the x-intecepts, we will solve the equation x  - 8x + 10 = 0

x  - 8x + 10 = 0

x = 

Quadratic formula

x = 

Substitute: a = 1, b = -8, c = 10

x = 

Simplify

x = 

x = 

x = 4  

x  6.45 or x  1.55

The x-intercepts are (6.45, 0) and (1.55, 0).

The graph is a parabola; since a > 0, the parabola opens upward. Since  a = 1 > 0, the parabola has the same shape

as f(x) = x .

c) We will find the vertex using the formula from part 1 above.

f(x) = 2x  + 12x + 17

x = -

Formula for x-coordinate of the vertex

x = - 

Substitute a = 2 and b = 12

The x-coordinate of the vertex is –3.

To find the y-intercept, we will calculate f(-3).

y = f(-3) = 2(-3)  + 12(-3) + 17 = 2(9) + 12(-3) + 17 = 18 – 36 + 17 = -1

So, the vertex is (-3,-1).

The axis of symmetry is x = -3.

The domain is (- , ). The range is [-1,  ).

To find the y-intercept, let x = 0.

f(0) = 2(0) + 12(0) + 17 = 17

So, the y-intercept is (0, 17)

To find the x-intercept, solve the equation 2x  + 12x + 17 = 0

2x  + 12x + 17 = 0

x = 

Quadratic formula

x = 

Substitute a = 2, b = 12, c = 17

x = 

x = 

x = 

x  -2.29 or x  -3.71

Simplify

The graph is a parabola; since a > 0, the parabola opens upward. Since  a =  2 = 2, the parabola is narrower than

the graph of f(x) = x .

Applications in which a quantity is to have a maximum or minimum value can often be modeled with quadratic functions. The vertex of the graph is the point of interest because it is at this point that the function has its maximum or minimum value.

Maximum and Minimum

For the function f(x) = ax  + bx + c,

1. If a > 0, then f has a minimum that occurs at x =  .

2. If a < 0, then f has a maximum that occurs at x =  .

Example. The average number of gallons of alcohol consumed by each adult in the United States can be modeled by

f(x) = -0.053x  + 1.17x + 35.6

where x represents the number of years after 1970. In what year was alcohol consumption at a maximum? What was the average consumption for that year?

Solution. f(x) = -0.053x  + 1.17x + 35.6

a = -0.053 and b = 1.17

Since a < 0, the function has a maximum when x =  .

Substituting the values of a and b into this formula, we obtain

x =   = 

This means that alcohol consumption was at a maximum approximately 11 years after 1970, or in 1981.

To obtain the average consumption for that year, we calculate f(11)

f(11) = -0.053(11)  + 1.17(11) = 35.6  42.06.

The average consumption for that year was approximately 42 gallons of alcohol by each adult.

Synthetic Division Process 

Synthetic division is a numerical method for dividing a polynomial by a binomial of the form x – c. The technique involves writing only the essential parts of a long division problem.

The following illustrates the use of synthetic division to divide a polynomial, written in descending order, by x – c , where c is a constant called the divider:

To divide a polynomial f(x) by x – c:Divide: 2x  - 7x  + 7x – 13 by x - 3

1. Write the divider, 3, and the coefficients of the dividend

2. Bring the first coefficient of the dividend down to the bottom row.

3. Multiply the first number in the bottom row by the divider; write this result in the 2nd column of the middle row

4. Add the 2nd column to obtain the 2nd number in the bottom row

5. Similarly, multiply the –1 in the bottom row by the divider; write this result in the 3rd column of the middle row; then add the 3rd column to obtain 4.

6. Finally, multiply 4 by the divider; write this result in the 4th column of the middle row; then add the 4th column and write the result in the bottom row.

7. The entries in row three, excluding the last entry, are the coefficients of the quotient. The degree of the quotient is one less than the degree of the dividend. The last entry is the remainder of the division.

The degree of the dividend is 3. So the degree of the quotient is 2.

Quotient: 2x - x + 4

Remainder: -1

2x - x + 4 - 

Example. Use synthetic division to perform the indicated operation: 

Solution. The divisor, x + 2, must be written as a difference, x – (-2), to determine that the divider is –2.

The Remainder Theorem

If a polynomial f(x) is divided by x – c , where c is a constant, then the remainder is equal to f(c).

Note: The Remainder Theorem applies only if the divisor is of the form x – c. Divisors of the form x + c must be changed to x – (-c).

The Remainder Theorem gives us a second way to evaluate a polynomial for x = c. We can evaluate f(c) directly, or we can use synthetic division.

Example. Use the Remainder Theorem and synthetic division to find f(k):

k = 4 ; f(x) = 

Solution. The Remainder Theorem tells us that if we use synthetic division and divide f(x) with divider 4, the remainder will be equal to f(4).

The dividend has no "x" term; the coefficient 0 must be used for this missing term.

The remainder is 127. So, f(4) = 127.

Example. Use synthetic division to decide whether -6 is a zero of the polynomial

f(x) =  .

Solution. We know that if -6 is a zero of f(x), then f(-6) = 0. So, we will use synthetic division and the Remainder Theorem to find f(-6).

Since the remainder is 0, f(-6) = 0, and –6 is a zero of the polynomial f(x).

Division Algorithm

If f(x) and d(x) are polynomials, with d(x)  0, and the degree of d(x) is less than or equal to the degree of f(x), then there exist unique polynomials q(x) and r(x) such that

f(x) = d(x)  q(x) + r(x)

Dividend = Divisor  Quotient + Remainder

If r(x) = 0, then d(x) divides evenly into f(x), and d(x) and q(x) are factors of f(x).

We will use the Division Algorithm to do the following problem:

Express the polynomial f(x) =   in the form f(x) = (x – k )q(x) + r for k = 2.

In this case, the divisor, d(x), is x – 2. We will use synthetic division to divide f(x) by

x – 2 .

The quotient, q(x), is  , and the remainder, r, is 6.

f(x) = (x – 2 )(  ) + 6

Zeros of a Polynomial Function

The Factor Theorem

For a polynomial f(x) and a constant c,

a. If f(c) = 0, then x – c is a factor of f(x).

b. If x – c is a factor of f(x), then f(c) = 0.

The Factor Theorem tells us that if we find a value of c such that f(c) = 0, then x – c is a factor of f(x). And, if x – c is a factor of f(x), then f(c) = 0.

Fundamental Theorem of Algebra

Every polynomial of degree 1 or more has at least one complex zero.

Number of Zeros Theorem

A polynomial of degree n has at most n distinct zeros.

Conjugate Zeros Theorem

If f(x) is a polynomial having only real coefficients and if a + bi is a zero of f(x), then a – bi is also a zero of f(x).

Example. Use the Factor Theorem to decide whether x + 1 is a factor of

f(x) = 2x  + x + 2.

Solution. According to the Factor Theorem, x + 1, which we can rewrite as (x – (-1)), is a factor of f(x) if f(-1) = 0. We can find out if f(-1) = 0 by using the Remainder Theorem and synthetic division:

The remainder is -1. So, f(-1) = -1. Since f(-1) is not equal to 0, x + 1 is not a factor of f(x).

Example. For each polynomial, one zero is given. Find the other zeros:

a. f(x) =   ; 1 is a zero

b. f(x) =   ; i is a zero

Solution.

a. Since 1 is a zero of f(x), we will begin by dividing f(x) by x – 1:

The quotient is x  + 5x + 5. Using the Division Algorithm, we factor f(x) as:

f(x) = (x – 1)(x  + 5x + 5)

We can find the remaining zeros by setting the quotient equal to 0 and solving the equation:

x  + 5x + 5 = 0

x = 

Use the quadratic formula:

x = 

with a = 1, b = 5, and c = 5

x = 

Simplify

x = 

The zeros of f(x) are 1,  , and 

b. We will begin by dividing f(x) by the given zero, i.

The Complex Conjugate Theorem tells us that since i is a zero of f(x), then its conjugate, -i, is also a zero of f(x). Therefore, we will divide the quotient obtained above by –i:

The quotient is x  + 10x + 26. We can find the remaining zeros by setting this quotient equal to 0 and solving:

x  + 10x + 26 = 0

x = 

Use the quadratic formula with a = 1, b = 10, and c = 26.

x = 

x = 

x = 

x = -5  i

Simplify

The four zeros of f(x) are i, -i, -5+i, and –5-i.

Example. Find a polynomial f(x) of degree 3 with only real coefficients that has zeros 2, -3, and 5; such that f(3) = 6.

Solution. From the Factor Theorem, we know that if 2, -3, and 5 are zeros of f(x), then

x – 2, x + 3, and x – 5 are factors of f(x). So, we can write f(x) in the following way:

f(x) = a(x – 2)(x + 3)(x – 5) where a is a constant.

To find a, we use the fact that f(3) = 6, and substitute 3 for x in f(x):

f(x) = a(x – 2)(x + 3)(x – 5)

f(3) = a(3 – 2)(3 + 3)(3 – 5) Replace x with 3

6 = a(1)(6)(-2)

6 = -12a

 = a

Replace f(3) with 6; simplify

So, f(x) =  (x – 2)(x + 3)(x – 5).

Example. Factor f(x) =   into linear factors, given that -5 is a zero of f(x).

Solution. Let f(x) be the dividend, and use synthetic division with divider -5:

The quotient, q(x) is   and the remainder, r(x), is 0. Using the division algorithm, we get that f(x) =

(x + 5)(  ). We can factor the quotient into two linear factors to obtain: f(x) = (x + 5)(6x – 1)(x – 2)

Example. For the polynomial f(x) =   , find all zeros and their multiplicities.

Solution. According to the Factor Theorem, since x + 1 is a factor of f(x), then -1 is a zero of f(x). Since the factor x + 1 is raised to the 2nd power, it is a factor two times and we say that -1 is a zero of multiplicity 2. Similarly, since x – 1 is a factor, 1 is a zero of f(x). Since x – 1 is a factor 3 times, 1 is a zero of multiplicity 3. Since the remaining

factor of f(x), x  - 10, is quadratic, we will set it equal to zero to find the remaining zeros:

x  - 10 = 0

x  = 10Add 10 to both sides

x =  Use the Square Root Property

 and -  are both zeros of multiplicity 1.

Example. Find a polynomial of lowest degree with only real coefficients having zeros -1 and 6 – 3i.

Solution. We know from the Conjugate Zeros Theorem that since the complex number

6 – 3i is a zero of f(x), then its conjugate, 6 + 3i, is also a zero. By the Factor Theorem, f(x) must have three factors: (x – (-1)), (x – (6-3i)), and (x – (6+3i)). We can express f(x) in the following way:

f(x) = (x – (-1))(x – (6-3i))(x – (6+3i))

f(x) = (x + 1)(x – 6 + 3i)(x – 6 - 3i) Use Distributive Property

f(x) = (x + 1)(x  - 6x – 3ix – 6x + 36 + 18i + 3ix –

18i – 9i )

Multiply the 2nd and 3rd factors

f(x) = (x + 1)(x  - 12x + 45)Simplify

f(x) = Multiply

f(x) = 

Exponential Functions

Definition of an Exponential Function

An exponential function can be written as f(x) = a , where a > 0, a  1, and x is any real number.

The two restrictions on a in the definition are important:

a  1 because 1  has a value of 1 for all values of x, and the function would

be the constant function f(x) = 1.

a > 0 so that the function can be defined for all real numbers.

The number e is an irrational number. The value of e , to nine decimal places, is

e  2.718281828. The number e occurs frequently in applications in science, engineering, business, and others.

Natural Base e

The number e is often the base of an exponential function, f(x) = e . This function is called the natural exponential function.

Property of Equivalent Exponents

For a > 0 and a  1, if  , then b = c.

We can use this property to solve an exponential equation. If we can write each side of an exponential equation as an exponential function with the same base, then the exponents are equal.

Example. Solve: a. 

b. 

c. 

d. 

Solution.

a.

Definition of negative exponent

Property of Exponents

-k = 2 Property of Equivalent Exponents

k = -2 Solve for k

b.

Express 8 as a power of 2

3 – y = 3 Property of Equivalent Exponents

-y = 0 Solve for y

y = 0

c.

Raise both sides to the power  , the reciprocal of

the power 

z = Definition of rational exponent

z = 2Simplify the radical

z = 4 Simplify

d.

Express 27 and 9 as powers of 3

Property of Exponents

12z = 2z + 2 Property of Equivalent Exponents

10z = 2 Solve for z

z = 

Compound interest is a situation that can be modeled by an exponential function.

Compound Interest

If P dollars is deposited in an account paying an annual rate of interest r compounded m times per year, then after t years the account will contain A dollars, where

A is sometimes called the future value, and P is called the present value.

Example. Use the formula for compound interest to find the future value: $8906.54 at 5% compounded semiannually for 9 years.

Solution. P is the amount deposited, so P = 8906.54; r is the interest rate (in decimal form), so r = 0.05; m is the number of times per year that interest is compounded, so m = 2; and t is the number of years, so t = 9.

Substituting these values into the compound interest formula;

Compound Interest formula

Substitution

Simplify

A = 13,891.16 Result if rounded to the nearest cent

The future value of this deposit is $13,891.16.

Example. Find the present value for the future value $25,000 if interest is 6% compounded quarterly for 11 quarters.

Solution. A is the future value, so A = 25,000; r = 0.06; m = 4; and t =   = 2.75

Compound Interest Formula

Substitution

Simplify

Divide both sides by (1.015)

P = 21,223.33 Result rounded to the nearest cent

The present value is $21,223.33.

Example. The exponential function P(t) = 67.38(1.026)  describes the population (in millions) of Mexico t years after 1980.

a. What was the population of Mexico in 1980?

b. Predict Mexico’s population in 2007 and 2034.

Solution.

a. t is the number of years after 1980. In 1980, t = 0. Substitute t = 0 into the function:

P(0) = 67.38(1.026)  = 67.38(1) = 67.38

The population of Mexico in 1980 was 67.38 million.

b. To find the value of t in 2007: 2007 – 1980 = 27. So, let t = 27;

P(27) = 67.38(1.026)   134.74

The population of Mexico in 2007 will be about 134.74 million.

In 2034, t = 2034 – 1980 = 54

P(54) = 67.38(1.026)   269.46

The population of Mexico in 2034 will be about 269.46 million.

Logarithmic Functions

Definition of Logarithm

For all real numbers y, and all positive numbers a and x, where a  1:

y = log x if and only if x = a .

A logarithm is an exponent. The logarithm of x with base b, written as log x, is equal to the exponent to which b

must be raised to get x. Thus, log 32 = 5 because 2 raised to the power 5 is equal to 32.

Example.

a. Write an equivalent statement in logarithmic form: 

b. Write an equivalent statement in exponential form: log 49 = 2

Solution.

a. We know from the definition of logarithm that if x = a , then y = log x. In our given exponential

expression,  , we have a = 27, y =  , and x = 9. It is equivalent to

 in logarithmic form.

b. We know from the definition of logarithm that if y = log x , then x = a . We are given log 49 = 2 , so a = 7, x = 49, and y = 2. So the equivalent exponential form is

.

We can often evaluate logarithms and solve equations involving logarithms by changing logarithmic forms to exponential forms.

Example. Find the value of each expression: a.   b. 

Solution.

a. Let y =  . We can use the definition of logarithm with a = 16 and x = 8:

y =   is equivalent to  . We will now solve this exponential equation:

Express 8 and 16 as powers of 2

Property of Exponents

3 = 4y Property of Equivalent Exponents

Solve for y

b. Let y =  . We can use definition of logarithm with a = 7 and x =  :

y =   is equivalent to   = 7 . To solve this equation:

 = 7

Definition of rational exponent

Property of Equivalent Exponents

Example. Solve: a. log 27 = 3 b. log x = 2

Solution.

a.

log 27 = 3

x  = 27Definition of logarithm

Raise both sides to the power 1/3

x =   = 3Definition of rational exponent

b.

log x = 2

Definition of logarithm

Properties of Logarithms

If x and y are any positive real numbers, r is any real number, and a is any positive real number, a  1, then

1. 

2. 

3. 

4. 

5. 

6. 

The properties of logarithms can be used to expand or condense a logarithmic expression. To expand a logarithmic expression, we write it as a sum, difference, or product of logarithms. For example:

Property 2 of logarithms

Property 1 of logarithms

Use exponential notation; use distributive property to remove parentheses

Property 3 of logarithms; log 36 = 2 because 6=36

To condense a logarithmic expression, we write it as a single logarithm. To do this, we reverse the procedure above:

We will condense this expression by writing it as a single logarithm

Property 3 of logarithms

Property 2 of logarithms

Property 1 of logarithms

Evaluating Logarithms; Change of Base

The logarithmic function with base 10 is called the common logarithm.

