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Algebra IIChapter 1.1 Expressions and Formulas
Objectives:β’ Use the order of operations to evaluate expressions.β’ Use formulas.
Recall from junior high mathematics, the order of operations is the acronym PEMDAS.
1. Evaluate everything in parenthesis.2. Evaluate all exponents.3. Evaluate all multiplication and/or division from left to right.4. Evaluate all addition and/or subtraction from left to right.5. There are few exceptions
Parenthesis can look like, () or [] and will usually alternate if parentheses are inside parentheses, e.g. 5(3 + [4 β 2]).
Once in a great while you may see, {}. The use of braces are usually reserved for denoting a set of numbers, e.g. {1,2,3β¦}
This notation makes it easier to read the order of operations and determine which ending parenthesis is paired with which beginning parenthesis.
Algebra IIChapter 1.1 Expressions and Formulas
Objectives:β’ Use the order of operations to evaluate expressions.β’ Use formulas.
Variables are often used in algebra. These variables are usually symbols or letters used to represent unknown values.
Variables also are used to represent any possible value.
Algebraic expressions are mathematical sentences that have at least one variable.
If the variable has a known value, then by substituting that value into the expression will allow us evaluate the expression. (see example 2)
Recall the use of the fraction bar, Ξπ¦
Ξπ₯. The fraction bar not only indicates the operation of
division, it is also used as a grouping symbol; separating the numerator and denominator.
This means, when working with fractions, the numerator and denominator must be evaluated prior to the division indicated by the fraction bar. (see example 3)
A formula is a mathematical sentence that expresses the relationship between certain quantities. If you know all values except one, then that value can be calculated. (see ex. 4)
Bookwork: page 9; problems 16-34 and 38-50 even
Algebra IIChapter 1.2 Properties of Real Numbers
Objectives:β’ Classify real numbers.β’ Use the properties of real numbers to evaluate expressions.
Every number you can think of and use every day is part of the real number system.
Real numbers can be classified as either rational or irrational.
A rational number is a number that can be expressed as a ratio π
π, where m and n are
integers and π β 0. The decimal form of a rational number is either terminating or repeating.
An irrational number is a number whose decimal form neither terminates or repeats.
Other sets of numbers include; natural numbers {1, 2, 3, 4,β¦}, whole numbers {0, 1, 2, 3, 4,β¦}, and integers {-3, -2, -1, 0, 1, 2,β¦}.
FYI: the square root of any whole number is either a whole number or an irrational
number. 36 is a whole number, but 35, lies between 5 and 6, must be irrational.
Algebra IIChapter 1.2 Properties of Real Numbers
Objectives:β’ Classify real numbers.β’ Use the properties of real numbers to evaluate expressions.
The real number system has certain properties that enable us to evaluate expressions.
Property Addition Multiplication
Commutative π + π = π + π π β π = π β π
Associative π + π + π = π + (π + π) π β π β π = π β (π β π)
Identity π + 0 = π = 0 + π π β 1 = π = 1 β π
Inverse π + βπ = 0 = βπ + ππ β 0, π β
1
π= 1 =
1
πβ π
Distributive π π + π = ππ + ππ π + π π = ππ + ππ
Notice there are no properties for subtraction or division.
3 β 4 β 4 β 3 πππ2
23
ππ πππ‘ ππππππ ππππ.
Algebra IIChapter 1.2 Properties of Real Numbers
Objectives:β’ Classify real numbers.β’ Use the properties of real numbers to evaluate expressions.
Simplify the following expression:
2 5π + π + 3(2π β 4π)
= 2 5π + 2 π + 3 2π β 3(4π) Distributive Property
= 10π + 2π + 6π β 12π Multiply
= 10π + 6π + 2π β 12π Commutative Property
= 10 + 6 π + 2 β 12 π Distributive Property
= 16π β 10π Simplify
Bookwork: page 15; problems 19-39 and 43-58; Look at the rest, they may show up as an assessment.
