Review Topics Ch R 1 in College Algebra

Review Topics (Ch R & 1 in College Algebra Book) • Exponents & Radical Expressions (P. 21 -25 and P. 72 -77) • Complex Numbers (P. 109 – 114) • Factoring (p. 49 – 55) • Quadratic Equations (P. 97 – 105) • Rational Expressions (P. 61 – 69) • Rational Equations & Clearing Fractions (P. 88 – 91) • Radical Equations (P. 118 – 123) • 1. 5: Solving Inequalities • 1. 6: Equations and Inequalities involving absolute value

Review of Exponents 82 =8 • 8 = 64 24 = 2 • 2 • 2 = 16 x 2 = x • x Base = x Exponent = 2 x 4 = x • x • x Base = x Exponent = 4 Exponents of 1 Anything to the 1 power is itself 51 = 5 Zero Exponents Anything to the zero power = 1 x 1 = x (xy)1 = xy 50 = 1 x 0 = 1 (xy)0 = 1 Negative Exponents 5 -2 = 1/(52) = 1/25 x-2 = 1/(x 2) xy-3 = x/(y 3) a-n = 1/an 1/a-n = an a-n/a-m = am/an (xy)-3 = 1/(xy)3 = 1/(x 3 y 3)

Powers with Base 10 100 = 1 101 = 10 102 = 100 103 = 1000 104 = 10000 The exponent is the same as the number of 0’s after the 1. 100 = 1 10 -1 = 1/101 10 -2 = 1/102 10 -3 = 1/103 10 -4 = 1/1000 = 1/10000 =. 1 =. 001 =. 0001 The exponent is the same as the number of digits after the decimal where 1 is placed Scientific Notation uses the concept of powers with base 10. Scientific Notation is of the form: __. ______ x 10 (** Note: Only 1 digit to the left of the decimal) You can change standard numbers to scientific notation You can change scientific notation numbers to standard numbers

Scientific Notation uses the concept of powers with base 10. 5 ______ 321 Scientific Notation is of the form: __. x 10 (** Note: Only 1 digit to the left of the decimal) -2 Changing a number from scientific notation to standard form 5. 321 Step 1: Write the number down without the 10 n part. Step 2: Find the decimal point Step 3: Move the decimal point n places in the ‘number-line’ direction of the sign of the exponent. . 05321 Step 4: Fillin any ‘empty moving spaces’ with 0. Changing a number from standard form to scientific notation Step 1: Locate the decimal point. Step 2: Move the decimal point so there is 1 digit to the left of the decimal. Step 3: Write new number adding a x 10 n where n is the # of digits moved left adding a x 10 -n where n is the #digits moved right. 0 5 3 2 1 = 5. 321 x 10 -2

Raising Quotients to Powers a b n Examples: an = bn a b 3 4 2 2 x y 3 = 32 42 = (2 x)3 y 3 -3 = -n a-n b-n = = bn an = b a y 3 (2 x)3 y 3 8 x 3 9 = 16 8 x 3 y 3 = (2 x)-3 y-3 = 1 y-3(2 x)3 = = n

Product Rule am • an = a(m+n) x 3 • x 5 = xxx • xxxxx = x 8 x-3 • x 5 = xxxxx = x 2 xxx 1 x 4 y 3 x-3 y 6 = xxxx • yyyyyy = xy 9 xxx 3 x 2 y 4 x-5 • 7 x = 3 xxyyyy • 7 x = 21 x-2 y 4 = 21 y 4 xxxxx x 2

Quotient Rule am = a(m-n) an 43 = 4 • 4 = 41 = 4 42 4 • 4 43 42 x 5 = xxxxx = x 3 x 2 xx x 5 = x(5 -2) = x 3 x 2 15 x 2 y 3 = 15 xx yyy = 3 y 2 5 x 4 y 5 xxxx y x 2 3 a-2 b 5 = 3 bbbbb = b 8 9 a 4 b-3 9 aaaa aa 3 a 6 = 64 16 15 x 2 y 3 5 x 4 y 3 a-2 b 5 9 a 4 b-3 = 8 2 = 4 = 3 • x -2 • y 2 = 3 y 2 x 2 = a(-2 -4)b(5 -(-3)) = a-6 b 8 = b 8 3 3 3 a 6

Powers to Powers (am)n = amn (a 2)3 a 2 • a 2 = aa aa aa = a 6 (24)-2 = 1 (24)2 (x 3)-2 = x – 6 (x -5)2 x – 10 = 1 =1 = 1/256 24 • 24 16 • 16 = x 10 = x 4 x 6 (24)-2 = 2 -8 = 1 28 256

Products to Powers (ab)n = anbn (6 y)2 = 62 y 2 = 36 y 2 (2 a 2 b-3)2 = 22 a 4 b-6 = 4 a 4 = a 4(ab 3)3 4 a 3 b 9 b 6 b 15 What about this problem? 5. 2 x 1014 3. 8 x 105 = 5. 2/3. 8 x 109 1. 37 x 109 Do you know how to do exponents on the calculator?

