Chapter 1 By: Matt Raimondi Algebra II Chapter 1 By: Matt Raimondi

1-1 Expressions and Formulas Order of operations First, always try to simplify expressions in brackets, parentheses, and fraction bars. Next, evaluate all numbers raised to a power or under a radical. Do all multiplication/ division from left to right. Then do all adding and subtracting from left to right. An easy way to remember this is by remembering: Please Excuse My Dear Aunt Sally Parentheses, Exponents, Multiplication, Division, Addition, Subtraction

1-1 Example First we have to simplify the brackets. We subtract 9-6 Multiply 3*2 Evaluate what is left in the brackets. Multiply –1 by 5 to get rid of the brackets. Evaluate 32 Subtract 5 and 9 from 2. 2+5{2-3*2+(9-6)} – 32 2+5{2-3*2+(3)} – 32 2+5{2-6+(3)} – 32 2+5{-1} – 32 2+(-5) – 32 2-5-9 = -12

1-1 Problems 1. 2*5-4+6/3 2. (12/4/3-1)100 3. 5-2(2+6/3) 1. 2*5-4+6/3 2. (12/4/3-1)100 3. 5-2(2+6/3) 4. 1+4*6+5-10/2 5. (5*2-5^2)+10 6. 2^2 +5*4/2-8 4. 25 5. -5 6. 6 1. 6 2. 0 3. -3 Answers:

1-2 Properties of Real Numbers Types of numbers Real Numbers – All numbers that you can encounter in real life. Rational – Rational numbers can be expressed as a fraction, repeating decimal, or a terminating decimal. ie .5, ½, 1.33333333….. Irrational – Numbers that cannot be expressed like rational numbers. ie π, √5, √2 Integers – { ….-3,-2,-1,0,1,2,3…..} Wholes – {0,1,2,3,4……} Naturals – {1,2,3,4,5…..}

1-2 Properties cont. Addition Multiplication Commutative A+B=B+A Associative (A+B)+C=A+(B+C) (A*B)*C=A*(B*C) Identity A+0=A A*1=A Inverse A+(-A)=0 If A ≠0 A* 1/A=1=1/A*A Distributive A(B+C)=AB+AC and (B+C)A=BA+CA

1-2 Examples Identify all of the groups the following numbers belong to. 1) √13 (reals and irrationals) 2) 7 (reals, rationals, integers, wholes, and naturals) 3) ½ (reals and rationals) 4) 0 (reals, rationals, integers, and wholes) 5) -1.978 (reals and rationals)

1-2 Problems 1) 28 2) -1 3) π 4) 2.79 5) 1/3 6) √12 7) √9 Identify all of the groups the following numbers belong to. 1) 28 2) -1 3) π 4) 2.79 5) 1/3 6) √12 7) √9 1 reals, rationals, integers, wholes, naturals 2)reals, rationals, integers 3)reals irrationals 4) reals, rationals 5) reals, rationals 6)reals, irrationals 7) reals, rationals, integers, wholes, naturals

1-2 Problems cont. Identify the following properties 1) 5+2=2+5 1) 5+2=2+5 2) (2*12)*3=2*(12*3) 3) 4*1=4 4) 5(a+b)=5a+5b 5) 6+0=6 6) ½ * 2=1 7) 10*5=5*10 8) 15+ (-15)=0 1) Commutative 2) Associative 3) Multiplicative identity 4) Distributive 5) Addition identity 6) Multiplicative inverse 7) Commutative 8) Addition inverse

1-3 Measures of Central Tendency The median of a set of numbers is the middle term. If there is an even number of terms, then add the two middle terms and divide by two. The mode of a set of numbers is the term that appears most frequently. The mean of a set of numbers is the average of the set of numbers.

1-3 Example Use the following set of test scores to find the mean, median and mode. 58,74,82,95,63,84,75,82,64,99,84,84,48 To find the median we should first rearrange the scores in order to see which term is the middle term. We then get 48,58,63,64,74,75,82,82,84,84,84,95,99 The middle term is 82, so the median is 82. To find the mode, look and see which term appears the most. 84 is the only score that appears 3 times. So, the mode is 84 To find the mean, we need the algebraic average of the terms. So we add up all of the terms and divide by how many there are. There are a total of 13 terms. The total divided by 13 gives up approximately 76%. So the mean of the test scores is 76%.

1-3 Problems Find the mean 10,8,2,17,3 2)12,14,5,2,6,40,5 10,8,2,17,3 2)12,14,5,2,6,40,5 12,8,5,5,32,7,10,3,7,11 Find the mode 4) 12,2,3,10,17,19,2,14 5) 3,10,9,3,3,10,4,18,10,9,3,16 Find the median 50,78,58,90,90,63,84,71,70,100 99,75,50,86,88,63,64,70,76,99,73 1) 8 2) 12 3) 10 4) 2 5)3 6) 74.5 7) 75

1-4 Practice Equations Solve for the unknown variable: 6) 30-10x=0 7) 2x+5=-15 8) 2x-8=5x+19 9) 6w-2(2-w)=5 10) 2x+.5=20-9x Evaluate the following: 1) X=5 2) x=2 9x+7-3x 2-x+10x 3) y=3 4) x=9 y(2+5y) (x+3)/(x-5) 5) x=2 y=4 xy+3y-2x 1) 37 2)20 3)51 4)3 5)16 6)x=3 7)x=-10 8)x=-9 9)w=9/8 10)x=20.5/11=41/22

1-5 Solving Absolute Value Equations The absolute value of a number is the number of units it is from zero on the number line. For example, the absolute value of -3 is 3 because it is 3 units away from zero on the number line. The absolute value of positive numbers and zero is always the number. The absolute value of negative numbers is the same as the number without the negative sign. The absolute value is never a negative number. This is because a distance cannot be negative.