Algebra IIChapter 1.3 Solving Equations
Objectives:β’ Translate verbal expressions into algebraic expressions and back again.β’ Solve equations using properties of equality.
In the real world, mathematical problems are given to us in verbal expressions. For example, we have 2200 lbs. of dirt that needs treated for metals.
In order to solve this mathematically, additional information is needed.
What percentage of the 2200 lbs. of dirt is metals? What metals are present?
Once this information is obtained, mathematical equations can be written.
Seven less than a number π β 7
Three times the square of a number 3π₯2
The cube of a number increased by four times the same number
π3 + 4π
Twice the sum of a number and five 2(π¦ + 5)
Algebra IIChapter 1.3 Solving Equations
Objectives:β’ Translate verbal expressions into algebraic expressions and back again.β’ Solve equations using properties of equality.
A mathematical sentence that contains one or more variables is called an open sentence. A mathematical sentence stating that two expressions are equal is called an equation.
10 = 12 β 2 Ten is equal to twelve minus two.
The sum of a number and negative eight is negative nine.π + β8 = β9A number divided by six is equal to that number squared.π
6= π2
Open sentences are neither true nor false until the variables have been replaced by numbers. If the replacement results in a true statement, the statement is called a solution of the open sentence.
To solve equations, we can use properties of equality. Though similar, do not confuse these properties with the properties of expressions from 1.2.
Algebra IIChapter 1.3 Solving Equations
Objectives:β’ Translate verbal expressions into algebraic expressions and back again.β’ Solve equations using properties of equality.
Property Symbols Example
Reflexive πΉππ πππ¦ ππππ ππ’ππππ π, π = π β7 + π = β7 + π
Symmetric πππ ππππ ππ’πππππ π πππ π,ππ π = π, π‘βππ π = π
ππ 3 = 5π₯ β 6,π‘βππ 5x β 6 = 3
Transitive πΉππ πππ ππππ ππ’πππππ π, π, πππ π,ππ π = π πππ π = π, π‘βππ π = π
ππ 2π₯ + 1 = 7 πππ 7= 5π₯ β 8, π‘βππ 2π₯ + 1 = 5π₯ β 8
Substitution If a = b, then a may be replaced by b and b may be replaced by a.
ππ π₯ = π¦ + 2 πππ 4π₯ + π¦= 18 , π‘βππ 4 π¦ + 2 + π¦ = 18
Addition and Subtraction
For any real numbers a, b, and c, if a = b, then a + c = b + c and a β c = b β c.
If x β 4 = 5, then x β 4 + 4 = 5 + 4.If n + 3 = -11, then n + 3 β 3 = -11 β 3.
Multiplication and Division
For any real numbers a, b, and c, if a =
b, then π β π = π β π ππππ
π=
π
π, π β 0
If π
4= 6, π‘βππ 4 β
π
4= 6 β 4
ππ β 3π¦ = 6, π‘βππβ3π¦
β3=
6
β3
Bookwork: page 24; problems 19-26, 30,32,34, 35-40, 42-62 even. Always look at the word problems 63-74.
Algebra IIChapter 1.4 Solving Absolute Value
EquationsObjectives:β’ Evaluate expressions involving absolute values.β’ Solve absolute equations.
When we look at two points on a number line, 10 and 21, the distance between these two points is, 21 β 10 = 11 π’πππ‘π .
What happens if the two points are -21 and -10? The distance between these two points is β10 β β21 = 11 units.
But wait a moment, the distance could be calculated β21 β β10 = β11 units.
Can we measure distance to be a negative number?
No, you can not travel a negative distance. The negative indicates a direction.
A distance travelled is always a positive number, look at a map.
If the distance between β21 and β10 is β11, then we must use the positive form of the number, 11.
The absolute value of a number is the distance from 0 on a number line. Since distance is non-negative, the absolute value of a number is always non-negative.
Algebra IIChapter 1.4 Solving Absolute Value
EquationsObjectives:β’ Evaluate expressions involving absolute values.β’ Solve absolute equations.
The symbol used to show absolute value is π₯ , which indicates the absolute value of x.
If a is a real number, the absolute value of a is a, if a is a positive or zero number. The absolute value of a is βπ if a is a negative number.
It is important to note that βπ is read as the opposite of a, not negative a.
When evaluating expressions with absolute values, the absolute value bars act as grouping symbols like parentheses. Recall PEMDAS.
Solving absolute value equations can have a second solution. Lets look at an example.
Solve π β ππ = π
The equation is stating that the absolute value of x minus 18 is 5.
This means the distance from π₯ β 18 is five. But what direction? Positive or negative?
Algebra IIChapter 1.4 Solving Absolute Value
EquationsObjectives:β’ Evaluate expressions involving absolute values.β’ Solve absolute equations.
Solve π β ππ = π
If π = π
π₯ β 18 = 5
π₯ β 18 + 18 = 5 + 18
π₯ = 23
If π = βπ
π₯ β 18 = β5
π₯ β 18 + 18 = β5 + 18
π₯ = 13
Lets Check
23 β 18 = 5
5 = 5
5 = 5
13 β 18 = 5
β5 = 5
5 = 5
Algebra IIChapter 1.4 Solving Absolute Value
EquationsObjectives:β’ Evaluate expressions involving absolute values.β’ Solve absolute equations.
Solve ππ β π + π = π
5π₯ β 6 + 9 = 0
5π₯ β 6 = β9
Can this ever be true? No, an absolute value must be a positive number.
Is there a solution? No, the solution set is an empty set, symbolized by ππ β .
Solve π + π = ππ β π
If π = π
π₯ + 6 = 3π₯ β 2
8 = 2π₯
π₯ = 4
If π = βπ
π₯ + 6 = β(3π₯ β 2)
4π₯ = β4
π₯ = β1
If we check our answers, β1 can not be a solution. We must always check our answers!
Bookwork: page30, problems 18-50 even and 51.
Algebra IIChapter 1.5 Solving Inequalities
Objectives:β’ Solve Inequalities
We use inequalities all the time. When we compare phone rate plans or compare sale prices at stores.
When we compare two real numbers a and b, one of three statements must be true:
π < π π = π π > π
This is known as the Trichotomy Property, or the property of order, of numbers.
The properties of addition, subtraction, multiplication, and division work for inequalities as they do for equalities.
Property Words Example
Addition For any real numbers, a, b, and c:If π > π, π‘βππ π + π > π + π.ππ π < π, π‘βππ π + π < π + π.
3 < 53 + β4 < 5 + β4
β1 < 1
Subtraction for any real numbers, a, b, and c:If π > π, π‘βππ π β π > π β π.ππ π < π, π‘βππ π β π < π β π.
2 > β72 β 8 > β7 β 8β6 > (β15)
Multiplication For any real numbers, a, b, and c:ππ π > 0, ππ π > π, π‘βππ π β π > π β π.
ππ π < π, π‘βππ π β π < π β πππ π < 0, ππ π > π, π‘βππ π β π < π β π.
ππ π < π, π‘βππ π β π > π β π
β2 < 34(β2) < 4(3)
5 > β1(β3)(5) < (β3)(β1)
Division For any real numbers, a, b, and c:
ππ π > 0, ππ π > π, π‘βπππ
π>π
π.
ππ π < π, π‘βπππ
π<π
π
ππ π < 0, ππ π > π, π‘βπππ
π<π
π.
ππ π < π, π‘βπππ
π>π
π
β18 < β9
β18
3<β9
3
12 > 8
12
β2<
8
β2
Algebra IIChapter 1.5 Solving Inequalities
Objectives:β’ Solve Inequalities
What the table shows us is that when we multiply or divide by a negative number, we must reverse the inequality sign.
These rules are also true for the inequalities β€ πππ β₯.
How do we express the solution set of an inequality?
First, we can graph the solution.
If π₯ > 2, π‘βππ
Notice that the inequality is not an equal sign; therefore, the starting point is an open point.