We write  as y = log x

A calculator with a log key can be used to find common logarithms of any positive number.

One mathematical model that contains a common logarithm is the formula for pH of a solution:

where  is the hydronium ion concentration in moles per liter. pH measures the acidity or alkalinity of solutions. An acid solution has pH < 7; solutions with pH > 7 are called basic. Water has pH = 7.

The hydronium concentration of milk is 3.97 x 10 . To find the pH of milk:

pH formula

Substitution

pH = - [ -6.4]

pH = 6.4

The pH of milk is about 6.4.

The magnitude R on the Richter scale of an earthquake of intensity I is given by

where   is the intensity of a zero-level earthquake.

Northern California’s 1989 earthquake was 10  times as intense as a zero-level earthquake; that is,   To find the magnitude on the Richter scale:

Formula for magnitude

Substitute for I

R = log 10Simplify

R = 7.1 Logarithm Property 6

Northern California’s 1989 earthquake registered 7.1 on the Richter scale.

The model that describes sound intensity is similar to the model for earthquake intensity. The loudness of sounds is measured in decibels. The loudness level of a sound, d, in decibels, is given by

where I is the intensity of the sound and   is the intensity of a sound barely audible to the human ear.

The cry of a blue whale can be heard nearly 500 miles away, reaching an intensity of

. To determine the decibel level of this sound:

Loudness formula

Substitute the given intensity for I

d = 10 log (6.3 x 10 )Simplify

d = 10(18.799) Find the logarithm

d = 187.99

The decibel level of the blue whale’s cry is 187.99.

The logarithmic function with base e is called the natural logarithmic function. We write:

Natural logarithms can be found with a calculator that has an ln key.

Calculators give the values of both common logarithms (base 10) and natural logarithms (base e). To find a logarithm with any other base, we can use the Change-of-Base Theorem:

For any positive real numbers x, a, and b, where a  1 and b  1:

This theorem is used to write a logarithm in terms of quantities that can be evaluated with a calculator. Because calculators have keys for common and natural logarithms, we will change the given base, a, to either base 10 or base e. If we change the base to 10, the theorem looks like this:

b = 10; 

If we change the base to e, we have:

b = e; 

Example. Use the Change-of-Base Theorem to find each logarithm to the nearest hundredth: a.   

b. 

Solution.

a.

Change of base formula with base 10.

Either base 10 or e can be used.

Substitution

 2.79 Evaluate logs; round to the nearest hundredth

b.

Change of base formula with base e. Again, either base 10 or e can be used.

Substitution

 5.61 Evaluate; round to hundredths

Exponential and Logarithmic Equations

Exponential Equations

An exponential equation is an equation containing a variable in an exponent. The following property of logarithms can be used to solve many exponential equations:

Property of Logarithms, Part 2

If x, y, and a are positive numbers, a  1, then

1. If x = y, then 

2. If  , then x = y.

Exponential equations can be solved by isolating the exponential expression; using Property of Logarithms, Part 2 to take the log of both sides; simplifying; and then solving for the variable.

Example. Solve: 

Solution.

Use Property of Logarithms, part 2, to take the log of both sides

x log 10 = log 5.71Property of Logarithms: 

x  1 = log 5.71Property of Logarithms: 

x  0.7566

Example. Solve: 

Solution.

Divide both sides by 7

Use Property of Logarithms, Part 2, to take the log of both sides

Property of Logarithms: 

ln e = 1

Divide both sides by 3

x  1.266

Example. Solve: 

Solution.

Use Property of Logarithms, Part 2, to take the log of both sides

(x + 2) ln 2 = (2x + 1) ln 3Property of Logarithms: 

x ln 2 + 2 ln 2 = 2x ln 3 + ln 3 Distributive Property

x ln 2 - 2x ln 3 = ln 3 – 2 ln 2 Isolate terms with the variable on one side of the equation

x(ln 2 – 2 ln 3) = ln 3 – 2 ln 2 Factor out the common factor, x

x  0.191

Divide both sides by ln 2 – 2 ln 3

Logarithmic Equations

Logarithmic equations contain logarithmic expressions and constants. When one side of the equation contains a single logarithm and the other side contains a constant, the equation can be solved by rewriting the equation as an equivalent exponential equation using the definition of logarithm. For example,

Definition of logarithm

16 = x + 3 Simplify

13 = x Solve for x

All solutions of logarithmic equations must be checked, because negative numbers do not have logarithms:

Check: 

Substitute the solution, 13, in place of x

Simplify

2 = 2 because 

If one side of a logarithmic equation contains more than one logarithm, use properties of logarithms to condense them into a single logarithm. For example:

Property of

Logarithms: 

Definition of Logarithm

8 = x - 7xSimplify

0 = x - 7x – 8Write quadratic equation in standard form

0 = (x – 8)(x + 1) Solve by factoring

x – 8 = 0 or x + 1 = 0

x = 8 or x = -1

Check:

Substitute the solution 8 for x

Substitute the solution –1 for x

Subtract Subtract

3 + 0 = 3because 

; 

because 

The number -1 does not check, since negative numbers do not have logarithms

3 = 3

The solution set is {8}.

In the next example, every term contains a logarithmic expression. We will solve this equation by using logarithmic properties to rewrite each side as a single logarithm. We then use Property of Logarithms, Part 2, and set the quantities equal to each other.