Square Roots & Cube Roots A number b is a square root of a number a if b 2 = a A number b is a cube root of a number a if b 3 = a 25 = 5 since 52 = 25 3 Notice that 25 breaks down into 5 • 5 So, 25 = 5 • 5 Notice that 8 breaks down into 3 2 • 2 So, 8 = 2 • 2 See a ‘group of 2’ -> bring it outside the radical (square root sign). See a ‘group of 3’ –> bring it outside the radical (the cube root sign) Example: 200 = 2 • 10 = 10 2 Note: -25 is not a real number since no number multiplied by itself will be negative 8 = 2 since 23 = 8 3 3 Example: 200 = 2 • 100 3 = 2 • 10 3 = 2 • 5 • 2 3 = 2 • 2 • 5 3 = 2 25 3 Note: -8 IS a real number (-2) since -2 • -2 = -8

Nth Root ‘Sign’ Examples 16 -16 not a real number = 4 or -4 Even radicals of positive numbers Have 2 roots. The principal root Is positive. Even radicals of negative numbers Are not real numbers. 4 -16 not a real number 5 -32 = -2 Odd radicals of negative numbers Have 1 negative root. 5 32 = 2 Odd radicals of positive numbers Have 1 positive root.

Exponent Rules (XY)m = xmym m X Y Xm = m Y

Examples to Work through

Product Rule and Quotient Rule Example

Some Rules for Simplifying Radical Expressions

Example Set 1

Example Set 2

Example Set 3

Operations on Radical Expressions • Addition and Subtraction (Combining LIKE Terms) • Multiplication and Division • Rationalizing the Denominator

Radical Operations with Numbers

Radical Operations with Variables

Multiplying Radicals (FOIL works with Radicals Too!)

Rationalizing the Denominator • Remove all radicals from the denominator

Rationalizing Continued… • Multiply by the conjugate

Complex Numbers REAL NUMBERS Rational Numbers (1/2 – 7/11, 7/9, . 33 Integers (-2, -1, 0, 1, 2, 3. . . ) Whole Numbers (0, 1, 2, 3, 4. . . ) Natural Numbers (1, 2, 3, 4. . . ) Irrational Numbers , 8, - 13 Imaginary Numbers

Complex Numbers (a + bi) Real Numbers a + bi with b = 0 Rational Numbers Irrational Numbers Integers Whole Numbers Natural Numbers Imaginary Numbers a + bi with b 0 i = -1 where i 2= -1

Simplifying Complex Numbers A complex number is simplified if it is in standard form: a + bi Addition & Subtraction) Ex 1: (5 – 11 i) + (7 + 4 i) = 12 – 7 i Ex 2: (-5 + 7 i) – (-11 – 6 i) = -5 + 7 i +11 + 6 i = 6 + 13 i Multiplication) Ex 3: 4 i(3 – 5 i) = 12 i – 20 i 2 = 12 i – 20(-1) = 12 i +20 = 20 + 12 i Ex 4: (7 – 3 i) (-2 – 5 i) [Use FOIL] -14 – 35 i +6 i +15 i 2 -14 – 29 i +15(-1) -14 – 29 i – 15 -29 – 29 i

Complex Conjugates The complex conjugate of (a + bi) is (a – bi) The complex conjugate of (a – bi) is (a + bi) (a – bi) = a 2 + b 2 Division 7 + 4 i 2 – 5 i 2 + 5 i = 2 + 5 i 14 + 35 i + 8 i + 20 i 2 = 14 + 43 i +20(-1) 4 + 10 i – 25 i 2 4 – 25(-1) 14 + 43 i – 20 = -6 + 43 i = -6 43 i + 4 + 25 29 29 29

Square Root of a Negative Number 25 4 = 100 = 10 -25 -4 = (-1)(25) (-1)(4) Optional Step = (i 2)(25) (i 2)(4) = i 25 i 4 = (5 i) (2 i) = 10 i 2 = 10(-1) = -10

Practice – Square Root of Negatives

Practice – Simplify Imaginary Numbers i 0 = i 1 = 1 i i 2 = -1 Another way to calculate in i 3 = -i Divide n by 4. If the remainder is r then in = ir i 4 = 1 Example: i 11 = _____ i 5 = i 11/4 = 2 remainder 3 i 6 -1 So, i 11 = i 3 = -i =

Practice – Simplify More Imaginary Numbers

Practice – Addition/Subtraction 10 +8 i -4 +10 i

Practice – Complex Conjugates • Find complex conjugate. 3 i => -4 i =>

Practice Division w/Complex Conjugates 4__ 2 i =

Adding & Subtracting Polynomials Combine Like Terms (2 x 2 – 3 x +7) + (3 x 2 + 4 x – 2) = 5 x 2 + x + 5 (5 x 2 – 6 x + 1) – (-5 x 2 + 3 x – 5) (5 x 2 – 6 x + 1) + (5 x 2 - 3 x + 5) 10 x 2 – 9 x + 6 Types of Polynomials f(x) = 3 f(x) = 5 x – 3 f(x) = x 2 – 2 x – 1 f(x) = 3 x 3 + 2 x 2 – 6 = = Degree 0 Degree 1 Degree 2 Degree 3 Constant Function Linear Quadratic Cubic

Multiplication of Polynomials Step 1: Using the distributive property, multiply every term in the 1 st polynomial by every term in the 2 nd polynomial Step 2: Combine Like Terms Step 3: Place in Decreasing Order of Exponent 4 x 2 (2 x 3 + 10 x 2 – 2 x – 5) = 8 x 5 + 40 x 4 – 8 x 3 – 20 x 2 (x + 5) (2 x 3 + 10 x 2 – 2 x – 5) = 2 x 4 + 10 x 3 – 2 x 2 – 5 x + 10 x 3 + 50 x 2 – 10 x – 25 = 2 x 4 + 20 x 3 + 48 x 2 – 15 x -25

Binomial Multiplication with FOIL (2 x + 3) (x - 7) F. (First) O. (Outside) I. (Inside) L. (Last) (2 x)(x) (2 x)(-7) (3)(x) (3)(-7) 3 x -21 2 x 2 -14 x 2 x 2 + - 11 x 3 x -21

Division by a Monomial 3 x 2 + x x 4 x 2 + 8 x – 12 4 x 2 15 A 2 – 8 A 2 + 12 4 A 5 x 3 – 15 x 2 15 x 5 x 2 y + 10 xy 2 5 xy 12 A 5 – 8 A 2 + 12 4 A

Review: Factoring Polynomials To factor a number such as 10, find out ‘what times what’ = 10 10 = 5(2) To factor a polynomial, follow a similar process. Factor: 3 x 4 – 9 x 3 +12 x 2 3 x 2 (x 2 – 3 x + 4) Another Example: Factor 2 x(x + 1) + 3 (x + 1)(2 x + 3)

Solving Polynomial Equations By Factoring Zero Product Property : If AB = 0 then A = 0 or B = 0 Solve the Equation: 2 x 2 + x = 0 Step 1: Factor x (2 x + 1) = 0 Step 2: Zero Product x=0 or 2 x + 1 = 0 Step 3: Solve for X x=0 or x= -½ Question: Why are there 2 values for x? ? ?

Factoring Trinomials To factor a trinomial means to find 2 binomials whose product gives you the trinomial back again. Consider the expression: x 2 – 7 x + 10 The factored form is: (x – 5) (x – 2) Using FOIL, you can multiply the 2 binomials and see that the product gives you the original trinomial expression. How to find the factors of a trinomial: Step 1: Write down 2 parentheses pairs. Step 2: Do the FIRSTS Step 3 : Do the SIGNS Step 4: Generate factor pairs for LASTS Step 5: Use trial and error and check with FOIL

Practice Factor: 1. y 2 + 7 y – 30 4. – 15 a 2 – 70 a + 120 2. 10 x 2 +3 x – 18 5. 3 m 4 + 6 m 3 – 27 m 2 3. 8 k 2 + 34 k +35 6. x 2 + 10 x + 25

Special Types of Factoring Square Minus a Square A 2 – B 2 = (A + B) (A – B) Cube minus Cube and Cube plus a Cube (A 3 – B 3) = (A – B) (A 2 + AB + B 2) (A 3 + B 3) = (A + B) (A 2 - AB + B 2) Perfect Squares A 2 + 2 AB + B 2 = (A + B)2 A 2 – 2 AB + B 2 = (A – B)2

Quadratic Equations General Form of Quadratic Equation ax 2 + bx + c = 0 a, b, c are real numbers & a 0 A quadratic Equation: x 2 – 7 x + 10 = 0 a = _____ 1 b = _____ -7 c = ______ 10 Methods & Tools for Solving Quadratic Equations 1. Factor 2. Apply zero product principle (If AB = 0 then A = 0 or B = 0) 3. Square root method Example 1: Example 2: 4. Completing the Square x 2 – 7 x + 10 = 0 4 x 2 – 2 x = 0 5. Quadratic Formula (x – 5) (x – 2) = 0 2 x (2 x – 1) = 0 x– 5 = 0 +5 +5 or x– 2 = 0 +2 +2 x = 5 or x= 2 2 x=0 or 2 x-1=0 2 2 +1 +1 2 x=1 x = 0 or x=1/2

Square Root Method If u 2 = d then u = d or u = - d. If u 2 = d then u = + d Solving a Quadratic Equation with the Square Root Method Example 1: Example 2: 4 x 2 = 20 (x – 2)2 = 6 4 4 x – 2 = + 6 x 2 = 5 +2 +2 x = + 5 So, x = 5 or - 5 x = 2 + 6 So, x = 2 + 6 or 2 - 6

Completing the Square (Example 1) If x 2 + bx is a binomial then by adding b 2 which is the square of half 2 the coefficient of x, a perfect square trinomial results: x 2 + bx + b 2 = x + b 2 2 2 Solving a quadratic equation with ‘completing the square’ method. Example: x 2 - 6 x + 2 = 0 -2 -2 x 2 - 6 x = -2 x 2 - 6 x + 9 = -2 + 9 (x – 3)2 = 7 x – 3 = + 7 Step 1: Isolate the Binomial x = (3 + 7 ) or (3 - 7 ) Step 3: Apply square root method Step 2: Find ½ the coefficient of x (-3 ) and square it (9) & add to both sides. Note: If the coefficient of x 2 is not 1 you must divide by the coefficient of x 2 before completing the square. ex: 3 x 2 – 2 x – 4 = 0 (Must divide by 3 before taking ½ coefficient of x)

(Completing the Square – Example 2) Step 1: Check the coefficient of the x 2 term. If 1 goto step 2 If not 1, divide both sides by the coefficient of the x 2 term. Step 2: Calculate the value of : (b/2)2 [In this example: (2/2)2 = (1)2 = 1] Step 3: Isolate the binomial by grouping the x 2 and x term together, then add (b/2)2 to both sides of he equation. 2 +4 x – 1 = 0 2 x Step 4: Factor & apply square root method 2 x 2 +4 x – 1 = 0 2 2 (x + 1) = 3/2 (x + 1)2 = 3/2 x 2 +2 x – 1/2 = 0 (x 2 +2 x )=½ (x 2 +2 x + 1 ) = 1/2 + 1 √(x + 1)2 = √ 3/2 x + 1 = +/- √ 6/2 x = √ 6/2 – 1 or - √ 6/2 - 1

Quadratic Formula General Form of Quadratic Equation: ax 2 + bx + c = 0 Quadratic Formula: x = -b + b 2 – 4 ac 2 a discriminant: b 2 – 4 ac if 0, one real solution if >0, two unequal real solutions if <0, imaginary solutions Solving a quadratic equation with the ‘Quadratic Formula’ 2 x 2 – 6 x + 1= 0 2 a = ______ -6 b = ______ x = - (-6) + (-6)2 – 4(2)(1) 2(2) = 6 + 36 – 8 4 = 6 + 2 7 4 = 2 (3 + 7 ) 4 = (3 + 7 ) 2 1 c = _______

Solving Higher Degree Equations x 3 = 4 x 2 x 3 + 2 x 2 - 12 x = 0 x 3 - 4 x = 0 x (x 2 – 4) = 0 x (x – 2)(x + 2) = 0 2 x (x 2 + x – 6) = 0 x=0 2 x (x + 3) (x – 2) = 0 x– 2=0 x+2=0 2 x = 0 or x + 3 = 0 or x – 2 = 0 x=2 x = -2 x=0 or x = -3 or x=2