If π₯ β€ β1, π‘βππ
Notice that the inequality has an quality sign; therefore, the starting point is a closed point.
Algebra IIChapter 1.5 Solving Inequalities
Objectives:β’ Solve Inequalities
Second, we can use the set-builder notation.
To express the solution π₯ > 9, we can write π₯ π₯ > 9 .
This is read, the set of all numbers x such that x is greater than 9.
This notation can be used for all inequalities.
Third, we can use interval notation.
To express π₯ < 2, we can use the infinity symbols, β.
If the endpoint is not to be included, we will use a parentheses, ( or ).
If the endpoint is to be included, we use a bracket, [ or ].
Parentheses are always used for ββ πππ + β.
I show you these notations so when you see them on state assessments, you will recognize them. Only through practice will you get good at using them.
Bookwork: page 37, problems 15-46, look at 47-53
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
In English class we learn that a compound sentence is a sentence with two sentences, usually combined with the word and or the word or.
A compound inequality is an expression with two inequalities joined by the word and or the word or.
To solve for an and inequality, we must solve for each part separately and graph the intersection. Compound inequalities with and are also called conjunctions. Remember that the word and is a conjunction word in English, meaning to join two ideas.
To solve for an or inequality, we must solve for the union of both of the inequalities. Compound inequalities with or are also called disjunctions. Remember that you can not be both inside the room and outside the room; however, you can be inside the room or outside the room.
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
βANDβ Compound Inequalities: A compound inequality with the word and, or a mathematical expression with the implied meaning of and, is true if and only if both expressions are true.
π₯ β₯ β1
π₯ < 2
π₯ β₯ β1 πππ π₯ < 2
An easier way of writing this inequality is β1 β€ π₯ < 2
Lets look at a second example.
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
ππ¨π₯π―π ππ < ππ + π β€ πππππ πππππ πππ ππππππππ.
6 < 2π₯
13< 2π₯ + 7 πππ 2π₯ + 7 β€ 17
3 < π₯ β€ 5
Method 1
Insert the βandβ, then solve each inequality .
3 < π₯
2π₯ β€ 10
π₯ β€ 5
Method 2
Solve both parts at the same time.
13 < 2π₯ + 7 β€ 17
6 < 2π₯ β€ 10
3 < π₯ β€ 5
The solution set is π₯ 3 < π₯ β€ 5
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
βORβ Compound Inequalities: A compound inequality with the word or, or a mathematical expression with the or, is true if one or more of the expressions are true.
π₯ β€ 1
π₯ > 4
π₯ β€ 1 ππ π₯ > 4
Unlike the and statement, there is no easier way to write the inequality.
Lets look at a second example.
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
ππ¨π₯π―π π β π > βπ ππ π + π β€ βπ πππ πππππ πππ ππππππππ.
π¦ β 2 > (β3)
Must solve each inequality separately.
π¦ > (β1)
y + 4 β€ (β3)
π¦ β€ (β7)
The solution set is π¦ π¦ > β1 πππ¦ β€ (β7)
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
What happens when you have absolute values?
If π > π, π‘βππ π > π ππ π < βπ.
π < 4 ππ β π < 4
Recall that π could be +a or -a.
If π < π, π‘βππ β π < π < π.
Solve π < π β Absolute Value and a (<)
This is equivalent to π < 4 πππ π > (β4)
Solve π > π - Absolute Value and a (>)
π > 4 ππ β π > 4; meaning, π > 4 ππ π < (β4).
Algebra IIChapter 1.6 Compound and Absolute
Value InequalitiesObjectives:β’ Solve compound inequalitiesβ’ Solve absolute value inequalities
ππ¨π₯π―π ππ β ππ β₯ π πππ πππππ πππ ππππππππ.
3π₯ β 12 β₯ 6 or 3π₯ β 12 β€ (β6)
An absolute value with a > sign means an or statement. Must solve each inequality separately.
3π₯ β₯ 18 3π₯ β€ 6
π₯ β₯ 6 π₯ β€ 2
Bookwork: page 44; problems 16-52 even.