Example. Solve: log (2x – 1) = log (4x – 3) – log x

Solution.

log (2x – 1) = log (4x – 3) – log x

Property of

Logarithms: 

Property of Logarithms, Part 2

x(2x – 1) = 4x – 3 Multiply both sides by x

2x  - x = 4x – 3Distributive Property

2x  - 5x + 3 = 0Write the quadratic equation in standard form

(2x – 3)(x – 1) = 0 Solve by factoring

2x - 3 = 0 or x – 1 = 0

2x = 3 or x = 1

x = 

Check: x = 

log (2x – 1) = log (4x – 3) – log x

Substitute the solution   for x

log 2 = log 3 – log 1.5 Simplify

log 2 = log 

Property of

Logarithms: 

log 2 = log 2

Check: x = 1

log (2x – 1) = log (4x – 3) – log x

log (2(1) – 1) = log (4(1) – 3) – log 1 Substitute the solution 1 in place of x

log 1 = log 1 – log 1

0 = 0

Both solutions check, so the solution set is  .

Exponential Growth or Decay

Exponential Growth or Decay Function

Let   be the amount or number present at time t = 0. Then, under certain conditions, the amount present at any

time t is given by   , where k is a constant.

When k > 0, the function describes growth.

When k < 0, the function describes decay.

Example. A sample of 300 grams of plutonium 241 decays according to the function

where t is time in years. Find the amount of the sample remaining after (a) 4 years (b) 8 years (c) 20 years, then (d) find the half-life.

Solution.

(a)   represents the initial amount, so   = 300 grams. t represents time in years, so

t = 4:

Given formula

Substitute 4 for t and 300 for A

Simplify

A(4)  243

243 grams remain after 4 years.

(b) t = 8

Given formula

t = 8

Simplify

A(8) )  196

196 grams remain after 8 years.

(c) t = 20

Given formula

t = 20

Simplify

A(20)  104

104 grams remain after 20 years.

(d) The half-life is the time it takes for half of a given amount to decay. Since the initial amount is 300 grams, the

half-life is the time it takes for this amount to decay to 150 grams. So, we have   = 300 and A(t) = 150:

Given formula

150 = 300eSubstitution

Divide both sides by 300

0.5 = eSimpllify

ln 0.5 = ln eTake ln of both sides

ln 0.5 = -0.053t Property of logarithms

Divide both sides by -0.053

Simplify

t  13

The half-life is approximately 13 years.

Example. A sample from a refuse deposit near the Strait of Magellan had 60% of the carbon 14 of a contemporary sample. How old was the sample?

Solution. The amount of carbon 14 present after t years is given by

where   is the amount present in living plants and animals. In this problem, the amount of carbon 14 present is

60% of that in a living sample, or  .

Given formula

Substitute   for y.

Divide both sides by y

Simplify

Take ln of both sides

Property of logarithms

Multiply both sides by 5700; divide both sides by -ln 2

t  4201

The sample is approximately 4201 years old.

Continuous Compounding

If P dollars is deposited at a rate of interest r compounded continuously for t years, the final amount on deposit is A

= Pe  dollars.

Example. Suppose $12,000 is deposited in an account paying 4% interest compounded continuously for 7 years. Find the total amount on deposit at the end of 7 years.

Solution. P = 12,000 ; r = 0.04 (always change the rate to decimal form); t = 7

A = PeGiven formula

A = 12000  e Substitution

A = 12000  e Simplify

A  15,877.56

The total amount on deposit after 7 years is $15,877.56.

Example. How long will it take for $5000 to grow to $8400 at an interest rate of 6% if interest is compounded continuously?

Solution. P = 5000; A = 8400; r = 0.06

A = PeGiven formula

8400 = 5000eSubstitution

Divide both sides by 5000

1.68 = eSimplify

ln 1.68 = ln eTake ln of both sides

ln 1.68 = 0.06t Property of logarithms

t  8.6

Divide both sides by 0.06

It will take approximately 8.6 years for the investment to grow to $8400.

Linear Systems of Equations 

Two or more equations considered simultaneously form a system of equations. A solution of a system of equations in two variables is an ordered pair (a,b) that satisfies all equations of the system.

One method of solving a system of equations is the substitution method. This method involves converting the system to one equation in one variable by an appropriate substitution. It is a particularly useful method when it is easy to solve one of the equations for one of the variables.

The Substitution Method

1. Solve one of the equations for one variable in terms of the other.

2. Substitute the expression found in step 1 into the other equation.

3. Solve the equation obtained in step 2.

4. Back-substitute the value found in step 3 into the equation from step 1 to find the

value of the remaining variable.

5. Check the solution in both of the system’s given equation.

Example. Solve the system:

2x – y = 9

3x – 4y = -24

Solution.

2x – y = 9

-y = 9 – 2x

y = 2x - 9

Solve the first equation for y

3x – 4y = -24

3x – 4(2x – 9) = -24

Substitute the expression 2x – 9 for y in the second equation

3x – 8x + 36 = -24

-5x + 36 = -24

-5x = -60

x = 12

Solve for x

y = 2x – 9

y = 2(12) – 9

y = 24 – 9

y = 15

Back-substitute x = 12 into the expression found in step 1

The solution is (12, 15).

Check: We will substitute the solution into both of the original equations to check:

2x – y = 9

2(12) – 15 = 9

24 – 15 = 9

9 = 9

3x – 4y = -24

3(12) – 4(15) = -24

36 – 60 = -24

-24 = -24

Another method for solving systems of equations is the addition method, sometime called the elimination method.

The Addition/Elimination Method

1. If necessary, rewrite both equations in the form Ax + By = C.

2. If necessary, multiply either equation or both equations by appropriate numbers so that

the coefficients of x and y will be opposites with a sum of 0.

3. Add the equations in step 2. The sum is an equation in one variable.

4. Solve the equation from step 3.

5. Back-substitute the value obtained in step 4 into either of the given equations and

solve for the other variable.

6. Check the solution in both of the original equations.

Example. Solve the system:

3x = 33 – 5y

3y = 4x – 15

Solution. Rewrite both equations in the form Ax + By = C:

3x = 33 – 5y

3x + 5y = 33

3y = 4x – 15

-4x + 3y = -15

To eliminate x, multiply the first equation by 4 and the second equation by 3:

4(3x + 5y) = 4(33)

12x + 20y = 132

3(-4x + 3y) = 3(-15)

-12x + 9y = -45

The coefficients of x are now opposites. Add the equations together; then solve for y:

12x + 20y = 132

-12x + 9y = -45

29y = 87

y = 3

Back-substitute y = 3 into the first given equation; then solve for x:

3x = 33 – 5y

3x = 33 – 5(3)

3x = 33 – 15

3x = 18

x = 6

Solution: (6, 3)

Check:

3x = 33 – 5y

3(6) = 33 – 5(3)

18 = 33 – 15

18 = 18

3y = 4x – 15

3(3) = 4(6) – 15

9 = 24 – 15

9 = 9

Inconsistent and Dependent System of Linear Equations

If both variables are eliminated when a system of linear equations is solved by substitution or addition/elimination:

1. There is no solution if the resulting statement is false. The system is inconsistent.

2. There are infinitely many solutions if the resulting statement is true. The system is

dependent.

Example. Solve:

8x – 2y = 8

12x – 3y = 12

Solution. Each equation is already written in the form Ax + By = C. So we will begin by multiplying the first equation by 3 and the second equation by -2:

8x – 2y = 8

3(8x – 2y) = 3(8)

24x – 6y = 24

12x – 3y = 12

-2(12x – 3y) = -2(12)

-24x + 6y = -24

We will now add the two resulting equations:

24x – 6y = 24

-24x + 6y = -24

0 = 0

Both variables have been eliminated, and the resulting statement, 0 = 0, is true for all values of x and y. This means that the equations of the given system represent two different ways of writing the equation of the same line. All points on the line have coordinates that satisfy the system. We can use either equation to solve for y:

8x – 2y = 8

-2y = 8 – 8x

y = 4x – 4

Any ordered pair of the form (x, 4x – 4) is a solution of the system.

Note: We could have solved for x instead of y:

8x – 2y = 8

8x = 2y + 8

The solution set could also be represented as any ordered pair of the form ( , y).

Example. Solve:

3x – 3y = -2

x – y = 5

Solution. We will use the substitution method, and begin by solving for x in the second equation:

x – y = 5

x = y + 5

We can now substitute this expression for x in the first equation:

3x – 3y = -2

3(y + 5) – 3y = -2

3y + 15 – 3y = -2

15 = -2

The false statement, 15 = -2, indicates that the system has no solution.

Solving a System in Three Variables

The method for solving a system of linear equations in three variables is similar to that used on systems of linear equations in two variables. We use addition to eliminate any variable, reducing the system to two equations in two variables.

A solution of a system of three linear equations in three variables is an ordered triple of real numbers that satisfies all equations of the system.

Example. Solve the system:

x – 4y – z = 6 Equation 1

2x – y + 3z = 0 Equation 2

-3x + 2y – z = -4 Equation 3

Solution. We first choose two equations and use the addition method to eliminate a variable. There are many ways to accomplish this, but we will eliminate x from equations 1 and 2:

x – 4y – z = 6 Equation 1

-2(x – 4y – z) = -2(6)

-2x + 8y + 2z = -12

Multiply both sides of Equation 1 by -2

-2x + 8y + 2z = -12

2x - y + 3z = 0

7y + 5z = -12 Equation 4

Add the result from the previous step to Equation 2; we will call the result Equation 4

We must now eliminate the same variable, x, from another pair of equations. This time we will use equations 1 and 3:

x – 4y – z = 6 Equation 1

3(x – 4y – z) = 3(6)

3x – 12y – 3z = 18

Multiply both sides of Equation 1 by 3

3x – 12y – 3z = 18

-3x + 2y - z = -4

-10y – 4z = 14 Equation 5

Add the result from the previous step to Equation 3; we will call the result Equation 5

We now have two equations, 4 and 5, which contain only the variables y and z. We will now consider these equations as a system of two equations in two variables:

7y + 5z = -12

4(7y + 5z) = 4(-12)

28y + 20z = -48

Multiply both sides of Equation 4 by 4

-10y – 4z = 14

5(-10y – 4z) = 5(14)

-50y - 20z = 70

Multiply both sides of Equation 5 by 5

28y + 20z = -48

-50y – 20z = 70

-22y = 22

y = -1

Add the two equations obtained above; solve for y

Back-substitute y = -1 into either Equation 4 or 5 to find z:

7y + 5z = -12

7(-1) + 5z = -12

-7 + 5z = -12

5z = -5

z = -1

Back-substitute y = -1 and z = -1 into one of the three original equations to find x:

x – 4y – z = 6

x – 4(-1) – (-1) = 6

x + 4 + 1 = 6

x + 5 = 6

x = 1

The solution is (1, -1, -1).

Nonlinear Systems of Equations 

An equation in which one or more terms have a variable of degree 2 or higher is called a nonlinear equation. A nonlinear system of equations contains at least one nonlinear equation.

When solving a system in which one equation is linear, it is usually easiest to use the substitution method. Solve the linear equation for either x or y, then substitute the resulting expression into the nonlinear equation.

Example. Solve the system:

2x + 3y = 12 Equation 1

 Equation 2

Solution.

2x + 3y = 12

2x = 12 – 3y

x = 6 - 

Solve Equation 1, the linear equation, for y

Substitute the expression obtained in step 1, 6

-  , in place of x in Equation 2; simplify; solve for y

144 – 72y + 9y  + 9y  = 144

18y  - 72y = 0

18y(y – 4) = 0

18y = 0 or y – 4 = 0

y = 0 or y = 4

Solve by factoring

x = 6 - 

x = 6 - 

x = 6

So, (6,0) is one solution

Back-substitute each value of y into the expression obtained in step 1:

x = 6 - 

x = 6 - 

x = 6 - 

x = 0

So, (0,4) is another solution.

The addition method works well on nonlinear systems when each equation is in the form

. If necessary, we multiply either equation or both equations by numbers so that the coefficients

of x  or y  will have the sum of 0. We then add the equations. The sum will be an equation in one variable.

Example. Solve:

 Equation 1

 Equation 2

Solution. We will multiply both sides of equation 1 by 3 and both sides of equation 2 by 4.

3(

 Equation 3

4(

 Equation 4

In equations 3 and 4 obtained above, the coefficients of y  are 12 and –12; when we add the equations together, y  will be eliminated.

 Equation 3

 Equation 4

17x  = 68

x  = 4

x =  2

Now we back-substitute each of these values of x into either one of the original equations:

x = 2

y =  1

This means that when x = 2, y can be either 1 or –1. So there are two solutions:

x = -2

y =  1

When x = -2, y can be either 1 or -1, giving us two additional solutions:

(2,1) and (2,-1). (-2, 1) and (-2, -1).

Example. Solve:

xy = 4 Equation 1

 Equation 2

Solution. We will solve this system using the substitution method, and begin by solving Equation 1 for y:

xy = 4

y = 

Substitute   for y in Equation 2:

Equation 2

Substitute   for y

Simplify

Multiply both sides of the equation by x  to eliminate the fraction

2x  + 16 = 18xDistributive Property

2x  - 18x + 16 = 0Solve by factoring

x  - 9x + 8 = 0Divide both sides by 2

(x  - 8)(x  - 1) = 0

x  - 8 = 0 or x  - 1 = 0

x  = 8 or x  = 1

x =   or x = 

x =   or x = 

Each of these values of x will be back-substituted into the expression obtained in step 1,

y =  .

 ;   is a solution

 ;   is a solution

 ;   is a solution

 ;   is a solution

Matrix Solution of Linear Systems 

A matrix (plural: matrices) is a rectangular array of numbers arranged in rows and columns and enclosed in brackets. Each number in the matrix is called an element   of the matrix.

Each of the following is an example of a matrix:

We are going to look at a technique for solving linear systems of equations using matrices.

A matrix derived from a linear system of equations, each in standard form, is called the augmented matrix of the system. The augmented matrix for a linear system is a matrix with the system’s variables and equal signs eliminated. For example:

System

x – 2y – z = 2

2x – y = 4

-x + y – 2z = -4

Augmented Matrix

Each row of the augmented matrix represents one equation of the system. In each row, the coefficient of x is in the first column, the coefficient of y is in the second column, the coefficient of z is in the third column, and the constant is in the fourth column. Notice that if there is a missing term, as the "z" term in the second equation, a zero represents its coefficient in the augmented matrix.

Matrix Row Transformations

For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system:

1. Any two rows may be interchanged.

2. The elements of any row may be multiplied (or divided) by the same nonzero

number.

3. Any row may be changed by adding to its elements a multiple of the

corresponding elements of another row.

To solve a system of equations using the augmented matrix, we will use matrix row transformations to convert the augmented matrix into triangular form.

A matrix is in triangular form if there are 1’s down the main diagonal and 0’s below the diagonal. The following matrix is in triangular form:

The system of equations represented by this matrix is:

x + Ay + Bz = Cy + Dz = E

z = F 

The system is easily solved from this form.

To solve a linear system using matrix row transformations, we begin with the augmented matrix. We then use row transformations to obtain an equivalent matrix with 1’s down the diagonal and 0’s below the diagonal. We will proceed column by column from left to right. In each column, we will start by obtaining a 1 in the diagonal position; then obtain 0’s below the 1. This method is called the Gauss-Jordan method.

Example. Solve:

x + y – z = -2

2x – y + z = 5

-x + 2y + 2z = 1

Solution.

We begin by writing the system as an augmented matrix

We already have a 1 in the diagonal position of first column. Now we want 0’s below the 1. The first 0 can be obtained by multiplying row 1 by -2 and adding the results to row 2 :

Row 1 is unchanged:

-2 times row 1 is added to row 2

Row 3 is unchanged

The second 0 can be obtained by adding row 1 to row 3:

Row 1 is unchanged

Row 2 is unchanged

Row 1 is added to Row 3

Moving to the second column, we want a 1 in the diagonal position (where there is now –3). We get this by dividing every element in row 2 by -3:

Row 1 is unchanged

Row 2 is divided by –3

Row 3 is unchanged

To obtain a 0 below the 1 , we multiply row 2 by -3 and add it to the third row:

Row 1 is unchanged

Row 2 is unchanged

-3 times row 2 is added to row 3

The last step is to obtain a 1 on the diagonal in column 3. We divide each element in Row 3 by 4 to accomplish this:

Row 1 is unchanged

Row 2 is unchanged

Each element in Row 3 is divided by 4

Converting back to a system of equations, we have

x + y – z = -2

y – z = -3

z = 2

We have z = 2. To find y, we back-substitute 2 for z in the second equation:

y – z = -3

y – 2 = -3

y = -1

Finally, we back substitute -1 for y and 2 for z in the first equation:

x + y – z = -2

x + (-1) – 2 = -2

x – 3 = -2

x = 1

The solution is the ordered triple (1, -1, 2).

Example. Solve:

3x – 4y + 4z = 7

x – y – 2z = 2

2x – 3y + 6z = 5

Solution: Start with the augmented matrix:

To obtain a 1 in the top position of the first column, interchange rows 1 and 2:

To get 0’s below the 1: multiply row 1 by -3 and add to row 2; multiply row 1 by –2 and add to row 3

To get a 1 in the diagonal position in the second column, multiply row 2 by –1:

Add row 2 to row 3:

Row 3 represents the equation 0 = 0 which is always true. This indicates that the system is dependent and has infinitely many solutions.

Row 2 represents the equation y – 10z = -1. Solving this equation for y gives us

y = 10z – 1.

Row 1 represents the equation x – y – 2z = 2. Substitute 10z – 1 for y and solve for x:

x – (10z – 1) – 2z = 2

x – 10z + 1 – 2z = 2

x – 12z + 1 = 2

x = 12z + 1

Therefore, every ordered triple of the form (12z + 1, 10z – 1, z) is a solution of the system.

Determinants

If a matrix has the same number of rows and columns, it is called a square matrix. There is a real number associated with every square matrix, and that number is called the determinant. The determinant of matrix A is written as  A .

Definition of the Determinant of a 2 x 2 Matrix

Note: Matrices are enclosed in brackets; determinants are denoted with vertical bars.

Example. Evaluate each determinant:

(a)   (b) 

Solution.

(a)   = 5(-3) – 7(6) = -15 – 42 = -57

(b)   = 2(-5) – (-3)(4) = -10 + 12 = 2

We can evaluate 3 x 3 determinants using a method called expansion by minors.

The minor of an element of a 3 x 3 determinant is the 2 x 2 determinant that remains after you delete the row and column in which the element appears. For example, in the following determinant, we will find the minor of the element 6:

Since the element 6 appears in the first row and the first column, we delete them.

This leaves the 2 x 2 determinant: 

Definition of the Determinant of a 3 x 3 Matrix

When evaluating a 3 x 3 determinant using this definition, keep in mind the following:

1. Each of the three terms is obtained by multiplying an element of the first row by its

minor.

2. A minus sign precedes the second term.

Example. Find the value of the determinant: 

Solution.

= 1(-4 - 2) – 2(8 – 6) + (-3)(2 – (-3))

= 1(-6) – 2(2) + (-3)(5)

= -6 – 4 – 15

= -25

Cramer’s Rule

Determinants can be used to solve a linear system of equations using Cramer’s Rule.

Cramer’s Rule for Two Equations in Two Variables

Given the system 

This system has the unique solution 

where 

When solving a system of equations using Cramer’s Rule, remember the following:

1. Three different determinants are used to find x and y. The determinants in the denominators are identical.

2. The elements of D, the determinant in the denominator, are the coefficients of the variables in the system; coefficients of x in the first column and coefficients of y in the second column.

3.  , the determinant in the numerator of x, is obtained by replacing the x-coefficients,  , in D with the

constants from the right sides of the equations,  .

4.  , the determinant in the numerator for y, is obtained by replacing the y-coefficients,  , in D with the

constants from the right side of the equation,  .

Example. Use Cramer’s Rule to solve the system:

5x – 4y = 2

6x – 5y = 1

Solution. We begin by setting up and evaluating the three determinants  :

From Cramer’s Rule, we have 

The solution is (6,7).

Cramer’s Rule does not apply if D = 0. When D = 0, the system is either inconsistent or dependent. Another method must be used to solve it.

Example. Solve the system:

3x + 6y = -1

2x + 4y = 3

Solution. We begin by finding D:

Since D = 0, Cramer’s Rule does not apply. We will use elimination to solve the system.

3x + 6y = -1

2x + 4y = 3

2(3x + 6y) = 2(-1)

-3(2x + 4y) = -3(3)

Multiply both sides of equation 1 by 2 and both sides of equation 2 by –3 to eliminate x

6x + 12y = -2

-6x – 12y = -9

Simplify

0 = -11 Add the equations

The false statement, 0 = -11, indicates that the system is inconsistent and has no solution.

Cramer’s Rule can be generalized to systems of linear equations with more than two variables. Suppose we are

given a system with the determinant of the coefficient matrix D. Let   denote the determinant of the matrix obtained by replacing the column containing the coefficients of "n" with the constants from the right sides of the equations. Then we have the following result:

If a linear system of equations with variables x, y, z, . . . has a unique solution given by the

formulas 

Example. Use Cramer’s Rule to solve the system:

4x - y + z = -5

2x + 2y + 3z = 10

5x – 2y + 6z = 1

Solution. We begin by setting up four determinants:  :

D consists of the coefficients of x, y, and z from the three equations

 is obtained by replacing the x-coefficients in the first column of D with the constants from the right sides of the equations.

 is obtained by replacing the y-coefficients in the second column of D with the constants from the right sides of the equations

 is obtained by replacing the z-coefficients in the third column of D with the constants from the right sides of the equations

Next, we evaluate the four determinants:

= 4(12 – (-6)) + 1(12 – 15) + 1(-4 – 10)

= 4(18) + 1(-3) + 1(-14)

= 72 – 3 – 14

= 55

= -5(12 – (-6)) + 1(60 – 3) + 1(-20 – 2)

= -5(18)+1(57) + 1(-22)

= -90 + 57 – 22

= -55

= 4(60 – 3) + 5(12 – 15) + 1(2 – 50)

= 4(57) + 5(-3) + 1(-48)

= 228 - 15 – 48

= 165

= 4(2 – (-20)) + 1(2 – 50) – 5(-4 – 10)

= 4(22) + 1(-48) – 5(-14)

= 88 – 48 + 70

= 110

Substitute these four values into the formula from Cramer’s Rule:

The solution is (-1, 3, 2).