Solving By Grouping x 3 – 5 x 2 – x + 5 = 0 (x 3 – 5 x 2) + (-x + 5) = 0 x 2 (x – 5) – 1 (x – 5) = 0 (x – 5)(x 2 – 1) = 0 (x – 5)(x – 1) (x + 1) = 0 x– 5=0 or x-1=0 or x+1=0 x=5 or x=1 or x = -1

Rational Expressions Rational Expression – an expression in which a polynomial is divided by another nonzero polynomial. Examples of rational expressions 4 x x 2 x – 5 Domain = {x | x 0} Domain = {x | x 5/2} 2 x– 5 Domain = {x | x 5}

Multiplication and Division of Rational Expressions A • C = A B • C B 9 x 3 x 2 5 y – 10 = 5 (y – 2) 10 y - 20 10 (y – 2) 2 z 2 – 3 z – 9 z 2 + 2 z – 15 A 2 – B 2 A+B 3 x = 5 =1 10 2 = (2 z + 3) (z – 3) (z + 5) (z – 3) = = (A + B)(A – B) (A + B) = 2 z + 3 z+5 = (A – B)

Negation/Multiplying by – 1 -y – 2 4 y + 8 = y+2 4 y + 8 OR -y - 2 -4 y - 8

Check Your Understanding Simplify: x 2 – 6 x – 7 x 2 -1 1 x-2 (x + 1) (x – 7) (x – 1) 1 x– 2 3 x 2 + x - 6 3 • • (x + 3) 3 (x + 3) (x – 2) 3

Addition of Rational Expressions Adding rational expressions is like adding fractions With LIKE denominators: 1 8 + x x+2 x 3 x 2 + 4 x - 4 2 8 + + = 3 x - 1 x+2 = 3 8 4 x - 1 x+2 2 (2 + x) = = 3 x 2 + 4 x -4 (3 x 2 + 4 x – 4) (3 x -2)(x + 2) = 1 (3 x – 2)

Adding with UN-Like Denominators 3 4 1 x 2 – 9 + 1 8 (3) (2) 8 + 6 8 1 8 + 7 8 1 8 + 2 x+3 1 (x + 3)(x – 3) + 2 (x + 3) 1 (x + 3)(x – 3) + 2 (x – 3) (x + 3)(x – 3) 1 + 2(x – 3) (x + 3) (x – 3) = 1 + 2 x – 6 (x + 3) (x – 3) = 2 x - 5 (x + 3) (x – 3)

Subtraction of Rational Expressions To subtract rational expressions: Step 1: Get a Common Denominator Step 2: Combine Fractions DISTRIBUTING the ‘negative sign’ BE CAREFUL!! 2 x x 2 – 1 = - x– 1 (x + 1)(x – 1) x+1 x 2 - 1 = = 2 x – (x + 1) x 2 -1 1 (x + 1) = 2 x 2 – x - 1 x -1

Check Your Understanding Simplify: b 2 b - 4 b-1 b-2 - b 2(b – 2) - b-1 b-2 b 2(b – 2) + -b+1 b-2 b 2(b – 2) + 2(-b+1) 2(b – 2) b – 2 b+2 2(b – 2) = -b + 2 2(b – 2) = -1(b – 2) 2(b – 2) = -1 2

Complex Fractions A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. Examples: 1 5 x x 2 – 16 1 x + 2 x 2 4 7 1 x-4 3 x - 1 x 2 7/20 x x+4 x+2 3 x - 1

Rational Equations (2 x – 1) 3 x 2 x – 1 (x - 2) = 3 3 x = 3(2 x – 1) 3 x = 6 x – 3 -3 x = -3 x+1 x– 2 (x + 1) = 3 x-2 6 x+1 =x x+1=3 6 = x (x + 1) x=2 6 = x 2 + x x=1 x 2 + x – 6 = 0 (x + 3 ) (x - 2 ) = 0 Careful! – What do You notice about the answer? x = -3 or x=2

Rational Equations Cont… To solve a rational equation: Step 1: Factor all polynomials Step 2: Find the common denominator Step 3: Multiply all terms by the common denominator Step 4: Solve (12 x) x+1 2 x - x– 1 4 x = 1 3 = 6 (x + 1) -3(x – 1) = 4 x 6 x + 6 – 3 x + 3 = 4 x 3 x + 9 = 4 x -3 x 9 = x

Other Rational Equation Examples 3 x– 2 (x + 2)(x – 2) 3 x– 2 (4 x 2) + 5 x+2 = 12 x 2 - 4 1 x + 5 x+2 = 12 (x + 2) (x – 2) 4 x + 4 = 3 x 2 + 1 = 3 x 2 4 3 x 2 - 4 x 3(x + 2) + 5(x – 2) = 3 x + 6 + 5 x – 10 = 8 x – 4 = 12 +4 +4 8 x = 16 x = 2 12 12 - 4 =0 (3 x + 2) (x – 2) = 0 3 x + 2 = 0 or x – 2 = 0 3 x = -2 or x=2 x = -2/3 or x=2

Check Your Understanding Simplify: x 1 + x 2 – 1 1 x– 2 1 x(x – 1) - 1 x-1 3 x + 1 2(x – 3) x(x – 2) x 2 – 1 - 2 x(x + 1) 3 x(x – 1)(x + 1) Solve 6 - 1 x 2 4 =1 3 2 x – 1 = 2 2 x– 1 + 3 5 x+1 x+2 = x x 2 + x - 2 -1/4 Try this one: Solve for p: 1 =1 + 1 F p q

Solving Radical Equations #1 #2 #3 #4 X 2 = 64

Radical Equations Continued… Example 2: Example 1: x + 26 – 11 x = 4 - x ( 26 – 11 x)2 = (4 – x)2 26 – 11 x = (4 -x) 26 - 11 x = 16 – 4 x +x 2 26 – 11 x = 16 – 8 x + x 2 -26 +11 x 0 = x 2 + 3 x -10 0 = (x - 2) (x + 5) x – 2 = 0 or x + 5 = 0 x=2 x = -5 3 x + 1 – x + 4 =1 3 x + 1 = x + 4 + 1 ( 3 x + 1)2 = ( x + 4 + 1)2 3 x + 1 = ( x + 4 + 1) 3 x + 1 = x + 4 + 1 3 x + 1 = x + 4 + 2 x + 4 + 1 3 x + 1 = x + 5 + 2 x + 4 -x -5 2 x - 4 = 2 x + 4 (2 x - 4)2 = (2 x + 4)2 4 x+16 4 x 2 – 16 x +16 = 4(x+4) 4 x 2 – 20 x = 0 4 x(x – 5) = 0, so… 4 x = 0 or x – 5 = 0 x = 0 or x = 5

1. 5 Inequality Set & Interval Notation Set Builder Notation {1, 5, 6} { } {x | x > -4} x such that x is greater than – 4 {x | x < 2} x such that x is less than or equal to 2 Interval (-4, ) Notation Graph -4 {6} {x | -2 < x < 7} x such that x is greater than – 2 and less than or equal to 7 (- , 2] (-2, 7] 2 -2 0 Question: How would you write the set of all real numbers? 7 (- , ) or R

Inequality Example Statement 7 x + 15 > 13 x + 51 Reason [Given] -6 x + 15 > 51 [-13 x] -6 x > 36 [-15] x < 6 [Divide by – 6, so must ‘flip’ the inequality sign Set Notation: {x | x < 6} Interval Notation: (- , 6] Graph: 6

Compound Inequality -3 < 2 x + 1 < 3 -1 -1 -1 -4 < 2 x < 2 2 Set Notation: {x | -2 < x < 1} Interval Notation: (-2, 1] Graph: -2 < x < 1 -2 0 1

Set Operations and Compound Inequalities Intersection ( ) – “AND” A B = {x | x A and x B} Set Notation: {x | X 8 and X 5} X+ 1 9 and X– 2 3 Interval Notation: (- , 8] [5, ) X 8 and X 5 0 [ ] 5 8 Union ( ) – “OR” A B = {x | x A or x B} Set Notation: {x | X -2 or X -3} -4 x + 1 9 or X -2 or Interval Notation: (- , -2] (- , -3] 5 X+ 3 12 X -3 -2

1. 6 Absolute Value Inequality | 2 x + 3| > 5 2 x + 3 > 5 or 2 x > 2 -(2 x + 3) > 5 -2 x - 3 > 5 x>1 -2 x -2 Set Notation: {x | x < -4 or x > 1} x < -4 Interval Notation: (- , -4] or [1, ) Graph: -4 > 0 1 8 -2

Absolute Value Equations | 2 x – 3| = 11 2 x – 3 = 11 or -(2 x – 3) = 11 2 x = 14 -2 x + 3 = 11 x=7 -2 x = 8 x = -4