1-5 Solving Absolute Value Equations Lets take a look at an example equation: Abs(x+3)=5 X+3=5 and x+3=-5 -3 -3 -3 -3 X=2 and x=-8 Now to check it we test each solution Abs(2+3) and abs(-8+3)=5 Abs(5)=5 and abs(-5)=5 Since both of the equations are true, 2 and –8 are both answers. Look at the absolute values of the following numbers: Abs(2)=2 Abs(17.94)=17.94 Abs(-1/2)=1/2 Abs(-90)=90 Abs(0)=0 Abs(22.2)=22.2 Abs(-x)=x

1-5 Problems Solve the equations Abs(-10)=x Abs(x)=4 6) Abs(3x)=6 Explain why this equation cannot be solved. Abs(x+2)=-7 1)-4,4 2)13,-11 3)5,-13 4)-4,4 5)10 6)2,-2 7)the absolute value can never be a negative number

1-6 Solving Inequalities Solving inequalities is not much different than solving regular equations. Numbers can be added and subtracted the same as in equalities. Multiplication and division is the same with only one exception. Whenever a term is multiplied or divided by a negative term, it reverses the sign. So, a greater than sign turns into a less then sign. After solving an inequality, make sure to test your answer and see if it is true. Errors often occur when changing signs.

1-6 Examples 1) 5x-5 < 10 5x < 15 x < 3 5(0) -5 <10 -5 <10 14 < 1-3x 13 < -3x -13/3 > x 14 < 1-3(-10) 14 < 1+30 14 < 31 After getting x is less than three, we should check it with a value less than three. Zero is an easy number to use. As you can see, zero is a possible solution to the inequality. This equation has a step with division by a negative number. Notice the sign change. Lets try –10 to check it.

1-6 Problems Solve the following inequalities 5+x < 2 2) 7x+2>16 5) 3x+9-5x > -41 6) 2x+17 < 53 1) x <-3 2) x >2 3) x > 17 4) x < -22/3 5) x < 25 6) x < 18

1-6 Graphing Inequality Examples After finding the solution to an inequality, graphing generally follows. Graphs of inequalities only lie on the real number line. Greater than and less than statements have open circles because they do not include the solution. Greater than or equal to and less than or equal to statements have closed circles because they include the solution. A little trick to remember which way the arrow goes is to make sure your variable is on the left side. Then just look which way the point of the inequality sign points and the graph goes that way. For example lets look at x > -2 and x ≤ 4

1-6 Graphing Inequality Problems Solve and graph the following: x ≥ 4 x ≤ -1 x < 5 -3 > x 2x-5 < 7 12 +x ≥ -3x - 4 1) Solid @ 4, line to the right 2)solid @ -1, line to the left 3) open @ 5, line to the left 4) open @ -3, line to the left 5) open @ 6, line to the left 6) solid @ -4, line to the right

1-7 Solving Absolute Value Inequalities When you solve an absolute value inequality, you will end up with two parts to your answer. If the inequality began as a “greater than” statement, you will have an “or” statement in your final answer. If it began as a “less than” statement, you will have an “and” statement in your final answer.

1-7 Examples Abs(x+2) < 8 X+2 < 8 and x+2 > -8 So x is greater than –10 and less than 6 Abs(x-5) > 12 X-5 > 12 or x-5 < -12 X > 17 or x <-7 So x is less than –7 or greater than 17. When an absolute value inequality has an “and” statement the solution is a contained set. In the first example, x is between –10 and 6. On the other hand, when an absolute value inequality is an “or” statement, the set of numbers is not contained. In the second example, x could be less than –7 or greater than 17.

1-7 Graphing Absolute Value Inequalities Graphing absolute value inequalities is the same as graphing regular inequalities but with two parts. If the solution is an “or” statement the graph will have an open area in the middle that the graph doesn’t touch. It will continue to the left and right on the number line. When the solution is an “and” statement it will have a closed set. The solution will be between two numbers on the line.

1-7 Graphing Examples Lets look at the first example from earlier. We found the answer to be: x < 6 and x > -10 So this will be a closed set between -10 and 6. Now lets look at the second example from before. We found the solution to be: X > 17 or x <-7 So this will be an open set containing numbers greater than 17 or less than -7.

1-7 Problems State if the following are “and” or “or” statements: Abs(x+4) > 5 2) Abs(2+x) ≤ 19 3) Abs(x) >2 Solve the following inequalities 4) Abs(4x) ≥ 10 5) Abs(x-4) < 12 6) Abs(2x+2) ≤ 18 1) or 2) and 3) or 4) x ≤-5/2 or x ≥5/2 5) x >-8 and x < 16 6) x≤8 and x ≥ -10

1-7 Graphing Problems Graph the problems from the last page 1) Open dots @ 1 and -9, open set 2) closed dots @ 17 and -21, closed set 3) open dots @ 2 and -2, open set 4) closed dots @ -5/2 and 5/2, open set 5) open dots @ -8 and 16, closed set 6) closed dots @ 8 and -10, closed set Answers: